🔬 Model Lab

Problem 1:
The problem asks for the ratio of the side length of a square to the radius of a circle when both have the same area. Let's denote the side length of the square as $s$ and the radius of the circle as $r$. The area of the square is $s^2$ and the area of the circle is $\pi r^2$. Since they have the same area, we have $s^2 = \pi r^2$. To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$. Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$. Let's examine the choices: A. $\sqrt{\pi}^2 = \pi$, B. $\sqrt{\pi}$, C. $\pi$, D. $2\pi$, E. $\pi^2$. The correct ratio is given by choice B.

ANSWER 1: B

Problem 2:
To find the number of digits in $4^5 \cdot 5^{10}$, let's simplify the expression. $4^5 = (2^2)^5 = 2^{10}$, so $4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$. $10^{10}$ is a 1 followed by 10 zeros, making it an 11-digit number. Let's check the choices: A. 8, B. 9, C. 10, D. 11, E. 12. The correct number of digits is given by choice D.

ANSWER 2: D

Problem 3:
The problem states that 8 friends agreed to share a bill equally, but because Judi forgot her money, each of her 7 friends paid an extra $2.50. Let's denote the total bill as $B$. The amount each friend was originally supposed to pay is $\frac{B}{8}$. When Judi's portion is covered by her 7 friends, each pays $\frac{B}{8} + $2.50$. The total amount paid by the 7 friends is $7(\frac{B}{8} + 2.50)$, which equals $B$. So, we have $7(\frac{B}{8} + 2.50) = B$. Simplifying this gives $\frac{7B}{8} + 17.50 = B$. Subtracting $\frac{7B}{8}$ from both sides gives $17.50 = \frac{B}{8}$. Multiplying both sides by 8 gives $B = 140$. Let's check the choices: A. $120, B. $128, C. $140, D. $144, E. $160. The total bill matches choice C.

ANSWER 3: C

Problem 4:
Carrie has a garden measuring 6 feet by 8 feet, so its area is $6 \times 8 = 48$ square feet. She plants 4 strawberry plants per square foot, so she plants $48 \times 4 = 192$ plants. With an average of 10 strawberries per plant, she can expect to harvest $192 \times 10 = 1920$ strawberries. Let's check the choices: A. 560, B. 960, C. 1120, D. 1920, E. 3840. The expected harvest matches choice D.

ANSWER 4: D

Problem 5:
To compare the given fractions, let's convert them to equivalent decimals or find a common denominator. The fractions are $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$. Converting them to equivalent fractions with a common denominator (24) gives $\frac{8}{24}$, $\frac{6}{24}$, $\frac{9}{24}$, $\frac{10}{24}$, and $\frac{7}{24}$. Comparing these, $\frac{10}{24}$ is the largest. So, $\frac{5}{12}$ is the largest fraction. Let's check the choices: A. $\frac{1}{3}$, B. $\frac{1}{4}$, C. $\frac{3}{8}$, D. $\frac{5}{12}$, E. $\frac{7}{24}$. The largest fraction is given by choice D.

ANSWER 5: D

Problem 6:
Let's denote the original length and width of the rectangle as $L$ and $W$, respectively. The original area is $L \times W$. After increasing the length by 10% and decreasing the width by 10%, the new length is $1.1L$ and the new width is $0.9W$. The new area is $1.1L \times 0.9W = 0.99LW$. To find what percent of the old area the new area is, we calculate $\frac{0.99LW}{LW} \times 100\% = 99\%$. Let's check the choices: A. 90, B. 99, C. 100, D. 101, E. 110. The new area is 99% of the old area, matching choice B.

ANSWER 6: B

Problem 7:
To evaluate $\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$, first simplify the numerator: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$. Then, divide by $\frac{4}{5}$, which is equivalent to multiplying by $\frac{5}{4}$. So, $\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$. Let's check the choices: A. 1, B. $\frac{25}{16}$, C. 2, D. $\frac{43}{20}$, E. $\frac{47}{16}$. The result matches choice B.

ANSWER 7: B

Problem 8:
To evaluate $\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$, convert each fraction to a decimal: $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$. Let's check the choices: A. 0.0026, B. 0.0197, C. 0.1997, D. 0.26, E. 1.997. The sum matches choice C.

ANSWER 8: C

Problem 9:
The sum $90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99$ is an arithmetic series with the first term $a = 90$, the last term $l = 99$, and $n = 10$ terms. The sum of an arithmetic series is given by $S = \frac{n}{2}(a + l)$. So, $S = \frac{10}{2}(90 + 99) = 5 \times 189 = 945$. Let's check the choices: A. 845, B. 945, C. 1005, D. 1025, E. 1045. The sum matches choice B.

ANSWER 9: B

Problem 10:
Let's analyze the number of lines of symmetry for each figure: A. An equilateral triangle has 3 lines of symmetry. B. A non-square rhombus has 2 lines of symmetry. C. A non-square rectangle has 2 lines of symmetry. D. An isosceles trapezoid has 1 line of symmetry. E. A square has 4 lines of symmetry. The figure with the greatest number of lines of symmetry is the square. Let's check the choices: the square is choice E.

ANSWER 10: E

Problem 11:
From the given information: Neither Jo nor Bo has as much money as Flo, so $Jo < Flo$ and $Bo < Flo$. Both Bo and Coe have more than Moe, so $Moe < Bo$ and $Moe < Coe$. Jo has more than Moe but less than Bo, so $Moe < Jo < Bo$. Combining these, we have $Moe < Jo < Bo < Flo$ and $Moe < Coe$. Since we don't have a direct comparison between Coe and the others except Moe, and we know $Bo < Flo$, the order for the least amount of money is clear for Moe. Let's check the choices: A. Bo, B. Coe, C. Flo, D. Jo, E. Moe. Moe has the least amount of money, matching choice E.

ANSWER 11: E

Problem 12:
Let's check each pair to see if their product equals 36: A. $(-4) \times (-9) = 36$, B. $(-3) \times (-12) = 36$, C. $\frac{1}{2} \times (-72) = -36$, D. $1 \times 36 = 36$, E. $\frac{3}{2} \times 24 = 36$. The pair that does NOT have a product equal to 36 is C. $\{\frac{1}{2}, -72\}$.

ANSWER 12: C

Problem 1:

We are asked for the ratio

\[
\frac{\text{side length of square}}{\text{radius of circle}}.
\]

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The square has area

\[
s^2.
\]

The circle has area

\[
\pi r^2.
\]

They have the same area, so

\[
s^2=\pi r^2.
\]

Taking the square root of both sides,

\[
s=r\sqrt{\pi}.
\]

So

\[
\frac{s}{r}=\sqrt{\pi}.
\]

The choices involving \(\pi\), \(2\pi\), or fractions of \(\pi\) are too large or have the wrong form. The correct ratio is \(\sqrt{\pi}\).

ANSWER 1: B

Problem 2:

We need the number of digits in

\[
4^5 \cdot 5^{10}.
\]

Rewrite \(4^5\):

\[
4^5=(2^2)^5=2^{10}.
\]

So

\[
4^5 \cdot 5^{10}=2^{10}\cdot 5^{10}.
\]

Since \(2^{10}\cdot 5^{10}=(2\cdot 5)^{10}\),

\[
2^{10}\cdot 5^{10}=10^{10}.
\]

Now,

\[
10^{10}=10,000,000,000,
\]

which is a 1 followed by 10 zeros.

So it has

\[
11
\]

digits.

ANSWER 2: D

Problem 3:

Eight friends were supposed to share the bill equally. Judi forgot her money, so the other 7 friends each paid an extra \(\$2.50\).

The total extra money paid by the 7 friends was

\[
7 \times 2.50 = 17.50.
\]

This extra money covered Judi’s share of the bill.

So Judi’s share was \(\$17.50\). Since there were 8 equal shares, the total bill was

\[
8 \times 17.50 = 140.
\]

Thus the total bill was \(\$140\).

ANSWER 3: C

Problem 4:

Carrie’s garden is a rectangle measuring 6 feet by 8 feet.

First find the area:

\[
6 \times 8 = 48
\]

square feet.

She plants 4 strawberry plants per square foot, so the number of plants is

\[
48 \times 4 = 192.
\]

Each plant produces an average of 10 strawberries, so the total expected harvest is

\[
192 \times 10 = 1920.
\]

So Carrie can expect to harvest 1920 strawberries.

ANSWER 4: D

Problem 5:

We need to find the largest fraction among:

\[
\frac13,\frac14,\frac38,\frac5{12},\frac7{24}.
\]

Use a common denominator of 24:

\[
\frac13=\frac8{24}
\]

\[
\frac14=\frac6{24}
\]

\[
\frac38=\frac9{24}
\]

\[
\frac5{12}=\frac{10}{24}
\]

\[
\frac7{24}=\frac7{24}
\]

Now compare:

\[
\frac6{24},\frac7{24},\frac8{24},\frac9{24},\frac{10}{24}.
\]

The largest is

\[
\frac{10}{24}=\frac5{12}.
\]

ANSWER 5: D

Problem 6:

Let the original length be \(L\), and let the original width be \(W\).

The original area is

\[
LW.
\]

The length is increased by 10%, so the new length is

\[
1.10L.
\]

The width is decreased by 10%, so the new width is

\[
0.90W.
\]

The new area is

\[
(1.10L)(0.90W)=0.99LW.
\]

So the new area is \(99\%\) of the old area.

A common mistake is to think the \(+10\%\) and \(-10\%\) cancel, but they are applied to different dimensions, so the area changes by the product \(1.10 \times 0.90=0.99\).

ANSWER 6: B

Problem 7:

The expression is

\[
\left(\frac38+\frac78\right)\div \frac45.
\]

First add the fractions in parentheses:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45.
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54 = \frac{25}{16}.
\]

So the value is

\[
\frac{25}{16}.
\]

ANSWER 7: B

Problem 8:

We need to compute

\[
\frac1{10}+\frac9{100}+\frac9{1000}+\frac7{10000}.
\]

Convert each fraction to a decimal:

\[
\frac1{10}=0.1
\]

\[
\frac9{100}=0.09
\]

\[
\frac9{1000}=0.009
\]

\[
\frac7{10000}=0.0007
\]

Now add:

\[
0.1+0.09=0.19
\]

\[
0.19+0.009=0.199
\]

\[
0.199+0.0007=0.1997.
\]

So the answer is

\[
0.1997.
\]

ANSWER 8: C

Problem 9:

We need to add

\[
90+91+92+93+94+95+96+97+98+99.
\]

There are 10 numbers.

The average of the first and last number is

\[
\frac{90+99}{2}=\frac{189}{2}=94.5.
\]

So the sum is

\[
10 \times 94.5 = 945.
\]

Alternatively, pair the numbers:

\[
90+99=189
\]

\[
91+98=189
\]

\[
92+97=189
\]

\[
93+96=189
\]

\[
94+95=189
\]

There are 5 pairs, so

\[
5 \times 189 = 945.
\]

ANSWER 9: B

Problem 10:

We compare the number of lines of symmetry for each figure.

- Equilateral triangle: 3 lines of symmetry.
- Non-square rhombus: 2 lines of symmetry.
- Non-square rectangle: 2 lines of symmetry.
- Isosceles trapezoid: 1 line of symmetry.
- Square: 4 lines of symmetry.

The square has the greatest number of lines of symmetry.

ANSWER 10: E

Problem 11:

We are told:

- Neither Jo nor Bo has as much money as Flo, so Flo has more than Jo and Bo.
- Both Bo and Coe have more than Moe.
- Jo has more than Moe, but less than Bo.

From this:

\[
Bo > Moe
\]

\[
Coe > Moe
\]

\[
Jo > Moe
\]

Also, since Flo has more than Jo,

\[
Flo > Jo > Moe.
\]

So Flo also has more than Moe.

Therefore, everyone has more money than Moe.

So Moe has the least amount of money.

ANSWER 11: E

Problem 12:

We need to find the pair whose product is NOT equal to 36.

Check each pair:

A. \(\{-4,-9\}\)

\[
(-4)(-9)=36
\]

So A works.

B. \(\{-3,-12\}\)

\[
(-3)(-12)=36
\]

So B works.

C. \(\left\{\frac12,-72\right\}\)

\[
\frac12 \cdot (-72)=-36
\]

This is not 36.

D. \(\{1,36\}\)

\[
1 \cdot 36=36
\]

So D works.

E. \(\left\{\frac32,24\right\}\)

\[
\frac32 \cdot 24=36
\]

So E works.

The only pair whose product is not 36 is choice C.

ANSWER 12: C

**Problem 1:**
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

**Step-by-step working:**
1. Let the side length of the square be $s$ and the radius of the circle be $r$.
2. The area of the square is $s^2$.
3. The area of the circle is $\pi r^2$.
4. Since their areas are equal, we can set up the equation: $s^2 = \pi r^2$.
5. To find the ratio of the side length to the radius ($s/r$), we first divide both sides by $r^2$: $\frac{s^2}{r^2} = \pi$.
6. Taking the square root of both sides gives us: $\frac{s}{r} = \sqrt{\pi}$.
7. This matches choice B.

ANSWER 1: B

**Problem 2:**
How many digits are in the product $4^5 \cdot 5^{10}$?

**Step-by-step working:**
1. We want to find the number of digits in the number $4^5 \cdot 5^{10}$.
2. First, express the base $4$ as a power of $2$: $4 = 2^2$.
3. Substitute this into the expression: $(2^2)^5 \cdot 5^{10}$.
4. Use the power of a power property $(a^m)^n = a^{m \cdot n}$: $2^{10} \cdot 5^{10}$.
5. Since the exponents are the same, we can combine the bases: $(2 \cdot 5)^{10} = 10^{10}$.
6. The number $10^{10}$ is written as a $1$ followed by $10$ zeros.
7. Therefore, it has $1 + 10 = 11$ digits.

ANSWER 2: D

**Problem 3:**
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

**Step-by-step working:**
1. Judi's portion of the bill was covered by her $7$ friends.
2. Each of the $7$ friends paid an extra $\$2.50$.
3. The total amount they paid to cover Judi's share is $7 \times \$2.50 = \$17.50$.
4. This means that an equal share of the bill for one person is $\$17.50$.
5. Since there are $8$ friends in total, the total bill is $8 \times \$17.50$.
6. $8 \times \$17.50 = \$140.00$.

ANSWER 3: C

**Problem 4:**
Carrie has a rectangular garden that measures 6 feet by 8 feet. She plants the entire garden with strawberry plants. Carrie is able to plant 4 strawberry plants per square foot, and she harvests an average of 10 strawberries per plant. How many strawberries can she expect to harvest?

**Step-by-step working:**
1. First, find the area of the rectangular garden: $6 \text{ feet} \times 8 \text{ feet} = 48 \text{ square feet}$.
2. Next, calculate the total number of strawberry plants. She plants $4$ plants per square foot: $48 \times 4 = 192 \text{ plants}$.
3. Finally, calculate the total expected strawberry harvest. Each plant yields an average of $10$ strawberries: $192 \times 10 = 1920 \text{ strawberries}$.

ANSWER 4: D

**Problem 5:**
Which of the following is the largest?

**Step-by-step working:**
1. The choices given are the fractions: $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$.
2. To compare them easily, find a common denominator. The least common multiple (LCM) of $3, 4, 8, 12,$ and $24$ is $24$.
3. Convert each fraction:
   - A. $\frac{1}{3} = \frac{8}{24}$
   - B. $\frac{1}{4} = \frac{6}{24}$
   - C. $\frac{3}{8} = \frac{9}{24}$
   - D. $\frac{5}{12} = \frac{10}{24}$
   - E. $\frac{7}{24} = \frac{7}{24}$
4. Comparing the numerators ($8, 6, 9, 10, 7$), the largest is $10$.
5. Therefore, $\frac{10}{24}$ (which is $\frac{5}{12}$) is the largest fraction.

ANSWER 5: D

**Problem 6:**
The length of a rectangle is increased by 10% and the width is decreased by 10%. What percent of the old area is the new area?

**Step-by-step working:**
1. Let the original length be $L$ and the original width be $W$.
2. The old area is $A_{\text{old}} = L \times W$.
3. The new length is increased by $10\%$, making it $1.10L$.
4. The new width is decreased by $10\%$, making it $0.90W$.
5. The new area is $A_{\text{new}} = (1.10L) \times (0.90W) = 0.99(L \times W)$.
6. This means the new area is $0.99$ times the old area, which corresponds to $99\%$.

ANSWER 6: B

**Problem 7:**
$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}} =$

**Step-by-step working:**
1. First, evaluate the numerator by adding the fractions: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8}$.
2. Simplify the fraction: $\frac{10}{8} = \frac{5}{4}$.
3. Now, divide this result by the denominator $\frac{4}{5}$: $\frac{5}{4} \div \frac{4}{5}$.
4. Dividing by a fraction is the same as multiplying by its reciprocal: $\frac{5}{4} \times \frac{5}{4}$.
5. Multiply the numerators and the denominators: $\frac{5 \times 5}{4 \times 4} = \frac{25}{16}$.

ANSWER 7: B

**Problem 8:**
$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000} =$

**Step-by-step working:**
1. Convert each fraction into its decimal equivalent:
   - $\frac{1}{10} = 0.1$
   - $\frac{9}{100} = 0.09$
   - $\frac{9}{1000} = 0.009$
   - $\frac{7}{10000} = 0.0007$
2. Add the decimals together:
   $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$.

ANSWER 8: C

**Problem 9:**
$90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =$

**Step-by-step working:**
1. We are adding an arithmetic sequence of $10$ consecutive integers from $90$ to $99$.
2. We can split the sum into the tens and the units:
   - The tens contribute $90 \times 10 = 900$.
   - The units contribute $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$.
3. The sum of the digits from $0$ to $9$ is $\frac{9 \times 10}{2} = 45$.
4. Adding these together gives $900 + 45 = 945$.
*(Alternatively, using the formula for the sum of an arithmetic sequence: $S = \frac{n(a_1 + a_n)}{2} = \frac{10(90 + 99)}{2} = 5 \times 189 = 945$.)*

ANSWER 9: B

**Problem 10:**
Which of the following figures has the greatest number of lines of symmetry?

**Step-by-step working:**
1. Let's count the lines of symmetry for each given figure:
   - A. Equilateral triangle: $3$ lines of symmetry (from each vertex to the midpoint of the opposite side).
   - B. Non-square rhombus: $2$ lines of symmetry (along its diagonals).
   - C. Non-square rectangle: $2$ lines of symmetry (connecting the midpoints of opposite sides).
   - D. Isosceles trapezoid: $1$ line of sy

Problem 1:
The problem asks for the ratio of the side length of a square to the radius of a circle when both have the same area. Let's denote the side length of the square as $s$ and the radius of the circle as $r$. The area of the square is $s^2$ and the area of the circle is $\pi r^2$. Since they have the same area, we have $s^2 = \pi r^2$. To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$. Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$. Let's examine the choices: A. $\sqrt{\pi}^2 = \pi$, B. $\sqrt{\pi}$, C. $\pi$, D. $2\pi$, E. $\pi^2$. The correct ratio is given by choice B.

ANSWER 1: B

Problem 2:
To find the number of digits in $4^5 \cdot 5^{10}$, let's simplify the expression. $4^5 = (2^2)^5 = 2^{10}$, so $4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$. $10^{10}$ is a 1 followed by 10 zeros, making it an 11-digit number. Let's check the choices: A. 8, B. 9, C. 10, D. 11, E. 12. The correct number of digits is given by choice D.

ANSWER 2: D

Problem 3:
The problem states that 8 friends agreed to share a bill equally, but because Judi forgot her money, each of her 7 friends paid an extra $2.50. Let's denote the total bill as $B$. The amount each friend was originally supposed to pay is $\frac{B}{8}$. When Judi's portion is covered by her 7 friends, each pays $\frac{B}{8} + $2.50$. The total amount paid by the 7 friends is $7(\frac{B}{8} + 2.50)$, which equals $B$. So, we have $7(\frac{B}{8} + 2.50) = B$. Simplifying this gives $\frac{7B}{8} + 17.50 = B$. Subtracting $\frac{7B}{8}$ from both sides gives $17.50 = \frac{B}{8}$. Multiplying both sides by 8 gives $B = 140$. Let's check the choices: A. $120, B. $128, C. $140, D. $144, E. $160. The total bill matches choice C.

ANSWER 3: C

Problem 4:
Carrie has a garden measuring 6 feet by 8 feet, so its area is $6 \times 8 = 48$ square feet. She plants 4 strawberry plants per square foot, so she plants $48 \times 4 = 192$ plants. With an average of 10 strawberries per plant, she can expect to harvest $192 \times 10 = 1920$ strawberries. Let's check the choices: A. 560, B. 960, C. 1120, D. 1920, E. 3840. The expected harvest matches choice D.

ANSWER 4: D

Problem 5:
To compare the given fractions, let's convert them to equivalent decimals or find a common denominator. The fractions are $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$. Converting them to equivalent fractions with a common denominator (24) gives $\frac{8}{24}$, $\frac{6}{24}$, $\frac{9}{24}$, $\frac{10}{24}$, and $\frac{7}{24}$. Comparing these, $\frac{10}{24}$ is the largest. So, $\frac{5}{12}$ is the largest fraction. Let's check the choices: A. $\frac{1}{3}$, B. $\frac{1}{4}$, C. $\frac{3}{8}$, D. $\frac{5}{12}$, E. $\frac{7}{24}$. The largest fraction is given by choice D.

ANSWER 5: D

Problem 6:
Let's denote the original length and width of the rectangle as $L$ and $W$, respectively. The original area is $L \times W$. After increasing the length by 10% and decreasing the width by 10%, the new length is $1.1L$ and the new width is $0.9W$. The new area is $1.1L \times 0.9W = 0.99LW$. To find what percent of the old area the new area is, we calculate $\frac{0.99LW}{LW} \times 100\% = 99\%$. Let's check the choices: A. 90, B. 99, C. 100, D. 101, E. 110. The new area is 99% of the old area, matching choice B.

ANSWER 6: B

Problem 7:
To evaluate $\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$, first simplify the numerator: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$. Then, divide by $\frac{4}{5}$, which is equivalent to multiplying by $\frac{5}{4}$. So, $\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$. Let's check the choices: A. 1, B. $\frac{25}{16}$, C. 2, D. $\frac{43}{20}$, E. $\frac{47}{16}$. The result matches choice B.

ANSWER 7: B

Problem 8:
To evaluate $\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$, convert each fraction to a decimal: $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$. Let's check the choices: A. 0.0026, B. 0.0197, C. 0.1997, D. 0.26, E. 1.997. The sum matches choice C.

ANSWER 8: C

Problem 9:
The sum $90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99$ is an arithmetic series with the first term $a = 90$, the last term $l = 99$, and $n = 10$ terms. The sum of an arithmetic series is given by $S = \frac{n}{2}(a + l)$. So, $S = \frac{10}{2}(90 + 99) = 5 \times 189 = 945$. Let's check the choices: A. 845, B. 945, C. 1005, D. 1025, E. 1045. The sum matches choice B.

ANSWER 9: B

Problem 10:
Let's analyze the number of lines of symmetry for each figure: A. An equilateral triangle has 3 lines of symmetry. B. A non-square rhombus has 2 lines of symmetry. C. A non-square rectangle has 2 lines of symmetry. D. An isosceles trapezoid has 1 line of symmetry. E. A square has 4 lines of symmetry. The figure with the greatest number of lines of symmetry is the square. Let's check the choices: the square is choice E.

ANSWER 10: E

Problem 11:
From the given information: Neither Jo nor Bo has as much money as Flo, so $Jo < Flo$ and $Bo < Flo$. Both Bo and Coe have more than Moe, so $Moe < Bo$ and $Moe < Coe$. Jo has more than Moe but less than Bo, so $Moe < Jo < Bo$. Combining these, we have $Moe < Jo < Bo < Flo$ and $Moe < Coe$. Since we don't have a direct comparison between Coe and the others except Moe, and we know $Bo < Flo$, the order for the least amount of money is clear for Moe. Let's check the choices: A. Bo, B. Coe, C. Flo, D. Jo, E. Moe. Moe has the least amount of money, matching choice E.

ANSWER 11: E

Problem 12:
Let's check each pair to see if their product equals 36: A. $(-4) \times (-9) = 36$, B. $(-3) \times (-12) = 36$, C. $\frac{1}{2} \times (-72) = -36$, D. $1 \times 36 = 36$, E. $\frac{3}{2} \times 24 = 36$. The pair that does NOT have a product equal to 36 is C. $\{\frac{1}{2}, -72\}$.

ANSWER 12: C

Problem 1:

We are asked for the ratio

\[
\frac{\text{side length of square}}{\text{radius of circle}}.
\]

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The square has area

\[
s^2.
\]

The circle has area

\[
\pi r^2.
\]

They have the same area, so

\[
s^2=\pi r^2.
\]

Taking the square root of both sides,

\[
s=r\sqrt{\pi}.
\]

So

\[
\frac{s}{r}=\sqrt{\pi}.
\]

The choices involving \(\pi\), \(2\pi\), or fractions of \(\pi\) are too large or have the wrong form. The correct ratio is \(\sqrt{\pi}\).

ANSWER 1: B

Problem 2:

We need the number of digits in

\[
4^5 \cdot 5^{10}.
\]

Rewrite \(4^5\):

\[
4^5=(2^2)^5=2^{10}.
\]

So

\[
4^5 \cdot 5^{10}=2^{10}\cdot 5^{10}.
\]

Since \(2^{10}\cdot 5^{10}=(2\cdot 5)^{10}\),

\[
2^{10}\cdot 5^{10}=10^{10}.
\]

Now,

\[
10^{10}=10,000,000,000,
\]

which is a 1 followed by 10 zeros.

So it has

\[
11
\]

digits.

ANSWER 2: D

Problem 3:

Eight friends were supposed to share the bill equally. Judi forgot her money, so the other 7 friends each paid an extra \(\$2.50\).

The total extra money paid by the 7 friends was

\[
7 \times 2.50 = 17.50.
\]

This extra money covered Judi’s share of the bill.

So Judi’s share was \(\$17.50\). Since there were 8 equal shares, the total bill was

\[
8 \times 17.50 = 140.
\]

Thus the total bill was \(\$140\).

ANSWER 3: C

Problem 4:

Carrie’s garden is a rectangle measuring 6 feet by 8 feet.

First find the area:

\[
6 \times 8 = 48
\]

square feet.

She plants 4 strawberry plants per square foot, so the number of plants is

\[
48 \times 4 = 192.
\]

Each plant produces an average of 10 strawberries, so the total expected harvest is

\[
192 \times 10 = 1920.
\]

So Carrie can expect to harvest 1920 strawberries.

ANSWER 4: D

Problem 5:

We need to find the largest fraction among:

\[
\frac13,\frac14,\frac38,\frac5{12},\frac7{24}.
\]

Use a common denominator of 24:

\[
\frac13=\frac8{24}
\]

\[
\frac14=\frac6{24}
\]

\[
\frac38=\frac9{24}
\]

\[
\frac5{12}=\frac{10}{24}
\]

\[
\frac7{24}=\frac7{24}
\]

Now compare:

\[
\frac6{24},\frac7{24},\frac8{24},\frac9{24},\frac{10}{24}.
\]

The largest is

\[
\frac{10}{24}=\frac5{12}.
\]

ANSWER 5: D

Problem 6:

Let the original length be \(L\), and let the original width be \(W\).

The original area is

\[
LW.
\]

The length is increased by 10%, so the new length is

\[
1.10L.
\]

The width is decreased by 10%, so the new width is

\[
0.90W.
\]

The new area is

\[
(1.10L)(0.90W)=0.99LW.
\]

So the new area is \(99\%\) of the old area.

A common mistake is to think the \(+10\%\) and \(-10\%\) cancel, but they are applied to different dimensions, so the area changes by the product \(1.10 \times 0.90=0.99\).

ANSWER 6: B

Problem 7:

The expression is

\[
\left(\frac38+\frac78\right)\div \frac45.
\]

First add the fractions in parentheses:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45.
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54 = \frac{25}{16}.
\]

So the value is

\[
\frac{25}{16}.
\]

ANSWER 7: B

Problem 8:

We need to compute

\[
\frac1{10}+\frac9{100}+\frac9{1000}+\frac7{10000}.
\]

Convert each fraction to a decimal:

\[
\frac1{10}=0.1
\]

\[
\frac9{100}=0.09
\]

\[
\frac9{1000}=0.009
\]

\[
\frac7{10000}=0.0007
\]

Now add:

\[
0.1+0.09=0.19
\]

\[
0.19+0.009=0.199
\]

\[
0.199+0.0007=0.1997.
\]

So the answer is

\[
0.1997.
\]

ANSWER 8: C

Problem 9:

We need to add

\[
90+91+92+93+94+95+96+97+98+99.
\]

There are 10 numbers.

The average of the first and last number is

\[
\frac{90+99}{2}=\frac{189}{2}=94.5.
\]

So the sum is

\[
10 \times 94.5 = 945.
\]

Alternatively, pair the numbers:

\[
90+99=189
\]

\[
91+98=189
\]

\[
92+97=189
\]

\[
93+96=189
\]

\[
94+95=189
\]

There are 5 pairs, so

\[
5 \times 189 = 945.
\]

ANSWER 9: B

Problem 10:

We compare the number of lines of symmetry for each figure.

- Equilateral triangle: 3 lines of symmetry.
- Non-square rhombus: 2 lines of symmetry.
- Non-square rectangle: 2 lines of symmetry.
- Isosceles trapezoid: 1 line of symmetry.
- Square: 4 lines of symmetry.

The square has the greatest number of lines of symmetry.

ANSWER 10: E

Problem 11:

We are told:

- Neither Jo nor Bo has as much money as Flo, so Flo has more than Jo and Bo.
- Both Bo and Coe have more than Moe.
- Jo has more than Moe, but less than Bo.

From this:

\[
Bo > Moe
\]

\[
Coe > Moe
\]

\[
Jo > Moe
\]

Also, since Flo has more than Jo,

\[
Flo > Jo > Moe.
\]

So Flo also has more than Moe.

Therefore, everyone has more money than Moe.

So Moe has the least amount of money.

ANSWER 11: E

Problem 12:

We need to find the pair whose product is NOT equal to 36.

Check each pair:

A. \(\{-4,-9\}\)

\[
(-4)(-9)=36
\]

So A works.

B. \(\{-3,-12\}\)

\[
(-3)(-12)=36
\]

So B works.

C. \(\left\{\frac12,-72\right\}\)

\[
\frac12 \cdot (-72)=-36
\]

This is not 36.

D. \(\{1,36\}\)

\[
1 \cdot 36=36
\]

So D works.

E. \(\left\{\frac32,24\right\}\)

\[
\frac32 \cdot 24=36
\]

So E works.

The only pair whose product is not 36 is choice C.

ANSWER 12: C

**Problem 1:**
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

**Step-by-step working:**
1. Let the side length of the square be $s$ and the radius of the circle be $r$.
2. The area of the square is $s^2$.
3. The area of the circle is $\pi r^2$.
4. Since their areas are equal, we can set up the equation: $s^2 = \pi r^2$.
5. To find the ratio of the side length to the radius ($s/r$), we first divide both sides by $r^2$: $\frac{s^2}{r^2} = \pi$.
6. Taking the square root of both sides gives us: $\frac{s}{r} = \sqrt{\pi}$.
7. This matches choice B.

ANSWER 1: B

**Problem 2:**
How many digits are in the product $4^5 \cdot 5^{10}$?

**Step-by-step working:**
1. We want to find the number of digits in the number $4^5 \cdot 5^{10}$.
2. First, express the base $4$ as a power of $2$: $4 = 2^2$.
3. Substitute this into the expression: $(2^2)^5 \cdot 5^{10}$.
4. Use the power of a power property $(a^m)^n = a^{m \cdot n}$: $2^{10} \cdot 5^{10}$.
5. Since the exponents are the same, we can combine the bases: $(2 \cdot 5)^{10} = 10^{10}$.
6. The number $10^{10}$ is written as a $1$ followed by $10$ zeros.
7. Therefore, it has $1 + 10 = 11$ digits.

ANSWER 2: D

**Problem 3:**
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

**Step-by-step working:**
1. Judi's portion of the bill was covered by her $7$ friends.
2. Each of the $7$ friends paid an extra $\$2.50$.
3. The total amount they paid to cover Judi's share is $7 \times \$2.50 = \$17.50$.
4. This means that an equal share of the bill for one person is $\$17.50$.
5. Since there are $8$ friends in total, the total bill is $8 \times \$17.50$.
6. $8 \times \$17.50 = \$140.00$.

ANSWER 3: C

**Problem 4:**
Carrie has a rectangular garden that measures 6 feet by 8 feet. She plants the entire garden with strawberry plants. Carrie is able to plant 4 strawberry plants per square foot, and she harvests an average of 10 strawberries per plant. How many strawberries can she expect to harvest?

**Step-by-step working:**
1. First, find the area of the rectangular garden: $6 \text{ feet} \times 8 \text{ feet} = 48 \text{ square feet}$.
2. Next, calculate the total number of strawberry plants. She plants $4$ plants per square foot: $48 \times 4 = 192 \text{ plants}$.
3. Finally, calculate the total expected strawberry harvest. Each plant yields an average of $10$ strawberries: $192 \times 10 = 1920 \text{ strawberries}$.

ANSWER 4: D

**Problem 5:**
Which of the following is the largest?

**Step-by-step working:**
1. The choices given are the fractions: $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$.
2. To compare them easily, find a common denominator. The least common multiple (LCM) of $3, 4, 8, 12,$ and $24$ is $24$.
3. Convert each fraction:
   - A. $\frac{1}{3} = \frac{8}{24}$
   - B. $\frac{1}{4} = \frac{6}{24}$
   - C. $\frac{3}{8} = \frac{9}{24}$
   - D. $\frac{5}{12} = \frac{10}{24}$
   - E. $\frac{7}{24} = \frac{7}{24}$
4. Comparing the numerators ($8, 6, 9, 10, 7$), the largest is $10$.
5. Therefore, $\frac{10}{24}$ (which is $\frac{5}{12}$) is the largest fraction.

ANSWER 5: D

**Problem 6:**
The length of a rectangle is increased by 10% and the width is decreased by 10%. What percent of the old area is the new area?

**Step-by-step working:**
1. Let the original length be $L$ and the original width be $W$.
2. The old area is $A_{\text{old}} = L \times W$.
3. The new length is increased by $10\%$, making it $1.10L$.
4. The new width is decreased by $10\%$, making it $0.90W$.
5. The new area is $A_{\text{new}} = (1.10L) \times (0.90W) = 0.99(L \times W)$.
6. This means the new area is $0.99$ times the old area, which corresponds to $99\%$.

ANSWER 6: B

**Problem 7:**
$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}} =$

**Step-by-step working:**
1. First, evaluate the numerator by adding the fractions: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8}$.
2. Simplify the fraction: $\frac{10}{8} = \frac{5}{4}$.
3. Now, divide this result by the denominator $\frac{4}{5}$: $\frac{5}{4} \div \frac{4}{5}$.
4. Dividing by a fraction is the same as multiplying by its reciprocal: $\frac{5}{4} \times \frac{5}{4}$.
5. Multiply the numerators and the denominators: $\frac{5 \times 5}{4 \times 4} = \frac{25}{16}$.

ANSWER 7: B

**Problem 8:**
$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000} =$

**Step-by-step working:**
1. Convert each fraction into its decimal equivalent:
   - $\frac{1}{10} = 0.1$
   - $\frac{9}{100} = 0.09$
   - $\frac{9}{1000} = 0.009$
   - $\frac{7}{10000} = 0.0007$
2. Add the decimals together:
   $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$.

ANSWER 8: C

**Problem 9:**
$90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =$

**Step-by-step working:**
1. We are adding an arithmetic sequence of $10$ consecutive integers from $90$ to $99$.
2. We can split the sum into the tens and the units:
   - The tens contribute $90 \times 10 = 900$.
   - The units contribute $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$.
3. The sum of the digits from $0$ to $9$ is $\frac{9 \times 10}{2} = 45$.
4. Adding these together gives $900 + 45 = 945$.
*(Alternatively, using the formula for the sum of an arithmetic sequence: $S = \frac{n(a_1 + a_n)}{2} = \frac{10(90 + 99)}{2} = 5 \times 189 = 945$.)*

ANSWER 9: B

**Problem 10:**
Which of the following figures has the greatest number of lines of symmetry?

**Step-by-step working:**
1. Let's count the lines of symmetry for each given figure:
   - A. Equilateral triangle: $3$ lines of symmetry (from each vertex to the midpoint of the opposite side).
   - B. Non-square rhombus: $2$ lines of symmetry (along its diagonals).
   - C. Non-square rectangle: $2$ lines of symmetry (connecting the midpoints of opposite sides).
   - D. Isosceles trapezoid: $1$ line of sy

Problem 1:
The problem asks for the ratio of the side length of a square to the radius of a circle when both have the same area. Let's denote the side length of the square as $s$ and the radius of the circle as $r$. The area of the square is $s^2$ and the area of the circle is $\pi r^2$. Since they have the same area, we have $s^2 = \pi r^2$. To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$. Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$. Let's examine the choices: A. $\sqrt{\pi}^2 = \pi$, B. $\sqrt{\pi}$, C. $\pi$, D. $2\pi$, E. $\pi^2$. The correct ratio is given by choice B.

ANSWER 1: B

Problem 2:
To find the number of digits in $4^5 \cdot 5^{10}$, let's simplify the expression. $4^5 = (2^2)^5 = 2^{10}$, so $4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$. $10^{10}$ is a 1 followed by 10 zeros, making it an 11-digit number. Let's check the choices: A. 8, B. 9, C. 10, D. 11, E. 12. The correct number of digits is given by choice D.

ANSWER 2: D

Problem 3:
The problem states that 8 friends agreed to share a bill equally, but because Judi forgot her money, each of her 7 friends paid an extra $2.50. Let's denote the total bill as $B$. The amount each friend was originally supposed to pay is $\frac{B}{8}$. When Judi's portion is covered by her 7 friends, each pays $\frac{B}{8} + $2.50$. The total amount paid by the 7 friends is $7(\frac{B}{8} + 2.50)$, which equals $B$. So, we have $7(\frac{B}{8} + 2.50) = B$. Simplifying this gives $\frac{7B}{8} + 17.50 = B$. Subtracting $\frac{7B}{8}$ from both sides gives $17.50 = \frac{B}{8}$. Multiplying both sides by 8 gives $B = 140$. Let's check the choices: A. $120, B. $128, C. $140, D. $144, E. $160. The total bill matches choice C.

ANSWER 3: C

Problem 4:
Carrie has a garden measuring 6 feet by 8 feet, so its area is $6 \times 8 = 48$ square feet. She plants 4 strawberry plants per square foot, so she plants $48 \times 4 = 192$ plants. With an average of 10 strawberries per plant, she can expect to harvest $192 \times 10 = 1920$ strawberries. Let's check the choices: A. 560, B. 960, C. 1120, D. 1920, E. 3840. The expected harvest matches choice D.

ANSWER 4: D

Problem 5:
To compare the given fractions, let's convert them to equivalent decimals or find a common denominator. The fractions are $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$. Converting them to equivalent fractions with a common denominator (24) gives $\frac{8}{24}$, $\frac{6}{24}$, $\frac{9}{24}$, $\frac{10}{24}$, and $\frac{7}{24}$. Comparing these, $\frac{10}{24}$ is the largest. So, $\frac{5}{12}$ is the largest fraction. Let's check the choices: A. $\frac{1}{3}$, B. $\frac{1}{4}$, C. $\frac{3}{8}$, D. $\frac{5}{12}$, E. $\frac{7}{24}$. The largest fraction is given by choice D.

ANSWER 5: D

Problem 6:
Let's denote the original length and width of the rectangle as $L$ and $W$, respectively. The original area is $L \times W$. After increasing the length by 10% and decreasing the width by 10%, the new length is $1.1L$ and the new width is $0.9W$. The new area is $1.1L \times 0.9W = 0.99LW$. To find what percent of the old area the new area is, we calculate $\frac{0.99LW}{LW} \times 100\% = 99\%$. Let's check the choices: A. 90, B. 99, C. 100, D. 101, E. 110. The new area is 99% of the old area, matching choice B.

ANSWER 6: B

Problem 7:
To evaluate $\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$, first simplify the numerator: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$. Then, divide by $\frac{4}{5}$, which is equivalent to multiplying by $\frac{5}{4}$. So, $\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$. Let's check the choices: A. 1, B. $\frac{25}{16}$, C. 2, D. $\frac{43}{20}$, E. $\frac{47}{16}$. The result matches choice B.

ANSWER 7: B

Problem 8:
To evaluate $\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$, convert each fraction to a decimal: $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$. Let's check the choices: A. 0.0026, B. 0.0197, C. 0.1997, D. 0.26, E. 1.997. The sum matches choice C.

ANSWER 8: C

Problem 9:
The sum $90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99$ is an arithmetic series with the first term $a = 90$, the last term $l = 99$, and $n = 10$ terms. The sum of an arithmetic series is given by $S = \frac{n}{2}(a + l)$. So, $S = \frac{10}{2}(90 + 99) = 5 \times 189 = 945$. Let's check the choices: A. 845, B. 945, C. 1005, D. 1025, E. 1045. The sum matches choice B.

ANSWER 9: B

Problem 10:
Let's analyze the number of lines of symmetry for each figure: A. An equilateral triangle has 3 lines of symmetry. B. A non-square rhombus has 2 lines of symmetry. C. A non-square rectangle has 2 lines of symmetry. D. An isosceles trapezoid has 1 line of symmetry. E. A square has 4 lines of symmetry. The figure with the greatest number of lines of symmetry is the square. Let's check the choices: the square is choice E.

ANSWER 10: E

Problem 11:
From the given information: Neither Jo nor Bo has as much money as Flo, so $Jo < Flo$ and $Bo < Flo$. Both Bo and Coe have more than Moe, so $Moe < Bo$ and $Moe < Coe$. Jo has more than Moe but less than Bo, so $Moe < Jo < Bo$. Combining these, we have $Moe < Jo < Bo < Flo$ and $Moe < Coe$. Since we don't have a direct comparison between Coe and the others except Moe, and we know $Bo < Flo$, the order for the least amount of money is clear for Moe. Let's check the choices: A. Bo, B. Coe, C. Flo, D. Jo, E. Moe. Moe has the least amount of money, matching choice E.

ANSWER 11: E

Problem 12:
Let's check each pair to see if their product equals 36: A. $(-4) \times (-9) = 36$, B. $(-3) \times (-12) = 36$, C. $\frac{1}{2} \times (-72) = -36$, D. $1 \times 36 = 36$, E. $\frac{3}{2} \times 24 = 36$. The pair that does NOT have a product equal to 36 is C. $\{\frac{1}{2}, -72\}$.

ANSWER 12: C

Problem 1:

We are asked for the ratio

\[
\frac{\text{side length of square}}{\text{radius of circle}}.
\]

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The square has area

\[
s^2.
\]

The circle has area

\[
\pi r^2.
\]

They have the same area, so

\[
s^2=\pi r^2.
\]

Taking the square root of both sides,

\[
s=r\sqrt{\pi}.
\]

So

\[
\frac{s}{r}=\sqrt{\pi}.
\]

The choices involving \(\pi\), \(2\pi\), or fractions of \(\pi\) are too large or have the wrong form. The correct ratio is \(\sqrt{\pi}\).

ANSWER 1: B

Problem 2:

We need the number of digits in

\[
4^5 \cdot 5^{10}.
\]

Rewrite \(4^5\):

\[
4^5=(2^2)^5=2^{10}.
\]

So

\[
4^5 \cdot 5^{10}=2^{10}\cdot 5^{10}.
\]

Since \(2^{10}\cdot 5^{10}=(2\cdot 5)^{10}\),

\[
2^{10}\cdot 5^{10}=10^{10}.
\]

Now,

\[
10^{10}=10,000,000,000,
\]

which is a 1 followed by 10 zeros.

So it has

\[
11
\]

digits.

ANSWER 2: D

Problem 3:

Eight friends were supposed to share the bill equally. Judi forgot her money, so the other 7 friends each paid an extra \(\$2.50\).

The total extra money paid by the 7 friends was

\[
7 \times 2.50 = 17.50.
\]

This extra money covered Judi’s share of the bill.

So Judi’s share was \(\$17.50\). Since there were 8 equal shares, the total bill was

\[
8 \times 17.50 = 140.
\]

Thus the total bill was \(\$140\).

ANSWER 3: C

Problem 4:

Carrie’s garden is a rectangle measuring 6 feet by 8 feet.

First find the area:

\[
6 \times 8 = 48
\]

square feet.

She plants 4 strawberry plants per square foot, so the number of plants is

\[
48 \times 4 = 192.
\]

Each plant produces an average of 10 strawberries, so the total expected harvest is

\[
192 \times 10 = 1920.
\]

So Carrie can expect to harvest 1920 strawberries.

ANSWER 4: D

Problem 5:

We need to find the largest fraction among:

\[
\frac13,\frac14,\frac38,\frac5{12},\frac7{24}.
\]

Use a common denominator of 24:

\[
\frac13=\frac8{24}
\]

\[
\frac14=\frac6{24}
\]

\[
\frac38=\frac9{24}
\]

\[
\frac5{12}=\frac{10}{24}
\]

\[
\frac7{24}=\frac7{24}
\]

Now compare:

\[
\frac6{24},\frac7{24},\frac8{24},\frac9{24},\frac{10}{24}.
\]

The largest is

\[
\frac{10}{24}=\frac5{12}.
\]

ANSWER 5: D

Problem 6:

Let the original length be \(L\), and let the original width be \(W\).

The original area is

\[
LW.
\]

The length is increased by 10%, so the new length is

\[
1.10L.
\]

The width is decreased by 10%, so the new width is

\[
0.90W.
\]

The new area is

\[
(1.10L)(0.90W)=0.99LW.
\]

So the new area is \(99\%\) of the old area.

A common mistake is to think the \(+10\%\) and \(-10\%\) cancel, but they are applied to different dimensions, so the area changes by the product \(1.10 \times 0.90=0.99\).

ANSWER 6: B

Problem 7:

The expression is

\[
\left(\frac38+\frac78\right)\div \frac45.
\]

First add the fractions in parentheses:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45.
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54 = \frac{25}{16}.
\]

So the value is

\[
\frac{25}{16}.
\]

ANSWER 7: B

Problem 8:

We need to compute

\[
\frac1{10}+\frac9{100}+\frac9{1000}+\frac7{10000}.
\]

Convert each fraction to a decimal:

\[
\frac1{10}=0.1
\]

\[
\frac9{100}=0.09
\]

\[
\frac9{1000}=0.009
\]

\[
\frac7{10000}=0.0007
\]

Now add:

\[
0.1+0.09=0.19
\]

\[
0.19+0.009=0.199
\]

\[
0.199+0.0007=0.1997.
\]

So the answer is

\[
0.1997.
\]

ANSWER 8: C

Problem 9:

We need to add

\[
90+91+92+93+94+95+96+97+98+99.
\]

There are 10 numbers.

The average of the first and last number is

\[
\frac{90+99}{2}=\frac{189}{2}=94.5.
\]

So the sum is

\[
10 \times 94.5 = 945.
\]

Alternatively, pair the numbers:

\[
90+99=189
\]

\[
91+98=189
\]

\[
92+97=189
\]

\[
93+96=189
\]

\[
94+95=189
\]

There are 5 pairs, so

\[
5 \times 189 = 945.
\]

ANSWER 9: B

Problem 10:

We compare the number of lines of symmetry for each figure.

- Equilateral triangle: 3 lines of symmetry.
- Non-square rhombus: 2 lines of symmetry.
- Non-square rectangle: 2 lines of symmetry.
- Isosceles trapezoid: 1 line of symmetry.
- Square: 4 lines of symmetry.

The square has the greatest number of lines of symmetry.

ANSWER 10: E

Problem 11:

We are told:

- Neither Jo nor Bo has as much money as Flo, so Flo has more than Jo and Bo.
- Both Bo and Coe have more than Moe.
- Jo has more than Moe, but less than Bo.

From this:

\[
Bo > Moe
\]

\[
Coe > Moe
\]

\[
Jo > Moe
\]

Also, since Flo has more than Jo,

\[
Flo > Jo > Moe.
\]

So Flo also has more than Moe.

Therefore, everyone has more money than Moe.

So Moe has the least amount of money.

ANSWER 11: E

Problem 12:

We need to find the pair whose product is NOT equal to 36.

Check each pair:

A. \(\{-4,-9\}\)

\[
(-4)(-9)=36
\]

So A works.

B. \(\{-3,-12\}\)

\[
(-3)(-12)=36
\]

So B works.

C. \(\left\{\frac12,-72\right\}\)

\[
\frac12 \cdot (-72)=-36
\]

This is not 36.

D. \(\{1,36\}\)

\[
1 \cdot 36=36
\]

So D works.

E. \(\left\{\frac32,24\right\}\)

\[
\frac32 \cdot 24=36
\]

So E works.

The only pair whose product is not 36 is choice C.

ANSWER 12: C

**Problem 1:**
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

**Step-by-step working:**
1. Let the side length of the square be $s$ and the radius of the circle be $r$.
2. The area of the square is $s^2$.
3. The area of the circle is $\pi r^2$.
4. Since their areas are equal, we can set up the equation: $s^2 = \pi r^2$.
5. To find the ratio of the side length to the radius ($s/r$), we first divide both sides by $r^2$: $\frac{s^2}{r^2} = \pi$.
6. Taking the square root of both sides gives us: $\frac{s}{r} = \sqrt{\pi}$.
7. This matches choice B.

ANSWER 1: B

**Problem 2:**
How many digits are in the product $4^5 \cdot 5^{10}$?

**Step-by-step working:**
1. We want to find the number of digits in the number $4^5 \cdot 5^{10}$.
2. First, express the base $4$ as a power of $2$: $4 = 2^2$.
3. Substitute this into the expression: $(2^2)^5 \cdot 5^{10}$.
4. Use the power of a power property $(a^m)^n = a^{m \cdot n}$: $2^{10} \cdot 5^{10}$.
5. Since the exponents are the same, we can combine the bases: $(2 \cdot 5)^{10} = 10^{10}$.
6. The number $10^{10}$ is written as a $1$ followed by $10$ zeros.
7. Therefore, it has $1 + 10 = 11$ digits.

ANSWER 2: D

**Problem 3:**
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

**Step-by-step working:**
1. Judi's portion of the bill was covered by her $7$ friends.
2. Each of the $7$ friends paid an extra $\$2.50$.
3. The total amount they paid to cover Judi's share is $7 \times \$2.50 = \$17.50$.
4. This means that an equal share of the bill for one person is $\$17.50$.
5. Since there are $8$ friends in total, the total bill is $8 \times \$17.50$.
6. $8 \times \$17.50 = \$140.00$.

ANSWER 3: C

**Problem 4:**
Carrie has a rectangular garden that measures 6 feet by 8 feet. She plants the entire garden with strawberry plants. Carrie is able to plant 4 strawberry plants per square foot, and she harvests an average of 10 strawberries per plant. How many strawberries can she expect to harvest?

**Step-by-step working:**
1. First, find the area of the rectangular garden: $6 \text{ feet} \times 8 \text{ feet} = 48 \text{ square feet}$.
2. Next, calculate the total number of strawberry plants. She plants $4$ plants per square foot: $48 \times 4 = 192 \text{ plants}$.
3. Finally, calculate the total expected strawberry harvest. Each plant yields an average of $10$ strawberries: $192 \times 10 = 1920 \text{ strawberries}$.

ANSWER 4: D

**Problem 5:**
Which of the following is the largest?

**Step-by-step working:**
1. The choices given are the fractions: $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$.
2. To compare them easily, find a common denominator. The least common multiple (LCM) of $3, 4, 8, 12,$ and $24$ is $24$.
3. Convert each fraction:
   - A. $\frac{1}{3} = \frac{8}{24}$
   - B. $\frac{1}{4} = \frac{6}{24}$
   - C. $\frac{3}{8} = \frac{9}{24}$
   - D. $\frac{5}{12} = \frac{10}{24}$
   - E. $\frac{7}{24} = \frac{7}{24}$
4. Comparing the numerators ($8, 6, 9, 10, 7$), the largest is $10$.
5. Therefore, $\frac{10}{24}$ (which is $\frac{5}{12}$) is the largest fraction.

ANSWER 5: D

**Problem 6:**
The length of a rectangle is increased by 10% and the width is decreased by 10%. What percent of the old area is the new area?

**Step-by-step working:**
1. Let the original length be $L$ and the original width be $W$.
2. The old area is $A_{\text{old}} = L \times W$.
3. The new length is increased by $10\%$, making it $1.10L$.
4. The new width is decreased by $10\%$, making it $0.90W$.
5. The new area is $A_{\text{new}} = (1.10L) \times (0.90W) = 0.99(L \times W)$.
6. This means the new area is $0.99$ times the old area, which corresponds to $99\%$.

ANSWER 6: B

**Problem 7:**
$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}} =$

**Step-by-step working:**
1. First, evaluate the numerator by adding the fractions: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8}$.
2. Simplify the fraction: $\frac{10}{8} = \frac{5}{4}$.
3. Now, divide this result by the denominator $\frac{4}{5}$: $\frac{5}{4} \div \frac{4}{5}$.
4. Dividing by a fraction is the same as multiplying by its reciprocal: $\frac{5}{4} \times \frac{5}{4}$.
5. Multiply the numerators and the denominators: $\frac{5 \times 5}{4 \times 4} = \frac{25}{16}$.

ANSWER 7: B

**Problem 8:**
$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000} =$

**Step-by-step working:**
1. Convert each fraction into its decimal equivalent:
   - $\frac{1}{10} = 0.1$
   - $\frac{9}{100} = 0.09$
   - $\frac{9}{1000} = 0.009$
   - $\frac{7}{10000} = 0.0007$
2. Add the decimals together:
   $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$.

ANSWER 8: C

**Problem 9:**
$90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =$

**Step-by-step working:**
1. We are adding an arithmetic sequence of $10$ consecutive integers from $90$ to $99$.
2. We can split the sum into the tens and the units:
   - The tens contribute $90 \times 10 = 900$.
   - The units contribute $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$.
3. The sum of the digits from $0$ to $9$ is $\frac{9 \times 10}{2} = 45$.
4. Adding these together gives $900 + 45 = 945$.
*(Alternatively, using the formula for the sum of an arithmetic sequence: $S = \frac{n(a_1 + a_n)}{2} = \frac{10(90 + 99)}{2} = 5 \times 189 = 945$.)*

ANSWER 9: B

**Problem 10:**
Which of the following figures has the greatest number of lines of symmetry?

**Step-by-step working:**
1. Let's count the lines of symmetry for each given figure:
   - A. Equilateral triangle: $3$ lines of symmetry (from each vertex to the midpoint of the opposite side).
   - B. Non-square rhombus: $2$ lines of symmetry (along its diagonals).
   - C. Non-square rectangle: $2$ lines of symmetry (connecting the midpoints of opposite sides).
   - D. Isosceles trapezoid: $1$ line of sy

Problem 1:
The problem asks for the ratio of the side length of a square to the radius of a circle when both have the same area. Let's denote the side length of the square as $s$ and the radius of the circle as $r$. The area of the square is $s^2$ and the area of the circle is $\pi r^2$. Since they have the same area, we have $s^2 = \pi r^2$. To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$. Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$. Let's examine the choices: A. $\sqrt{\pi}^2 = \pi$, B. $\sqrt{\pi}$, C. $\pi$, D. $2\pi$, E. $\pi^2$. The correct ratio is given by choice B.

ANSWER 1: B

Problem 2:
To find the number of digits in $4^5 \cdot 5^{10}$, let's simplify the expression. $4^5 = (2^2)^5 = 2^{10}$, so $4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$. $10^{10}$ is a 1 followed by 10 zeros, making it an 11-digit number. Let's check the choices: A. 8, B. 9, C. 10, D. 11, E. 12. The correct number of digits is given by choice D.

ANSWER 2: D

Problem 3:
The problem states that 8 friends agreed to share a bill equally, but because Judi forgot her money, each of her 7 friends paid an extra $2.50. Let's denote the total bill as $B$. The amount each friend was originally supposed to pay is $\frac{B}{8}$. When Judi's portion is covered by her 7 friends, each pays $\frac{B}{8} + $2.50$. The total amount paid by the 7 friends is $7(\frac{B}{8} + 2.50)$, which equals $B$. So, we have $7(\frac{B}{8} + 2.50) = B$. Simplifying this gives $\frac{7B}{8} + 17.50 = B$. Subtracting $\frac{7B}{8}$ from both sides gives $17.50 = \frac{B}{8}$. Multiplying both sides by 8 gives $B = 140$. Let's check the choices: A. $120, B. $128, C. $140, D. $144, E. $160. The total bill matches choice C.

ANSWER 3: C

Problem 4:
Carrie has a garden measuring 6 feet by 8 feet, so its area is $6 \times 8 = 48$ square feet. She plants 4 strawberry plants per square foot, so she plants $48 \times 4 = 192$ plants. With an average of 10 strawberries per plant, she can expect to harvest $192 \times 10 = 1920$ strawberries. Let's check the choices: A. 560, B. 960, C. 1120, D. 1920, E. 3840. The expected harvest matches choice D.

ANSWER 4: D

Problem 5:
To compare the given fractions, let's convert them to equivalent decimals or find a common denominator. The fractions are $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$. Converting them to equivalent fractions with a common denominator (24) gives $\frac{8}{24}$, $\frac{6}{24}$, $\frac{9}{24}$, $\frac{10}{24}$, and $\frac{7}{24}$. Comparing these, $\frac{10}{24}$ is the largest. So, $\frac{5}{12}$ is the largest fraction. Let's check the choices: A. $\frac{1}{3}$, B. $\frac{1}{4}$, C. $\frac{3}{8}$, D. $\frac{5}{12}$, E. $\frac{7}{24}$. The largest fraction is given by choice D.

ANSWER 5: D

Problem 6:
Let's denote the original length and width of the rectangle as $L$ and $W$, respectively. The original area is $L \times W$. After increasing the length by 10% and decreasing the width by 10%, the new length is $1.1L$ and the new width is $0.9W$. The new area is $1.1L \times 0.9W = 0.99LW$. To find what percent of the old area the new area is, we calculate $\frac{0.99LW}{LW} \times 100\% = 99\%$. Let's check the choices: A. 90, B. 99, C. 100, D. 101, E. 110. The new area is 99% of the old area, matching choice B.

ANSWER 6: B

Problem 7:
To evaluate $\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$, first simplify the numerator: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$. Then, divide by $\frac{4}{5}$, which is equivalent to multiplying by $\frac{5}{4}$. So, $\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$. Let's check the choices: A. 1, B. $\frac{25}{16}$, C. 2, D. $\frac{43}{20}$, E. $\frac{47}{16}$. The result matches choice B.

ANSWER 7: B

Problem 8:
To evaluate $\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$, convert each fraction to a decimal: $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$. Let's check the choices: A. 0.0026, B. 0.0197, C. 0.1997, D. 0.26, E. 1.997. The sum matches choice C.

ANSWER 8: C

Problem 9:
The sum $90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99$ is an arithmetic series with the first term $a = 90$, the last term $l = 99$, and $n = 10$ terms. The sum of an arithmetic series is given by $S = \frac{n}{2}(a + l)$. So, $S = \frac{10}{2}(90 + 99) = 5 \times 189 = 945$. Let's check the choices: A. 845, B. 945, C. 1005, D. 1025, E. 1045. The sum matches choice B.

ANSWER 9: B

Problem 10:
Let's analyze the number of lines of symmetry for each figure: A. An equilateral triangle has 3 lines of symmetry. B. A non-square rhombus has 2 lines of symmetry. C. A non-square rectangle has 2 lines of symmetry. D. An isosceles trapezoid has 1 line of symmetry. E. A square has 4 lines of symmetry. The figure with the greatest number of lines of symmetry is the square. Let's check the choices: the square is choice E.

ANSWER 10: E

Problem 11:
From the given information: Neither Jo nor Bo has as much money as Flo, so $Jo < Flo$ and $Bo < Flo$. Both Bo and Coe have more than Moe, so $Moe < Bo$ and $Moe < Coe$. Jo has more than Moe but less than Bo, so $Moe < Jo < Bo$. Combining these, we have $Moe < Jo < Bo < Flo$ and $Moe < Coe$. Since we don't have a direct comparison between Coe and the others except Moe, and we know $Bo < Flo$, the order for the least amount of money is clear for Moe. Let's check the choices: A. Bo, B. Coe, C. Flo, D. Jo, E. Moe. Moe has the least amount of money, matching choice E.

ANSWER 11: E

Problem 12:
Let's check each pair to see if their product equals 36: A. $(-4) \times (-9) = 36$, B. $(-3) \times (-12) = 36$, C. $\frac{1}{2} \times (-72) = -36$, D. $1 \times 36 = 36$, E. $\frac{3}{2} \times 24 = 36$. The pair that does NOT have a product equal to 36 is C. $\{\frac{1}{2}, -72\}$.

ANSWER 12: C

Problem 1:

We are asked for the ratio

\[
\frac{\text{side length of square}}{\text{radius of circle}}.
\]

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The square has area

\[
s^2.
\]

The circle has area

\[
\pi r^2.
\]

They have the same area, so

\[
s^2=\pi r^2.
\]

Taking the square root of both sides,

\[
s=r\sqrt{\pi}.
\]

So

\[
\frac{s}{r}=\sqrt{\pi}.
\]

The choices involving \(\pi\), \(2\pi\), or fractions of \(\pi\) are too large or have the wrong form. The correct ratio is \(\sqrt{\pi}\).

ANSWER 1: B

Problem 2:

We need the number of digits in

\[
4^5 \cdot 5^{10}.
\]

Rewrite \(4^5\):

\[
4^5=(2^2)^5=2^{10}.
\]

So

\[
4^5 \cdot 5^{10}=2^{10}\cdot 5^{10}.
\]

Since \(2^{10}\cdot 5^{10}=(2\cdot 5)^{10}\),

\[
2^{10}\cdot 5^{10}=10^{10}.
\]

Now,

\[
10^{10}=10,000,000,000,
\]

which is a 1 followed by 10 zeros.

So it has

\[
11
\]

digits.

ANSWER 2: D

Problem 3:

Eight friends were supposed to share the bill equally. Judi forgot her money, so the other 7 friends each paid an extra \(\$2.50\).

The total extra money paid by the 7 friends was

\[
7 \times 2.50 = 17.50.
\]

This extra money covered Judi’s share of the bill.

So Judi’s share was \(\$17.50\). Since there were 8 equal shares, the total bill was

\[
8 \times 17.50 = 140.
\]

Thus the total bill was \(\$140\).

ANSWER 3: C

Problem 4:

Carrie’s garden is a rectangle measuring 6 feet by 8 feet.

First find the area:

\[
6 \times 8 = 48
\]

square feet.

She plants 4 strawberry plants per square foot, so the number of plants is

\[
48 \times 4 = 192.
\]

Each plant produces an average of 10 strawberries, so the total expected harvest is

\[
192 \times 10 = 1920.
\]

So Carrie can expect to harvest 1920 strawberries.

ANSWER 4: D

Problem 5:

We need to find the largest fraction among:

\[
\frac13,\frac14,\frac38,\frac5{12},\frac7{24}.
\]

Use a common denominator of 24:

\[
\frac13=\frac8{24}
\]

\[
\frac14=\frac6{24}
\]

\[
\frac38=\frac9{24}
\]

\[
\frac5{12}=\frac{10}{24}
\]

\[
\frac7{24}=\frac7{24}
\]

Now compare:

\[
\frac6{24},\frac7{24},\frac8{24},\frac9{24},\frac{10}{24}.
\]

The largest is

\[
\frac{10}{24}=\frac5{12}.
\]

ANSWER 5: D

Problem 6:

Let the original length be \(L\), and let the original width be \(W\).

The original area is

\[
LW.
\]

The length is increased by 10%, so the new length is

\[
1.10L.
\]

The width is decreased by 10%, so the new width is

\[
0.90W.
\]

The new area is

\[
(1.10L)(0.90W)=0.99LW.
\]

So the new area is \(99\%\) of the old area.

A common mistake is to think the \(+10\%\) and \(-10\%\) cancel, but they are applied to different dimensions, so the area changes by the product \(1.10 \times 0.90=0.99\).

ANSWER 6: B

Problem 7:

The expression is

\[
\left(\frac38+\frac78\right)\div \frac45.
\]

First add the fractions in parentheses:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45.
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54 = \frac{25}{16}.
\]

So the value is

\[
\frac{25}{16}.
\]

ANSWER 7: B

Problem 8:

We need to compute

\[
\frac1{10}+\frac9{100}+\frac9{1000}+\frac7{10000}.
\]

Convert each fraction to a decimal:

\[
\frac1{10}=0.1
\]

\[
\frac9{100}=0.09
\]

\[
\frac9{1000}=0.009
\]

\[
\frac7{10000}=0.0007
\]

Now add:

\[
0.1+0.09=0.19
\]

\[
0.19+0.009=0.199
\]

\[
0.199+0.0007=0.1997.
\]

So the answer is

\[
0.1997.
\]

ANSWER 8: C

Problem 9:

We need to add

\[
90+91+92+93+94+95+96+97+98+99.
\]

There are 10 numbers.

The average of the first and last number is

\[
\frac{90+99}{2}=\frac{189}{2}=94.5.
\]

So the sum is

\[
10 \times 94.5 = 945.
\]

Alternatively, pair the numbers:

\[
90+99=189
\]

\[
91+98=189
\]

\[
92+97=189
\]

\[
93+96=189
\]

\[
94+95=189
\]

There are 5 pairs, so

\[
5 \times 189 = 945.
\]

ANSWER 9: B

Problem 10:

We compare the number of lines of symmetry for each figure.

- Equilateral triangle: 3 lines of symmetry.
- Non-square rhombus: 2 lines of symmetry.
- Non-square rectangle: 2 lines of symmetry.
- Isosceles trapezoid: 1 line of symmetry.
- Square: 4 lines of symmetry.

The square has the greatest number of lines of symmetry.

ANSWER 10: E

Problem 11:

We are told:

- Neither Jo nor Bo has as much money as Flo, so Flo has more than Jo and Bo.
- Both Bo and Coe have more than Moe.
- Jo has more than Moe, but less than Bo.

From this:

\[
Bo > Moe
\]

\[
Coe > Moe
\]

\[
Jo > Moe
\]

Also, since Flo has more than Jo,

\[
Flo > Jo > Moe.
\]

So Flo also has more than Moe.

Therefore, everyone has more money than Moe.

So Moe has the least amount of money.

ANSWER 11: E

Problem 12:

We need to find the pair whose product is NOT equal to 36.

Check each pair:

A. \(\{-4,-9\}\)

\[
(-4)(-9)=36
\]

So A works.

B. \(\{-3,-12\}\)

\[
(-3)(-12)=36
\]

So B works.

C. \(\left\{\frac12,-72\right\}\)

\[
\frac12 \cdot (-72)=-36
\]

This is not 36.

D. \(\{1,36\}\)

\[
1 \cdot 36=36
\]

So D works.

E. \(\left\{\frac32,24\right\}\)

\[
\frac32 \cdot 24=36
\]

So E works.

The only pair whose product is not 36 is choice C.

ANSWER 12: C

**Problem 1:**
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

**Step-by-step working:**
1. Let the side length of the square be $s$ and the radius of the circle be $r$.
2. The area of the square is $s^2$.
3. The area of the circle is $\pi r^2$.
4. Since their areas are equal, we can set up the equation: $s^2 = \pi r^2$.
5. To find the ratio of the side length to the radius ($s/r$), we first divide both sides by $r^2$: $\frac{s^2}{r^2} = \pi$.
6. Taking the square root of both sides gives us: $\frac{s}{r} = \sqrt{\pi}$.
7. This matches choice B.

ANSWER 1: B

**Problem 2:**
How many digits are in the product $4^5 \cdot 5^{10}$?

**Step-by-step working:**
1. We want to find the number of digits in the number $4^5 \cdot 5^{10}$.
2. First, express the base $4$ as a power of $2$: $4 = 2^2$.
3. Substitute this into the expression: $(2^2)^5 \cdot 5^{10}$.
4. Use the power of a power property $(a^m)^n = a^{m \cdot n}$: $2^{10} \cdot 5^{10}$.
5. Since the exponents are the same, we can combine the bases: $(2 \cdot 5)^{10} = 10^{10}$.
6. The number $10^{10}$ is written as a $1$ followed by $10$ zeros.
7. Therefore, it has $1 + 10 = 11$ digits.

ANSWER 2: D

**Problem 3:**
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

**Step-by-step working:**
1. Judi's portion of the bill was covered by her $7$ friends.
2. Each of the $7$ friends paid an extra $\$2.50$.
3. The total amount they paid to cover Judi's share is $7 \times \$2.50 = \$17.50$.
4. This means that an equal share of the bill for one person is $\$17.50$.
5. Since there are $8$ friends in total, the total bill is $8 \times \$17.50$.
6. $8 \times \$17.50 = \$140.00$.

ANSWER 3: C

**Problem 4:**
Carrie has a rectangular garden that measures 6 feet by 8 feet. She plants the entire garden with strawberry plants. Carrie is able to plant 4 strawberry plants per square foot, and she harvests an average of 10 strawberries per plant. How many strawberries can she expect to harvest?

**Step-by-step working:**
1. First, find the area of the rectangular garden: $6 \text{ feet} \times 8 \text{ feet} = 48 \text{ square feet}$.
2. Next, calculate the total number of strawberry plants. She plants $4$ plants per square foot: $48 \times 4 = 192 \text{ plants}$.
3. Finally, calculate the total expected strawberry harvest. Each plant yields an average of $10$ strawberries: $192 \times 10 = 1920 \text{ strawberries}$.

ANSWER 4: D

**Problem 5:**
Which of the following is the largest?

**Step-by-step working:**
1. The choices given are the fractions: $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$.
2. To compare them easily, find a common denominator. The least common multiple (LCM) of $3, 4, 8, 12,$ and $24$ is $24$.
3. Convert each fraction:
   - A. $\frac{1}{3} = \frac{8}{24}$
   - B. $\frac{1}{4} = \frac{6}{24}$
   - C. $\frac{3}{8} = \frac{9}{24}$
   - D. $\frac{5}{12} = \frac{10}{24}$
   - E. $\frac{7}{24} = \frac{7}{24}$
4. Comparing the numerators ($8, 6, 9, 10, 7$), the largest is $10$.
5. Therefore, $\frac{10}{24}$ (which is $\frac{5}{12}$) is the largest fraction.

ANSWER 5: D

**Problem 6:**
The length of a rectangle is increased by 10% and the width is decreased by 10%. What percent of the old area is the new area?

**Step-by-step working:**
1. Let the original length be $L$ and the original width be $W$.
2. The old area is $A_{\text{old}} = L \times W$.
3. The new length is increased by $10\%$, making it $1.10L$.
4. The new width is decreased by $10\%$, making it $0.90W$.
5. The new area is $A_{\text{new}} = (1.10L) \times (0.90W) = 0.99(L \times W)$.
6. This means the new area is $0.99$ times the old area, which corresponds to $99\%$.

ANSWER 6: B

**Problem 7:**
$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}} =$

**Step-by-step working:**
1. First, evaluate the numerator by adding the fractions: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8}$.
2. Simplify the fraction: $\frac{10}{8} = \frac{5}{4}$.
3. Now, divide this result by the denominator $\frac{4}{5}$: $\frac{5}{4} \div \frac{4}{5}$.
4. Dividing by a fraction is the same as multiplying by its reciprocal: $\frac{5}{4} \times \frac{5}{4}$.
5. Multiply the numerators and the denominators: $\frac{5 \times 5}{4 \times 4} = \frac{25}{16}$.

ANSWER 7: B

**Problem 8:**
$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000} =$

**Step-by-step working:**
1. Convert each fraction into its decimal equivalent:
   - $\frac{1}{10} = 0.1$
   - $\frac{9}{100} = 0.09$
   - $\frac{9}{1000} = 0.009$
   - $\frac{7}{10000} = 0.0007$
2. Add the decimals together:
   $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$.

ANSWER 8: C

**Problem 9:**
$90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =$

**Step-by-step working:**
1. We are adding an arithmetic sequence of $10$ consecutive integers from $90$ to $99$.
2. We can split the sum into the tens and the units:
   - The tens contribute $90 \times 10 = 900$.
   - The units contribute $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$.
3. The sum of the digits from $0$ to $9$ is $\frac{9 \times 10}{2} = 45$.
4. Adding these together gives $900 + 45 = 945$.
*(Alternatively, using the formula for the sum of an arithmetic sequence: $S = \frac{n(a_1 + a_n)}{2} = \frac{10(90 + 99)}{2} = 5 \times 189 = 945$.)*

ANSWER 9: B

**Problem 10:**
Which of the following figures has the greatest number of lines of symmetry?

**Step-by-step working:**
1. Let's count the lines of symmetry for each given figure:
   - A. Equilateral triangle: $3$ lines of symmetry (from each vertex to the midpoint of the opposite side).
   - B. Non-square rhombus: $2$ lines of symmetry (along its diagonals).
   - C. Non-square rectangle: $2$ lines of symmetry (connecting the midpoints of opposite sides).
   - D. Isosceles trapezoid: $1$ line of sy

Problem 1:
The problem asks for the ratio of the side length of a square to the radius of a circle when both have the same area. Let's denote the side length of the square as $s$ and the radius of the circle as $r$. The area of the square is $s^2$ and the area of the circle is $\pi r^2$. Since they have the same area, we have $s^2 = \pi r^2$. To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$. Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$. Let's examine the choices: A. $\sqrt{\pi}^2 = \pi$, B. $\sqrt{\pi}$, C. $\pi$, D. $2\pi$, E. $\pi^2$. The correct ratio is given by choice B.

ANSWER 1: B

Problem 2:
To find the number of digits in $4^5 \cdot 5^{10}$, let's simplify the expression. $4^5 = (2^2)^5 = 2^{10}$, so $4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$. $10^{10}$ is a 1 followed by 10 zeros, making it an 11-digit number. Let's check the choices: A. 8, B. 9, C. 10, D. 11, E. 12. The correct number of digits is given by choice D.

ANSWER 2: D

Problem 3:
The problem states that 8 friends agreed to share a bill equally, but because Judi forgot her money, each of her 7 friends paid an extra $2.50. Let's denote the total bill as $B$. The amount each friend was originally supposed to pay is $\frac{B}{8}$. When Judi's portion is covered by her 7 friends, each pays $\frac{B}{8} + $2.50$. The total amount paid by the 7 friends is $7(\frac{B}{8} + 2.50)$, which equals $B$. So, we have $7(\frac{B}{8} + 2.50) = B$. Simplifying this gives $\frac{7B}{8} + 17.50 = B$. Subtracting $\frac{7B}{8}$ from both sides gives $17.50 = \frac{B}{8}$. Multiplying both sides by 8 gives $B = 140$. Let's check the choices: A. $120, B. $128, C. $140, D. $144, E. $160. The total bill matches choice C.

ANSWER 3: C

Problem 4:
Carrie has a garden measuring 6 feet by 8 feet, so its area is $6 \times 8 = 48$ square feet. She plants 4 strawberry plants per square foot, so she plants $48 \times 4 = 192$ plants. With an average of 10 strawberries per plant, she can expect to harvest $192 \times 10 = 1920$ strawberries. Let's check the choices: A. 560, B. 960, C. 1120, D. 1920, E. 3840. The expected harvest matches choice D.

ANSWER 4: D

Problem 5:
To compare the given fractions, let's convert them to equivalent decimals or find a common denominator. The fractions are $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$. Converting them to equivalent fractions with a common denominator (24) gives $\frac{8}{24}$, $\frac{6}{24}$, $\frac{9}{24}$, $\frac{10}{24}$, and $\frac{7}{24}$. Comparing these, $\frac{10}{24}$ is the largest. So, $\frac{5}{12}$ is the largest fraction. Let's check the choices: A. $\frac{1}{3}$, B. $\frac{1}{4}$, C. $\frac{3}{8}$, D. $\frac{5}{12}$, E. $\frac{7}{24}$. The largest fraction is given by choice D.

ANSWER 5: D

Problem 6:
Let's denote the original length and width of the rectangle as $L$ and $W$, respectively. The original area is $L \times W$. After increasing the length by 10% and decreasing the width by 10%, the new length is $1.1L$ and the new width is $0.9W$. The new area is $1.1L \times 0.9W = 0.99LW$. To find what percent of the old area the new area is, we calculate $\frac{0.99LW}{LW} \times 100\% = 99\%$. Let's check the choices: A. 90, B. 99, C. 100, D. 101, E. 110. The new area is 99% of the old area, matching choice B.

ANSWER 6: B

Problem 7:
To evaluate $\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$, first simplify the numerator: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$. Then, divide by $\frac{4}{5}$, which is equivalent to multiplying by $\frac{5}{4}$. So, $\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$. Let's check the choices: A. 1, B. $\frac{25}{16}$, C. 2, D. $\frac{43}{20}$, E. $\frac{47}{16}$. The result matches choice B.

ANSWER 7: B

Problem 8:
To evaluate $\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$, convert each fraction to a decimal: $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$. Let's check the choices: A. 0.0026, B. 0.0197, C. 0.1997, D. 0.26, E. 1.997. The sum matches choice C.

ANSWER 8: C

Problem 9:
The sum $90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99$ is an arithmetic series with the first term $a = 90$, the last term $l = 99$, and $n = 10$ terms. The sum of an arithmetic series is given by $S = \frac{n}{2}(a + l)$. So, $S = \frac{10}{2}(90 + 99) = 5 \times 189 = 945$. Let's check the choices: A. 845, B. 945, C. 1005, D. 1025, E. 1045. The sum matches choice B.

ANSWER 9: B

Problem 10:
Let's analyze the number of lines of symmetry for each figure: A. An equilateral triangle has 3 lines of symmetry. B. A non-square rhombus has 2 lines of symmetry. C. A non-square rectangle has 2 lines of symmetry. D. An isosceles trapezoid has 1 line of symmetry. E. A square has 4 lines of symmetry. The figure with the greatest number of lines of symmetry is the square. Let's check the choices: the square is choice E.

ANSWER 10: E

Problem 11:
From the given information: Neither Jo nor Bo has as much money as Flo, so $Jo < Flo$ and $Bo < Flo$. Both Bo and Coe have more than Moe, so $Moe < Bo$ and $Moe < Coe$. Jo has more than Moe but less than Bo, so $Moe < Jo < Bo$. Combining these, we have $Moe < Jo < Bo < Flo$ and $Moe < Coe$. Since we don't have a direct comparison between Coe and the others except Moe, and we know $Bo < Flo$, the order for the least amount of money is clear for Moe. Let's check the choices: A. Bo, B. Coe, C. Flo, D. Jo, E. Moe. Moe has the least amount of money, matching choice E.

ANSWER 11: E

Problem 12:
Let's check each pair to see if their product equals 36: A. $(-4) \times (-9) = 36$, B. $(-3) \times (-12) = 36$, C. $\frac{1}{2} \times (-72) = -36$, D. $1 \times 36 = 36$, E. $\frac{3}{2} \times 24 = 36$. The pair that does NOT have a product equal to 36 is C. $\{\frac{1}{2}, -72\}$.

ANSWER 12: C

Problem 1:

We are asked for the ratio

\[
\frac{\text{side length of square}}{\text{radius of circle}}.
\]

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The square has area

\[
s^2.
\]

The circle has area

\[
\pi r^2.
\]

They have the same area, so

\[
s^2=\pi r^2.
\]

Taking the square root of both sides,

\[
s=r\sqrt{\pi}.
\]

So

\[
\frac{s}{r}=\sqrt{\pi}.
\]

The choices involving \(\pi\), \(2\pi\), or fractions of \(\pi\) are too large or have the wrong form. The correct ratio is \(\sqrt{\pi}\).

ANSWER 1: B

Problem 2:

We need the number of digits in

\[
4^5 \cdot 5^{10}.
\]

Rewrite \(4^5\):

\[
4^5=(2^2)^5=2^{10}.
\]

So

\[
4^5 \cdot 5^{10}=2^{10}\cdot 5^{10}.
\]

Since \(2^{10}\cdot 5^{10}=(2\cdot 5)^{10}\),

\[
2^{10}\cdot 5^{10}=10^{10}.
\]

Now,

\[
10^{10}=10,000,000,000,
\]

which is a 1 followed by 10 zeros.

So it has

\[
11
\]

digits.

ANSWER 2: D

Problem 3:

Eight friends were supposed to share the bill equally. Judi forgot her money, so the other 7 friends each paid an extra \(\$2.50\).

The total extra money paid by the 7 friends was

\[
7 \times 2.50 = 17.50.
\]

This extra money covered Judi’s share of the bill.

So Judi’s share was \(\$17.50\). Since there were 8 equal shares, the total bill was

\[
8 \times 17.50 = 140.
\]

Thus the total bill was \(\$140\).

ANSWER 3: C

Problem 4:

Carrie’s garden is a rectangle measuring 6 feet by 8 feet.

First find the area:

\[
6 \times 8 = 48
\]

square feet.

She plants 4 strawberry plants per square foot, so the number of plants is

\[
48 \times 4 = 192.
\]

Each plant produces an average of 10 strawberries, so the total expected harvest is

\[
192 \times 10 = 1920.
\]

So Carrie can expect to harvest 1920 strawberries.

ANSWER 4: D

Problem 5:

We need to find the largest fraction among:

\[
\frac13,\frac14,\frac38,\frac5{12},\frac7{24}.
\]

Use a common denominator of 24:

\[
\frac13=\frac8{24}
\]

\[
\frac14=\frac6{24}
\]

\[
\frac38=\frac9{24}
\]

\[
\frac5{12}=\frac{10}{24}
\]

\[
\frac7{24}=\frac7{24}
\]

Now compare:

\[
\frac6{24},\frac7{24},\frac8{24},\frac9{24},\frac{10}{24}.
\]

The largest is

\[
\frac{10}{24}=\frac5{12}.
\]

ANSWER 5: D

Problem 6:

Let the original length be \(L\), and let the original width be \(W\).

The original area is

\[
LW.
\]

The length is increased by 10%, so the new length is

\[
1.10L.
\]

The width is decreased by 10%, so the new width is

\[
0.90W.
\]

The new area is

\[
(1.10L)(0.90W)=0.99LW.
\]

So the new area is \(99\%\) of the old area.

A common mistake is to think the \(+10\%\) and \(-10\%\) cancel, but they are applied to different dimensions, so the area changes by the product \(1.10 \times 0.90=0.99\).

ANSWER 6: B

Problem 7:

The expression is

\[
\left(\frac38+\frac78\right)\div \frac45.
\]

First add the fractions in parentheses:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45.
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54 = \frac{25}{16}.
\]

So the value is

\[
\frac{25}{16}.
\]

ANSWER 7: B

Problem 8:

We need to compute

\[
\frac1{10}+\frac9{100}+\frac9{1000}+\frac7{10000}.
\]

Convert each fraction to a decimal:

\[
\frac1{10}=0.1
\]

\[
\frac9{100}=0.09
\]

\[
\frac9{1000}=0.009
\]

\[
\frac7{10000}=0.0007
\]

Now add:

\[
0.1+0.09=0.19
\]

\[
0.19+0.009=0.199
\]

\[
0.199+0.0007=0.1997.
\]

So the answer is

\[
0.1997.
\]

ANSWER 8: C

Problem 9:

We need to add

\[
90+91+92+93+94+95+96+97+98+99.
\]

There are 10 numbers.

The average of the first and last number is

\[
\frac{90+99}{2}=\frac{189}{2}=94.5.
\]

So the sum is

\[
10 \times 94.5 = 945.
\]

Alternatively, pair the numbers:

\[
90+99=189
\]

\[
91+98=189
\]

\[
92+97=189
\]

\[
93+96=189
\]

\[
94+95=189
\]

There are 5 pairs, so

\[
5 \times 189 = 945.
\]

ANSWER 9: B

Problem 10:

We compare the number of lines of symmetry for each figure.

- Equilateral triangle: 3 lines of symmetry.
- Non-square rhombus: 2 lines of symmetry.
- Non-square rectangle: 2 lines of symmetry.
- Isosceles trapezoid: 1 line of symmetry.
- Square: 4 lines of symmetry.

The square has the greatest number of lines of symmetry.

ANSWER 10: E

Problem 11:

We are told:

- Neither Jo nor Bo has as much money as Flo, so Flo has more than Jo and Bo.
- Both Bo and Coe have more than Moe.
- Jo has more than Moe, but less than Bo.

From this:

\[
Bo > Moe
\]

\[
Coe > Moe
\]

\[
Jo > Moe
\]

Also, since Flo has more than Jo,

\[
Flo > Jo > Moe.
\]

So Flo also has more than Moe.

Therefore, everyone has more money than Moe.

So Moe has the least amount of money.

ANSWER 11: E

Problem 12:

We need to find the pair whose product is NOT equal to 36.

Check each pair:

A. \(\{-4,-9\}\)

\[
(-4)(-9)=36
\]

So A works.

B. \(\{-3,-12\}\)

\[
(-3)(-12)=36
\]

So B works.

C. \(\left\{\frac12,-72\right\}\)

\[
\frac12 \cdot (-72)=-36
\]

This is not 36.

D. \(\{1,36\}\)

\[
1 \cdot 36=36
\]

So D works.

E. \(\left\{\frac32,24\right\}\)

\[
\frac32 \cdot 24=36
\]

So E works.

The only pair whose product is not 36 is choice C.

ANSWER 12: C

**Problem 1:**
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

**Step-by-step working:**
1. Let the side length of the square be $s$ and the radius of the circle be $r$.
2. The area of the square is $s^2$.
3. The area of the circle is $\pi r^2$.
4. Since their areas are equal, we can set up the equation: $s^2 = \pi r^2$.
5. To find the ratio of the side length to the radius ($s/r$), we first divide both sides by $r^2$: $\frac{s^2}{r^2} = \pi$.
6. Taking the square root of both sides gives us: $\frac{s}{r} = \sqrt{\pi}$.
7. This matches choice B.

ANSWER 1: B

**Problem 2:**
How many digits are in the product $4^5 \cdot 5^{10}$?

**Step-by-step working:**
1. We want to find the number of digits in the number $4^5 \cdot 5^{10}$.
2. First, express the base $4$ as a power of $2$: $4 = 2^2$.
3. Substitute this into the expression: $(2^2)^5 \cdot 5^{10}$.
4. Use the power of a power property $(a^m)^n = a^{m \cdot n}$: $2^{10} \cdot 5^{10}$.
5. Since the exponents are the same, we can combine the bases: $(2 \cdot 5)^{10} = 10^{10}$.
6. The number $10^{10}$ is written as a $1$ followed by $10$ zeros.
7. Therefore, it has $1 + 10 = 11$ digits.

ANSWER 2: D

**Problem 3:**
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

**Step-by-step working:**
1. Judi's portion of the bill was covered by her $7$ friends.
2. Each of the $7$ friends paid an extra $\$2.50$.
3. The total amount they paid to cover Judi's share is $7 \times \$2.50 = \$17.50$.
4. This means that an equal share of the bill for one person is $\$17.50$.
5. Since there are $8$ friends in total, the total bill is $8 \times \$17.50$.
6. $8 \times \$17.50 = \$140.00$.

ANSWER 3: C

**Problem 4:**
Carrie has a rectangular garden that measures 6 feet by 8 feet. She plants the entire garden with strawberry plants. Carrie is able to plant 4 strawberry plants per square foot, and she harvests an average of 10 strawberries per plant. How many strawberries can she expect to harvest?

**Step-by-step working:**
1. First, find the area of the rectangular garden: $6 \text{ feet} \times 8 \text{ feet} = 48 \text{ square feet}$.
2. Next, calculate the total number of strawberry plants. She plants $4$ plants per square foot: $48 \times 4 = 192 \text{ plants}$.
3. Finally, calculate the total expected strawberry harvest. Each plant yields an average of $10$ strawberries: $192 \times 10 = 1920 \text{ strawberries}$.

ANSWER 4: D

**Problem 5:**
Which of the following is the largest?

**Step-by-step working:**
1. The choices given are the fractions: $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$.
2. To compare them easily, find a common denominator. The least common multiple (LCM) of $3, 4, 8, 12,$ and $24$ is $24$.
3. Convert each fraction:
   - A. $\frac{1}{3} = \frac{8}{24}$
   - B. $\frac{1}{4} = \frac{6}{24}$
   - C. $\frac{3}{8} = \frac{9}{24}$
   - D. $\frac{5}{12} = \frac{10}{24}$
   - E. $\frac{7}{24} = \frac{7}{24}$
4. Comparing the numerators ($8, 6, 9, 10, 7$), the largest is $10$.
5. Therefore, $\frac{10}{24}$ (which is $\frac{5}{12}$) is the largest fraction.

ANSWER 5: D

**Problem 6:**
The length of a rectangle is increased by 10% and the width is decreased by 10%. What percent of the old area is the new area?

**Step-by-step working:**
1. Let the original length be $L$ and the original width be $W$.
2. The old area is $A_{\text{old}} = L \times W$.
3. The new length is increased by $10\%$, making it $1.10L$.
4. The new width is decreased by $10\%$, making it $0.90W$.
5. The new area is $A_{\text{new}} = (1.10L) \times (0.90W) = 0.99(L \times W)$.
6. This means the new area is $0.99$ times the old area, which corresponds to $99\%$.

ANSWER 6: B

**Problem 7:**
$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}} =$

**Step-by-step working:**
1. First, evaluate the numerator by adding the fractions: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8}$.
2. Simplify the fraction: $\frac{10}{8} = \frac{5}{4}$.
3. Now, divide this result by the denominator $\frac{4}{5}$: $\frac{5}{4} \div \frac{4}{5}$.
4. Dividing by a fraction is the same as multiplying by its reciprocal: $\frac{5}{4} \times \frac{5}{4}$.
5. Multiply the numerators and the denominators: $\frac{5 \times 5}{4 \times 4} = \frac{25}{16}$.

ANSWER 7: B

**Problem 8:**
$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000} =$

**Step-by-step working:**
1. Convert each fraction into its decimal equivalent:
   - $\frac{1}{10} = 0.1$
   - $\frac{9}{100} = 0.09$
   - $\frac{9}{1000} = 0.009$
   - $\frac{7}{10000} = 0.0007$
2. Add the decimals together:
   $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$.

ANSWER 8: C

**Problem 9:**
$90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =$

**Step-by-step working:**
1. We are adding an arithmetic sequence of $10$ consecutive integers from $90$ to $99$.
2. We can split the sum into the tens and the units:
   - The tens contribute $90 \times 10 = 900$.
   - The units contribute $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$.
3. The sum of the digits from $0$ to $9$ is $\frac{9 \times 10}{2} = 45$.
4. Adding these together gives $900 + 45 = 945$.
*(Alternatively, using the formula for the sum of an arithmetic sequence: $S = \frac{n(a_1 + a_n)}{2} = \frac{10(90 + 99)}{2} = 5 \times 189 = 945$.)*

ANSWER 9: B

**Problem 10:**
Which of the following figures has the greatest number of lines of symmetry?

**Step-by-step working:**
1. Let's count the lines of symmetry for each given figure:
   - A. Equilateral triangle: $3$ lines of symmetry (from each vertex to the midpoint of the opposite side).
   - B. Non-square rhombus: $2$ lines of symmetry (along its diagonals).
   - C. Non-square rectangle: $2$ lines of symmetry (connecting the midpoints of opposite sides).
   - D. Isosceles trapezoid: $1$ line of sy

Problem 1:
The problem asks for the ratio of the side length of a square to the radius of a circle when both have the same area. Let's denote the side length of the square as $s$ and the radius of the circle as $r$. The area of the square is $s^2$ and the area of the circle is $\pi r^2$. Since they have the same area, we have $s^2 = \pi r^2$. To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$. Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$. Let's examine the choices: A. $\sqrt{\pi}^2 = \pi$, B. $\sqrt{\pi}$, C. $\pi$, D. $2\pi$, E. $\pi^2$. The correct ratio is given by choice B.

ANSWER 1: B

Problem 2:
To find the number of digits in $4^5 \cdot 5^{10}$, let's simplify the expression. $4^5 = (2^2)^5 = 2^{10}$, so $4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$. $10^{10}$ is a 1 followed by 10 zeros, making it an 11-digit number. Let's check the choices: A. 8, B. 9, C. 10, D. 11, E. 12. The correct number of digits is given by choice D.

ANSWER 2: D

Problem 3:
The problem states that 8 friends agreed to share a bill equally, but because Judi forgot her money, each of her 7 friends paid an extra $2.50. Let's denote the total bill as $B$. The amount each friend was originally supposed to pay is $\frac{B}{8}$. When Judi's portion is covered by her 7 friends, each pays $\frac{B}{8} + $2.50$. The total amount paid by the 7 friends is $7(\frac{B}{8} + 2.50)$, which equals $B$. So, we have $7(\frac{B}{8} + 2.50) = B$. Simplifying this gives $\frac{7B}{8} + 17.50 = B$. Subtracting $\frac{7B}{8}$ from both sides gives $17.50 = \frac{B}{8}$. Multiplying both sides by 8 gives $B = 140$. Let's check the choices: A. $120, B. $128, C. $140, D. $144, E. $160. The total bill matches choice C.

ANSWER 3: C

Problem 4:
Carrie has a garden measuring 6 feet by 8 feet, so its area is $6 \times 8 = 48$ square feet. She plants 4 strawberry plants per square foot, so she plants $48 \times 4 = 192$ plants. With an average of 10 strawberries per plant, she can expect to harvest $192 \times 10 = 1920$ strawberries. Let's check the choices: A. 560, B. 960, C. 1120, D. 1920, E. 3840. The expected harvest matches choice D.

ANSWER 4: D

Problem 5:
To compare the given fractions, let's convert them to equivalent decimals or find a common denominator. The fractions are $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$. Converting them to equivalent fractions with a common denominator (24) gives $\frac{8}{24}$, $\frac{6}{24}$, $\frac{9}{24}$, $\frac{10}{24}$, and $\frac{7}{24}$. Comparing these, $\frac{10}{24}$ is the largest. So, $\frac{5}{12}$ is the largest fraction. Let's check the choices: A. $\frac{1}{3}$, B. $\frac{1}{4}$, C. $\frac{3}{8}$, D. $\frac{5}{12}$, E. $\frac{7}{24}$. The largest fraction is given by choice D.

ANSWER 5: D

Problem 6:
Let's denote the original length and width of the rectangle as $L$ and $W$, respectively. The original area is $L \times W$. After increasing the length by 10% and decreasing the width by 10%, the new length is $1.1L$ and the new width is $0.9W$. The new area is $1.1L \times 0.9W = 0.99LW$. To find what percent of the old area the new area is, we calculate $\frac{0.99LW}{LW} \times 100\% = 99\%$. Let's check the choices: A. 90, B. 99, C. 100, D. 101, E. 110. The new area is 99% of the old area, matching choice B.

ANSWER 6: B

Problem 7:
To evaluate $\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$, first simplify the numerator: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$. Then, divide by $\frac{4}{5}$, which is equivalent to multiplying by $\frac{5}{4}$. So, $\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$. Let's check the choices: A. 1, B. $\frac{25}{16}$, C. 2, D. $\frac{43}{20}$, E. $\frac{47}{16}$. The result matches choice B.

ANSWER 7: B

Problem 8:
To evaluate $\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$, convert each fraction to a decimal: $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$. Let's check the choices: A. 0.0026, B. 0.0197, C. 0.1997, D. 0.26, E. 1.997. The sum matches choice C.

ANSWER 8: C

Problem 9:
The sum $90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99$ is an arithmetic series with the first term $a = 90$, the last term $l = 99$, and $n = 10$ terms. The sum of an arithmetic series is given by $S = \frac{n}{2}(a + l)$. So, $S = \frac{10}{2}(90 + 99) = 5 \times 189 = 945$. Let's check the choices: A. 845, B. 945, C. 1005, D. 1025, E. 1045. The sum matches choice B.

ANSWER 9: B

Problem 10:
Let's analyze the number of lines of symmetry for each figure: A. An equilateral triangle has 3 lines of symmetry. B. A non-square rhombus has 2 lines of symmetry. C. A non-square rectangle has 2 lines of symmetry. D. An isosceles trapezoid has 1 line of symmetry. E. A square has 4 lines of symmetry. The figure with the greatest number of lines of symmetry is the square. Let's check the choices: the square is choice E.

ANSWER 10: E

Problem 11:
From the given information: Neither Jo nor Bo has as much money as Flo, so $Jo < Flo$ and $Bo < Flo$. Both Bo and Coe have more than Moe, so $Moe < Bo$ and $Moe < Coe$. Jo has more than Moe but less than Bo, so $Moe < Jo < Bo$. Combining these, we have $Moe < Jo < Bo < Flo$ and $Moe < Coe$. Since we don't have a direct comparison between Coe and the others except Moe, and we know $Bo < Flo$, the order for the least amount of money is clear for Moe. Let's check the choices: A. Bo, B. Coe, C. Flo, D. Jo, E. Moe. Moe has the least amount of money, matching choice E.

ANSWER 11: E

Problem 12:
Let's check each pair to see if their product equals 36: A. $(-4) \times (-9) = 36$, B. $(-3) \times (-12) = 36$, C. $\frac{1}{2} \times (-72) = -36$, D. $1 \times 36 = 36$, E. $\frac{3}{2} \times 24 = 36$. The pair that does NOT have a product equal to 36 is C. $\{\frac{1}{2}, -72\}$.

ANSWER 12: C

Problem 1:

We are asked for the ratio

\[
\frac{\text{side length of square}}{\text{radius of circle}}.
\]

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The square has area

\[
s^2.
\]

The circle has area

\[
\pi r^2.
\]

They have the same area, so

\[
s^2=\pi r^2.
\]

Taking the square root of both sides,

\[
s=r\sqrt{\pi}.
\]

So

\[
\frac{s}{r}=\sqrt{\pi}.
\]

The choices involving \(\pi\), \(2\pi\), or fractions of \(\pi\) are too large or have the wrong form. The correct ratio is \(\sqrt{\pi}\).

ANSWER 1: B

Problem 2:

We need the number of digits in

\[
4^5 \cdot 5^{10}.
\]

Rewrite \(4^5\):

\[
4^5=(2^2)^5=2^{10}.
\]

So

\[
4^5 \cdot 5^{10}=2^{10}\cdot 5^{10}.
\]

Since \(2^{10}\cdot 5^{10}=(2\cdot 5)^{10}\),

\[
2^{10}\cdot 5^{10}=10^{10}.
\]

Now,

\[
10^{10}=10,000,000,000,
\]

which is a 1 followed by 10 zeros.

So it has

\[
11
\]

digits.

ANSWER 2: D

Problem 3:

Eight friends were supposed to share the bill equally. Judi forgot her money, so the other 7 friends each paid an extra \(\$2.50\).

The total extra money paid by the 7 friends was

\[
7 \times 2.50 = 17.50.
\]

This extra money covered Judi’s share of the bill.

So Judi’s share was \(\$17.50\). Since there were 8 equal shares, the total bill was

\[
8 \times 17.50 = 140.
\]

Thus the total bill was \(\$140\).

ANSWER 3: C

Problem 4:

Carrie’s garden is a rectangle measuring 6 feet by 8 feet.

First find the area:

\[
6 \times 8 = 48
\]

square feet.

She plants 4 strawberry plants per square foot, so the number of plants is

\[
48 \times 4 = 192.
\]

Each plant produces an average of 10 strawberries, so the total expected harvest is

\[
192 \times 10 = 1920.
\]

So Carrie can expect to harvest 1920 strawberries.

ANSWER 4: D

Problem 5:

We need to find the largest fraction among:

\[
\frac13,\frac14,\frac38,\frac5{12},\frac7{24}.
\]

Use a common denominator of 24:

\[
\frac13=\frac8{24}
\]

\[
\frac14=\frac6{24}
\]

\[
\frac38=\frac9{24}
\]

\[
\frac5{12}=\frac{10}{24}
\]

\[
\frac7{24}=\frac7{24}
\]

Now compare:

\[
\frac6{24},\frac7{24},\frac8{24},\frac9{24},\frac{10}{24}.
\]

The largest is

\[
\frac{10}{24}=\frac5{12}.
\]

ANSWER 5: D

Problem 6:

Let the original length be \(L\), and let the original width be \(W\).

The original area is

\[
LW.
\]

The length is increased by 10%, so the new length is

\[
1.10L.
\]

The width is decreased by 10%, so the new width is

\[
0.90W.
\]

The new area is

\[
(1.10L)(0.90W)=0.99LW.
\]

So the new area is \(99\%\) of the old area.

A common mistake is to think the \(+10\%\) and \(-10\%\) cancel, but they are applied to different dimensions, so the area changes by the product \(1.10 \times 0.90=0.99\).

ANSWER 6: B

Problem 7:

The expression is

\[
\left(\frac38+\frac78\right)\div \frac45.
\]

First add the fractions in parentheses:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45.
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54 = \frac{25}{16}.
\]

So the value is

\[
\frac{25}{16}.
\]

ANSWER 7: B

Problem 8:

We need to compute

\[
\frac1{10}+\frac9{100}+\frac9{1000}+\frac7{10000}.
\]

Convert each fraction to a decimal:

\[
\frac1{10}=0.1
\]

\[
\frac9{100}=0.09
\]

\[
\frac9{1000}=0.009
\]

\[
\frac7{10000}=0.0007
\]

Now add:

\[
0.1+0.09=0.19
\]

\[
0.19+0.009=0.199
\]

\[
0.199+0.0007=0.1997.
\]

So the answer is

\[
0.1997.
\]

ANSWER 8: C

Problem 9:

We need to add

\[
90+91+92+93+94+95+96+97+98+99.
\]

There are 10 numbers.

The average of the first and last number is

\[
\frac{90+99}{2}=\frac{189}{2}=94.5.
\]

So the sum is

\[
10 \times 94.5 = 945.
\]

Alternatively, pair the numbers:

\[
90+99=189
\]

\[
91+98=189
\]

\[
92+97=189
\]

\[
93+96=189
\]

\[
94+95=189
\]

There are 5 pairs, so

\[
5 \times 189 = 945.
\]

ANSWER 9: B

Problem 10:

We compare the number of lines of symmetry for each figure.

- Equilateral triangle: 3 lines of symmetry.
- Non-square rhombus: 2 lines of symmetry.
- Non-square rectangle: 2 lines of symmetry.
- Isosceles trapezoid: 1 line of symmetry.
- Square: 4 lines of symmetry.

The square has the greatest number of lines of symmetry.

ANSWER 10: E

Problem 11:

We are told:

- Neither Jo nor Bo has as much money as Flo, so Flo has more than Jo and Bo.
- Both Bo and Coe have more than Moe.
- Jo has more than Moe, but less than Bo.

From this:

\[
Bo > Moe
\]

\[
Coe > Moe
\]

\[
Jo > Moe
\]

Also, since Flo has more than Jo,

\[
Flo > Jo > Moe.
\]

So Flo also has more than Moe.

Therefore, everyone has more money than Moe.

So Moe has the least amount of money.

ANSWER 11: E

Problem 12:

We need to find the pair whose product is NOT equal to 36.

Check each pair:

A. \(\{-4,-9\}\)

\[
(-4)(-9)=36
\]

So A works.

B. \(\{-3,-12\}\)

\[
(-3)(-12)=36
\]

So B works.

C. \(\left\{\frac12,-72\right\}\)

\[
\frac12 \cdot (-72)=-36
\]

This is not 36.

D. \(\{1,36\}\)

\[
1 \cdot 36=36
\]

So D works.

E. \(\left\{\frac32,24\right\}\)

\[
\frac32 \cdot 24=36
\]

So E works.

The only pair whose product is not 36 is choice C.

ANSWER 12: C

**Problem 1:**
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

**Step-by-step working:**
1. Let the side length of the square be $s$ and the radius of the circle be $r$.
2. The area of the square is $s^2$.
3. The area of the circle is $\pi r^2$.
4. Since their areas are equal, we can set up the equation: $s^2 = \pi r^2$.
5. To find the ratio of the side length to the radius ($s/r$), we first divide both sides by $r^2$: $\frac{s^2}{r^2} = \pi$.
6. Taking the square root of both sides gives us: $\frac{s}{r} = \sqrt{\pi}$.
7. This matches choice B.

ANSWER 1: B

**Problem 2:**
How many digits are in the product $4^5 \cdot 5^{10}$?

**Step-by-step working:**
1. We want to find the number of digits in the number $4^5 \cdot 5^{10}$.
2. First, express the base $4$ as a power of $2$: $4 = 2^2$.
3. Substitute this into the expression: $(2^2)^5 \cdot 5^{10}$.
4. Use the power of a power property $(a^m)^n = a^{m \cdot n}$: $2^{10} \cdot 5^{10}$.
5. Since the exponents are the same, we can combine the bases: $(2 \cdot 5)^{10} = 10^{10}$.
6. The number $10^{10}$ is written as a $1$ followed by $10$ zeros.
7. Therefore, it has $1 + 10 = 11$ digits.

ANSWER 2: D

**Problem 3:**
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

**Step-by-step working:**
1. Judi's portion of the bill was covered by her $7$ friends.
2. Each of the $7$ friends paid an extra $\$2.50$.
3. The total amount they paid to cover Judi's share is $7 \times \$2.50 = \$17.50$.
4. This means that an equal share of the bill for one person is $\$17.50$.
5. Since there are $8$ friends in total, the total bill is $8 \times \$17.50$.
6. $8 \times \$17.50 = \$140.00$.

ANSWER 3: C

**Problem 4:**
Carrie has a rectangular garden that measures 6 feet by 8 feet. She plants the entire garden with strawberry plants. Carrie is able to plant 4 strawberry plants per square foot, and she harvests an average of 10 strawberries per plant. How many strawberries can she expect to harvest?

**Step-by-step working:**
1. First, find the area of the rectangular garden: $6 \text{ feet} \times 8 \text{ feet} = 48 \text{ square feet}$.
2. Next, calculate the total number of strawberry plants. She plants $4$ plants per square foot: $48 \times 4 = 192 \text{ plants}$.
3. Finally, calculate the total expected strawberry harvest. Each plant yields an average of $10$ strawberries: $192 \times 10 = 1920 \text{ strawberries}$.

ANSWER 4: D

**Problem 5:**
Which of the following is the largest?

**Step-by-step working:**
1. The choices given are the fractions: $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$.
2. To compare them easily, find a common denominator. The least common multiple (LCM) of $3, 4, 8, 12,$ and $24$ is $24$.
3. Convert each fraction:
   - A. $\frac{1}{3} = \frac{8}{24}$
   - B. $\frac{1}{4} = \frac{6}{24}$
   - C. $\frac{3}{8} = \frac{9}{24}$
   - D. $\frac{5}{12} = \frac{10}{24}$
   - E. $\frac{7}{24} = \frac{7}{24}$
4. Comparing the numerators ($8, 6, 9, 10, 7$), the largest is $10$.
5. Therefore, $\frac{10}{24}$ (which is $\frac{5}{12}$) is the largest fraction.

ANSWER 5: D

**Problem 6:**
The length of a rectangle is increased by 10% and the width is decreased by 10%. What percent of the old area is the new area?

**Step-by-step working:**
1. Let the original length be $L$ and the original width be $W$.
2. The old area is $A_{\text{old}} = L \times W$.
3. The new length is increased by $10\%$, making it $1.10L$.
4. The new width is decreased by $10\%$, making it $0.90W$.
5. The new area is $A_{\text{new}} = (1.10L) \times (0.90W) = 0.99(L \times W)$.
6. This means the new area is $0.99$ times the old area, which corresponds to $99\%$.

ANSWER 6: B

**Problem 7:**
$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}} =$

**Step-by-step working:**
1. First, evaluate the numerator by adding the fractions: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8}$.
2. Simplify the fraction: $\frac{10}{8} = \frac{5}{4}$.
3. Now, divide this result by the denominator $\frac{4}{5}$: $\frac{5}{4} \div \frac{4}{5}$.
4. Dividing by a fraction is the same as multiplying by its reciprocal: $\frac{5}{4} \times \frac{5}{4}$.
5. Multiply the numerators and the denominators: $\frac{5 \times 5}{4 \times 4} = \frac{25}{16}$.

ANSWER 7: B

**Problem 8:**
$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000} =$

**Step-by-step working:**
1. Convert each fraction into its decimal equivalent:
   - $\frac{1}{10} = 0.1$
   - $\frac{9}{100} = 0.09$
   - $\frac{9}{1000} = 0.009$
   - $\frac{7}{10000} = 0.0007$
2. Add the decimals together:
   $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$.

ANSWER 8: C

**Problem 9:**
$90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =$

**Step-by-step working:**
1. We are adding an arithmetic sequence of $10$ consecutive integers from $90$ to $99$.
2. We can split the sum into the tens and the units:
   - The tens contribute $90 \times 10 = 900$.
   - The units contribute $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$.
3. The sum of the digits from $0$ to $9$ is $\frac{9 \times 10}{2} = 45$.
4. Adding these together gives $900 + 45 = 945$.
*(Alternatively, using the formula for the sum of an arithmetic sequence: $S = \frac{n(a_1 + a_n)}{2} = \frac{10(90 + 99)}{2} = 5 \times 189 = 945$.)*

ANSWER 9: B

**Problem 10:**
Which of the following figures has the greatest number of lines of symmetry?

**Step-by-step working:**
1. Let's count the lines of symmetry for each given figure:
   - A. Equilateral triangle: $3$ lines of symmetry (from each vertex to the midpoint of the opposite side).
   - B. Non-square rhombus: $2$ lines of symmetry (along its diagonals).
   - C. Non-square rectangle: $2$ lines of symmetry (connecting the midpoints of opposite sides).
   - D. Isosceles trapezoid: $1$ line of sy

Problem 1:
The problem asks for the ratio of the side length of a square to the radius of a circle when both have the same area. Let's denote the side length of the square as $s$ and the radius of the circle as $r$. The area of the square is $s^2$ and the area of the circle is $\pi r^2$. Since they have the same area, we have $s^2 = \pi r^2$. To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$. Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$. Let's examine the choices: A. $\sqrt{\pi}^2 = \pi$, B. $\sqrt{\pi}$, C. $\pi$, D. $2\pi$, E. $\pi^2$. The correct ratio is given by choice B.

ANSWER 1: B

Problem 2:
To find the number of digits in $4^5 \cdot 5^{10}$, let's simplify the expression. $4^5 = (2^2)^5 = 2^{10}$, so $4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$. $10^{10}$ is a 1 followed by 10 zeros, making it an 11-digit number. Let's check the choices: A. 8, B. 9, C. 10, D. 11, E. 12. The correct number of digits is given by choice D.

ANSWER 2: D

Problem 3:
The problem states that 8 friends agreed to share a bill equally, but because Judi forgot her money, each of her 7 friends paid an extra $2.50. Let's denote the total bill as $B$. The amount each friend was originally supposed to pay is $\frac{B}{8}$. When Judi's portion is covered by her 7 friends, each pays $\frac{B}{8} + $2.50$. The total amount paid by the 7 friends is $7(\frac{B}{8} + 2.50)$, which equals $B$. So, we have $7(\frac{B}{8} + 2.50) = B$. Simplifying this gives $\frac{7B}{8} + 17.50 = B$. Subtracting $\frac{7B}{8}$ from both sides gives $17.50 = \frac{B}{8}$. Multiplying both sides by 8 gives $B = 140$. Let's check the choices: A. $120, B. $128, C. $140, D. $144, E. $160. The total bill matches choice C.

ANSWER 3: C

Problem 4:
Carrie has a garden measuring 6 feet by 8 feet, so its area is $6 \times 8 = 48$ square feet. She plants 4 strawberry plants per square foot, so she plants $48 \times 4 = 192$ plants. With an average of 10 strawberries per plant, she can expect to harvest $192 \times 10 = 1920$ strawberries. Let's check the choices: A. 560, B. 960, C. 1120, D. 1920, E. 3840. The expected harvest matches choice D.

ANSWER 4: D

Problem 5:
To compare the given fractions, let's convert them to equivalent decimals or find a common denominator. The fractions are $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$. Converting them to equivalent fractions with a common denominator (24) gives $\frac{8}{24}$, $\frac{6}{24}$, $\frac{9}{24}$, $\frac{10}{24}$, and $\frac{7}{24}$. Comparing these, $\frac{10}{24}$ is the largest. So, $\frac{5}{12}$ is the largest fraction. Let's check the choices: A. $\frac{1}{3}$, B. $\frac{1}{4}$, C. $\frac{3}{8}$, D. $\frac{5}{12}$, E. $\frac{7}{24}$. The largest fraction is given by choice D.

ANSWER 5: D

Problem 6:
Let's denote the original length and width of the rectangle as $L$ and $W$, respectively. The original area is $L \times W$. After increasing the length by 10% and decreasing the width by 10%, the new length is $1.1L$ and the new width is $0.9W$. The new area is $1.1L \times 0.9W = 0.99LW$. To find what percent of the old area the new area is, we calculate $\frac{0.99LW}{LW} \times 100\% = 99\%$. Let's check the choices: A. 90, B. 99, C. 100, D. 101, E. 110. The new area is 99% of the old area, matching choice B.

ANSWER 6: B

Problem 7:
To evaluate $\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$, first simplify the numerator: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$. Then, divide by $\frac{4}{5}$, which is equivalent to multiplying by $\frac{5}{4}$. So, $\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$. Let's check the choices: A. 1, B. $\frac{25}{16}$, C. 2, D. $\frac{43}{20}$, E. $\frac{47}{16}$. The result matches choice B.

ANSWER 7: B

Problem 8:
To evaluate $\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$, convert each fraction to a decimal: $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$. Let's check the choices: A. 0.0026, B. 0.0197, C. 0.1997, D. 0.26, E. 1.997. The sum matches choice C.

ANSWER 8: C

Problem 9:
The sum $90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99$ is an arithmetic series with the first term $a = 90$, the last term $l = 99$, and $n = 10$ terms. The sum of an arithmetic series is given by $S = \frac{n}{2}(a + l)$. So, $S = \frac{10}{2}(90 + 99) = 5 \times 189 = 945$. Let's check the choices: A. 845, B. 945, C. 1005, D. 1025, E. 1045. The sum matches choice B.

ANSWER 9: B

Problem 10:
Let's analyze the number of lines of symmetry for each figure: A. An equilateral triangle has 3 lines of symmetry. B. A non-square rhombus has 2 lines of symmetry. C. A non-square rectangle has 2 lines of symmetry. D. An isosceles trapezoid has 1 line of symmetry. E. A square has 4 lines of symmetry. The figure with the greatest number of lines of symmetry is the square. Let's check the choices: the square is choice E.

ANSWER 10: E

Problem 11:
From the given information: Neither Jo nor Bo has as much money as Flo, so $Jo < Flo$ and $Bo < Flo$. Both Bo and Coe have more than Moe, so $Moe < Bo$ and $Moe < Coe$. Jo has more than Moe but less than Bo, so $Moe < Jo < Bo$. Combining these, we have $Moe < Jo < Bo < Flo$ and $Moe < Coe$. Since we don't have a direct comparison between Coe and the others except Moe, and we know $Bo < Flo$, the order for the least amount of money is clear for Moe. Let's check the choices: A. Bo, B. Coe, C. Flo, D. Jo, E. Moe. Moe has the least amount of money, matching choice E.

ANSWER 11: E

Problem 12:
Let's check each pair to see if their product equals 36: A. $(-4) \times (-9) = 36$, B. $(-3) \times (-12) = 36$, C. $\frac{1}{2} \times (-72) = -36$, D. $1 \times 36 = 36$, E. $\frac{3}{2} \times 24 = 36$. The pair that does NOT have a product equal to 36 is C. $\{\frac{1}{2}, -72\}$.

ANSWER 12: C

Problem 1:

We are asked for the ratio

\[
\frac{\text{side length of square}}{\text{radius of circle}}.
\]

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The square has area

\[
s^2.
\]

The circle has area

\[
\pi r^2.
\]

They have the same area, so

\[
s^2=\pi r^2.
\]

Taking the square root of both sides,

\[
s=r\sqrt{\pi}.
\]

So

\[
\frac{s}{r}=\sqrt{\pi}.
\]

The choices involving \(\pi\), \(2\pi\), or fractions of \(\pi\) are too large or have the wrong form. The correct ratio is \(\sqrt{\pi}\).

ANSWER 1: B

Problem 2:

We need the number of digits in

\[
4^5 \cdot 5^{10}.
\]

Rewrite \(4^5\):

\[
4^5=(2^2)^5=2^{10}.
\]

So

\[
4^5 \cdot 5^{10}=2^{10}\cdot 5^{10}.
\]

Since \(2^{10}\cdot 5^{10}=(2\cdot 5)^{10}\),

\[
2^{10}\cdot 5^{10}=10^{10}.
\]

Now,

\[
10^{10}=10,000,000,000,
\]

which is a 1 followed by 10 zeros.

So it has

\[
11
\]

digits.

ANSWER 2: D

Problem 3:

Eight friends were supposed to share the bill equally. Judi forgot her money, so the other 7 friends each paid an extra \(\$2.50\).

The total extra money paid by the 7 friends was

\[
7 \times 2.50 = 17.50.
\]

This extra money covered Judi’s share of the bill.

So Judi’s share was \(\$17.50\). Since there were 8 equal shares, the total bill was

\[
8 \times 17.50 = 140.
\]

Thus the total bill was \(\$140\).

ANSWER 3: C

Problem 4:

Carrie’s garden is a rectangle measuring 6 feet by 8 feet.

First find the area:

\[
6 \times 8 = 48
\]

square feet.

She plants 4 strawberry plants per square foot, so the number of plants is

\[
48 \times 4 = 192.
\]

Each plant produces an average of 10 strawberries, so the total expected harvest is

\[
192 \times 10 = 1920.
\]

So Carrie can expect to harvest 1920 strawberries.

ANSWER 4: D

Problem 5:

We need to find the largest fraction among:

\[
\frac13,\frac14,\frac38,\frac5{12},\frac7{24}.
\]

Use a common denominator of 24:

\[
\frac13=\frac8{24}
\]

\[
\frac14=\frac6{24}
\]

\[
\frac38=\frac9{24}
\]

\[
\frac5{12}=\frac{10}{24}
\]

\[
\frac7{24}=\frac7{24}
\]

Now compare:

\[
\frac6{24},\frac7{24},\frac8{24},\frac9{24},\frac{10}{24}.
\]

The largest is

\[
\frac{10}{24}=\frac5{12}.
\]

ANSWER 5: D

Problem 6:

Let the original length be \(L\), and let the original width be \(W\).

The original area is

\[
LW.
\]

The length is increased by 10%, so the new length is

\[
1.10L.
\]

The width is decreased by 10%, so the new width is

\[
0.90W.
\]

The new area is

\[
(1.10L)(0.90W)=0.99LW.
\]

So the new area is \(99\%\) of the old area.

A common mistake is to think the \(+10\%\) and \(-10\%\) cancel, but they are applied to different dimensions, so the area changes by the product \(1.10 \times 0.90=0.99\).

ANSWER 6: B

Problem 7:

The expression is

\[
\left(\frac38+\frac78\right)\div \frac45.
\]

First add the fractions in parentheses:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45.
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54 = \frac{25}{16}.
\]

So the value is

\[
\frac{25}{16}.
\]

ANSWER 7: B

Problem 8:

We need to compute

\[
\frac1{10}+\frac9{100}+\frac9{1000}+\frac7{10000}.
\]

Convert each fraction to a decimal:

\[
\frac1{10}=0.1
\]

\[
\frac9{100}=0.09
\]

\[
\frac9{1000}=0.009
\]

\[
\frac7{10000}=0.0007
\]

Now add:

\[
0.1+0.09=0.19
\]

\[
0.19+0.009=0.199
\]

\[
0.199+0.0007=0.1997.
\]

So the answer is

\[
0.1997.
\]

ANSWER 8: C

Problem 9:

We need to add

\[
90+91+92+93+94+95+96+97+98+99.
\]

There are 10 numbers.

The average of the first and last number is

\[
\frac{90+99}{2}=\frac{189}{2}=94.5.
\]

So the sum is

\[
10 \times 94.5 = 945.
\]

Alternatively, pair the numbers:

\[
90+99=189
\]

\[
91+98=189
\]

\[
92+97=189
\]

\[
93+96=189
\]

\[
94+95=189
\]

There are 5 pairs, so

\[
5 \times 189 = 945.
\]

ANSWER 9: B

Problem 10:

We compare the number of lines of symmetry for each figure.

- Equilateral triangle: 3 lines of symmetry.
- Non-square rhombus: 2 lines of symmetry.
- Non-square rectangle: 2 lines of symmetry.
- Isosceles trapezoid: 1 line of symmetry.
- Square: 4 lines of symmetry.

The square has the greatest number of lines of symmetry.

ANSWER 10: E

Problem 11:

We are told:

- Neither Jo nor Bo has as much money as Flo, so Flo has more than Jo and Bo.
- Both Bo and Coe have more than Moe.
- Jo has more than Moe, but less than Bo.

From this:

\[
Bo > Moe
\]

\[
Coe > Moe
\]

\[
Jo > Moe
\]

Also, since Flo has more than Jo,

\[
Flo > Jo > Moe.
\]

So Flo also has more than Moe.

Therefore, everyone has more money than Moe.

So Moe has the least amount of money.

ANSWER 11: E

Problem 12:

We need to find the pair whose product is NOT equal to 36.

Check each pair:

A. \(\{-4,-9\}\)

\[
(-4)(-9)=36
\]

So A works.

B. \(\{-3,-12\}\)

\[
(-3)(-12)=36
\]

So B works.

C. \(\left\{\frac12,-72\right\}\)

\[
\frac12 \cdot (-72)=-36
\]

This is not 36.

D. \(\{1,36\}\)

\[
1 \cdot 36=36
\]

So D works.

E. \(\left\{\frac32,24\right\}\)

\[
\frac32 \cdot 24=36
\]

So E works.

The only pair whose product is not 36 is choice C.

ANSWER 12: C

**Problem 1:**
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

**Step-by-step working:**
1. Let the side length of the square be $s$ and the radius of the circle be $r$.
2. The area of the square is $s^2$.
3. The area of the circle is $\pi r^2$.
4. Since their areas are equal, we can set up the equation: $s^2 = \pi r^2$.
5. To find the ratio of the side length to the radius ($s/r$), we first divide both sides by $r^2$: $\frac{s^2}{r^2} = \pi$.
6. Taking the square root of both sides gives us: $\frac{s}{r} = \sqrt{\pi}$.
7. This matches choice B.

ANSWER 1: B

**Problem 2:**
How many digits are in the product $4^5 \cdot 5^{10}$?

**Step-by-step working:**
1. We want to find the number of digits in the number $4^5 \cdot 5^{10}$.
2. First, express the base $4$ as a power of $2$: $4 = 2^2$.
3. Substitute this into the expression: $(2^2)^5 \cdot 5^{10}$.
4. Use the power of a power property $(a^m)^n = a^{m \cdot n}$: $2^{10} \cdot 5^{10}$.
5. Since the exponents are the same, we can combine the bases: $(2 \cdot 5)^{10} = 10^{10}$.
6. The number $10^{10}$ is written as a $1$ followed by $10$ zeros.
7. Therefore, it has $1 + 10 = 11$ digits.

ANSWER 2: D

**Problem 3:**
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

**Step-by-step working:**
1. Judi's portion of the bill was covered by her $7$ friends.
2. Each of the $7$ friends paid an extra $\$2.50$.
3. The total amount they paid to cover Judi's share is $7 \times \$2.50 = \$17.50$.
4. This means that an equal share of the bill for one person is $\$17.50$.
5. Since there are $8$ friends in total, the total bill is $8 \times \$17.50$.
6. $8 \times \$17.50 = \$140.00$.

ANSWER 3: C

**Problem 4:**
Carrie has a rectangular garden that measures 6 feet by 8 feet. She plants the entire garden with strawberry plants. Carrie is able to plant 4 strawberry plants per square foot, and she harvests an average of 10 strawberries per plant. How many strawberries can she expect to harvest?

**Step-by-step working:**
1. First, find the area of the rectangular garden: $6 \text{ feet} \times 8 \text{ feet} = 48 \text{ square feet}$.
2. Next, calculate the total number of strawberry plants. She plants $4$ plants per square foot: $48 \times 4 = 192 \text{ plants}$.
3. Finally, calculate the total expected strawberry harvest. Each plant yields an average of $10$ strawberries: $192 \times 10 = 1920 \text{ strawberries}$.

ANSWER 4: D

**Problem 5:**
Which of the following is the largest?

**Step-by-step working:**
1. The choices given are the fractions: $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$.
2. To compare them easily, find a common denominator. The least common multiple (LCM) of $3, 4, 8, 12,$ and $24$ is $24$.
3. Convert each fraction:
   - A. $\frac{1}{3} = \frac{8}{24}$
   - B. $\frac{1}{4} = \frac{6}{24}$
   - C. $\frac{3}{8} = \frac{9}{24}$
   - D. $\frac{5}{12} = \frac{10}{24}$
   - E. $\frac{7}{24} = \frac{7}{24}$
4. Comparing the numerators ($8, 6, 9, 10, 7$), the largest is $10$.
5. Therefore, $\frac{10}{24}$ (which is $\frac{5}{12}$) is the largest fraction.

ANSWER 5: D

**Problem 6:**
The length of a rectangle is increased by 10% and the width is decreased by 10%. What percent of the old area is the new area?

**Step-by-step working:**
1. Let the original length be $L$ and the original width be $W$.
2. The old area is $A_{\text{old}} = L \times W$.
3. The new length is increased by $10\%$, making it $1.10L$.
4. The new width is decreased by $10\%$, making it $0.90W$.
5. The new area is $A_{\text{new}} = (1.10L) \times (0.90W) = 0.99(L \times W)$.
6. This means the new area is $0.99$ times the old area, which corresponds to $99\%$.

ANSWER 6: B

**Problem 7:**
$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}} =$

**Step-by-step working:**
1. First, evaluate the numerator by adding the fractions: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8}$.
2. Simplify the fraction: $\frac{10}{8} = \frac{5}{4}$.
3. Now, divide this result by the denominator $\frac{4}{5}$: $\frac{5}{4} \div \frac{4}{5}$.
4. Dividing by a fraction is the same as multiplying by its reciprocal: $\frac{5}{4} \times \frac{5}{4}$.
5. Multiply the numerators and the denominators: $\frac{5 \times 5}{4 \times 4} = \frac{25}{16}$.

ANSWER 7: B

**Problem 8:**
$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000} =$

**Step-by-step working:**
1. Convert each fraction into its decimal equivalent:
   - $\frac{1}{10} = 0.1$
   - $\frac{9}{100} = 0.09$
   - $\frac{9}{1000} = 0.009$
   - $\frac{7}{10000} = 0.0007$
2. Add the decimals together:
   $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$.

ANSWER 8: C

**Problem 9:**
$90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =$

**Step-by-step working:**
1. We are adding an arithmetic sequence of $10$ consecutive integers from $90$ to $99$.
2. We can split the sum into the tens and the units:
   - The tens contribute $90 \times 10 = 900$.
   - The units contribute $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$.
3. The sum of the digits from $0$ to $9$ is $\frac{9 \times 10}{2} = 45$.
4. Adding these together gives $900 + 45 = 945$.
*(Alternatively, using the formula for the sum of an arithmetic sequence: $S = \frac{n(a_1 + a_n)}{2} = \frac{10(90 + 99)}{2} = 5 \times 189 = 945$.)*

ANSWER 9: B

**Problem 10:**
Which of the following figures has the greatest number of lines of symmetry?

**Step-by-step working:**
1. Let's count the lines of symmetry for each given figure:
   - A. Equilateral triangle: $3$ lines of symmetry (from each vertex to the midpoint of the opposite side).
   - B. Non-square rhombus: $2$ lines of symmetry (along its diagonals).
   - C. Non-square rectangle: $2$ lines of symmetry (connecting the midpoints of opposite sides).
   - D. Isosceles trapezoid: $1$ line of sy

Problem 1:
The problem asks for the ratio of the side length of a square to the radius of a circle when both have the same area. Let's denote the side length of the square as $s$ and the radius of the circle as $r$. The area of the square is $s^2$ and the area of the circle is $\pi r^2$. Since they have the same area, we have $s^2 = \pi r^2$. To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$. Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$. Let's examine the choices: A. $\sqrt{\pi}^2 = \pi$, B. $\sqrt{\pi}$, C. $\pi$, D. $2\pi$, E. $\pi^2$. The correct ratio is given by choice B.

ANSWER 1: B

Problem 2:
To find the number of digits in $4^5 \cdot 5^{10}$, let's simplify the expression. $4^5 = (2^2)^5 = 2^{10}$, so $4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$. $10^{10}$ is a 1 followed by 10 zeros, making it an 11-digit number. Let's check the choices: A. 8, B. 9, C. 10, D. 11, E. 12. The correct number of digits is given by choice D.

ANSWER 2: D

Problem 3:
The problem states that 8 friends agreed to share a bill equally, but because Judi forgot her money, each of her 7 friends paid an extra $2.50. Let's denote the total bill as $B$. The amount each friend was originally supposed to pay is $\frac{B}{8}$. When Judi's portion is covered by her 7 friends, each pays $\frac{B}{8} + $2.50$. The total amount paid by the 7 friends is $7(\frac{B}{8} + 2.50)$, which equals $B$. So, we have $7(\frac{B}{8} + 2.50) = B$. Simplifying this gives $\frac{7B}{8} + 17.50 = B$. Subtracting $\frac{7B}{8}$ from both sides gives $17.50 = \frac{B}{8}$. Multiplying both sides by 8 gives $B = 140$. Let's check the choices: A. $120, B. $128, C. $140, D. $144, E. $160. The total bill matches choice C.

ANSWER 3: C

Problem 4:
Carrie has a garden measuring 6 feet by 8 feet, so its area is $6 \times 8 = 48$ square feet. She plants 4 strawberry plants per square foot, so she plants $48 \times 4 = 192$ plants. With an average of 10 strawberries per plant, she can expect to harvest $192 \times 10 = 1920$ strawberries. Let's check the choices: A. 560, B. 960, C. 1120, D. 1920, E. 3840. The expected harvest matches choice D.

ANSWER 4: D

Problem 5:
To compare the given fractions, let's convert them to equivalent decimals or find a common denominator. The fractions are $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$. Converting them to equivalent fractions with a common denominator (24) gives $\frac{8}{24}$, $\frac{6}{24}$, $\frac{9}{24}$, $\frac{10}{24}$, and $\frac{7}{24}$. Comparing these, $\frac{10}{24}$ is the largest. So, $\frac{5}{12}$ is the largest fraction. Let's check the choices: A. $\frac{1}{3}$, B. $\frac{1}{4}$, C. $\frac{3}{8}$, D. $\frac{5}{12}$, E. $\frac{7}{24}$. The largest fraction is given by choice D.

ANSWER 5: D

Problem 6:
Let's denote the original length and width of the rectangle as $L$ and $W$, respectively. The original area is $L \times W$. After increasing the length by 10% and decreasing the width by 10%, the new length is $1.1L$ and the new width is $0.9W$. The new area is $1.1L \times 0.9W = 0.99LW$. To find what percent of the old area the new area is, we calculate $\frac{0.99LW}{LW} \times 100\% = 99\%$. Let's check the choices: A. 90, B. 99, C. 100, D. 101, E. 110. The new area is 99% of the old area, matching choice B.

ANSWER 6: B

Problem 7:
To evaluate $\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$, first simplify the numerator: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$. Then, divide by $\frac{4}{5}$, which is equivalent to multiplying by $\frac{5}{4}$. So, $\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$. Let's check the choices: A. 1, B. $\frac{25}{16}$, C. 2, D. $\frac{43}{20}$, E. $\frac{47}{16}$. The result matches choice B.

ANSWER 7: B

Problem 8:
To evaluate $\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$, convert each fraction to a decimal: $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$. Let's check the choices: A. 0.0026, B. 0.0197, C. 0.1997, D. 0.26, E. 1.997. The sum matches choice C.

ANSWER 8: C

Problem 9:
The sum $90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99$ is an arithmetic series with the first term $a = 90$, the last term $l = 99$, and $n = 10$ terms. The sum of an arithmetic series is given by $S = \frac{n}{2}(a + l)$. So, $S = \frac{10}{2}(90 + 99) = 5 \times 189 = 945$. Let's check the choices: A. 845, B. 945, C. 1005, D. 1025, E. 1045. The sum matches choice B.

ANSWER 9: B

Problem 10:
Let's analyze the number of lines of symmetry for each figure: A. An equilateral triangle has 3 lines of symmetry. B. A non-square rhombus has 2 lines of symmetry. C. A non-square rectangle has 2 lines of symmetry. D. An isosceles trapezoid has 1 line of symmetry. E. A square has 4 lines of symmetry. The figure with the greatest number of lines of symmetry is the square. Let's check the choices: the square is choice E.

ANSWER 10: E

Problem 11:
From the given information: Neither Jo nor Bo has as much money as Flo, so $Jo < Flo$ and $Bo < Flo$. Both Bo and Coe have more than Moe, so $Moe < Bo$ and $Moe < Coe$. Jo has more than Moe but less than Bo, so $Moe < Jo < Bo$. Combining these, we have $Moe < Jo < Bo < Flo$ and $Moe < Coe$. Since we don't have a direct comparison between Coe and the others except Moe, and we know $Bo < Flo$, the order for the least amount of money is clear for Moe. Let's check the choices: A. Bo, B. Coe, C. Flo, D. Jo, E. Moe. Moe has the least amount of money, matching choice E.

ANSWER 11: E

Problem 12:
Let's check each pair to see if their product equals 36: A. $(-4) \times (-9) = 36$, B. $(-3) \times (-12) = 36$, C. $\frac{1}{2} \times (-72) = -36$, D. $1 \times 36 = 36$, E. $\frac{3}{2} \times 24 = 36$. The pair that does NOT have a product equal to 36 is C. $\{\frac{1}{2}, -72\}$.

ANSWER 12: C

Problem 1:

We are asked for the ratio

\[
\frac{\text{side length of square}}{\text{radius of circle}}.
\]

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The square has area

\[
s^2.
\]

The circle has area

\[
\pi r^2.
\]

They have the same area, so

\[
s^2=\pi r^2.
\]

Taking the square root of both sides,

\[
s=r\sqrt{\pi}.
\]

So

\[
\frac{s}{r}=\sqrt{\pi}.
\]

The choices involving \(\pi\), \(2\pi\), or fractions of \(\pi\) are too large or have the wrong form. The correct ratio is \(\sqrt{\pi}\).

ANSWER 1: B

Problem 2:

We need the number of digits in

\[
4^5 \cdot 5^{10}.
\]

Rewrite \(4^5\):

\[
4^5=(2^2)^5=2^{10}.
\]

So

\[
4^5 \cdot 5^{10}=2^{10}\cdot 5^{10}.
\]

Since \(2^{10}\cdot 5^{10}=(2\cdot 5)^{10}\),

\[
2^{10}\cdot 5^{10}=10^{10}.
\]

Now,

\[
10^{10}=10,000,000,000,
\]

which is a 1 followed by 10 zeros.

So it has

\[
11
\]

digits.

ANSWER 2: D

Problem 3:

Eight friends were supposed to share the bill equally. Judi forgot her money, so the other 7 friends each paid an extra \(\$2.50\).

The total extra money paid by the 7 friends was

\[
7 \times 2.50 = 17.50.
\]

This extra money covered Judi’s share of the bill.

So Judi’s share was \(\$17.50\). Since there were 8 equal shares, the total bill was

\[
8 \times 17.50 = 140.
\]

Thus the total bill was \(\$140\).

ANSWER 3: C

Problem 4:

Carrie’s garden is a rectangle measuring 6 feet by 8 feet.

First find the area:

\[
6 \times 8 = 48
\]

square feet.

She plants 4 strawberry plants per square foot, so the number of plants is

\[
48 \times 4 = 192.
\]

Each plant produces an average of 10 strawberries, so the total expected harvest is

\[
192 \times 10 = 1920.
\]

So Carrie can expect to harvest 1920 strawberries.

ANSWER 4: D

Problem 5:

We need to find the largest fraction among:

\[
\frac13,\frac14,\frac38,\frac5{12},\frac7{24}.
\]

Use a common denominator of 24:

\[
\frac13=\frac8{24}
\]

\[
\frac14=\frac6{24}
\]

\[
\frac38=\frac9{24}
\]

\[
\frac5{12}=\frac{10}{24}
\]

\[
\frac7{24}=\frac7{24}
\]

Now compare:

\[
\frac6{24},\frac7{24},\frac8{24},\frac9{24},\frac{10}{24}.
\]

The largest is

\[
\frac{10}{24}=\frac5{12}.
\]

ANSWER 5: D

Problem 6:

Let the original length be \(L\), and let the original width be \(W\).

The original area is

\[
LW.
\]

The length is increased by 10%, so the new length is

\[
1.10L.
\]

The width is decreased by 10%, so the new width is

\[
0.90W.
\]

The new area is

\[
(1.10L)(0.90W)=0.99LW.
\]

So the new area is \(99\%\) of the old area.

A common mistake is to think the \(+10\%\) and \(-10\%\) cancel, but they are applied to different dimensions, so the area changes by the product \(1.10 \times 0.90=0.99\).

ANSWER 6: B

Problem 7:

The expression is

\[
\left(\frac38+\frac78\right)\div \frac45.
\]

First add the fractions in parentheses:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45.
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54 = \frac{25}{16}.
\]

So the value is

\[
\frac{25}{16}.
\]

ANSWER 7: B

Problem 8:

We need to compute

\[
\frac1{10}+\frac9{100}+\frac9{1000}+\frac7{10000}.
\]

Convert each fraction to a decimal:

\[
\frac1{10}=0.1
\]

\[
\frac9{100}=0.09
\]

\[
\frac9{1000}=0.009
\]

\[
\frac7{10000}=0.0007
\]

Now add:

\[
0.1+0.09=0.19
\]

\[
0.19+0.009=0.199
\]

\[
0.199+0.0007=0.1997.
\]

So the answer is

\[
0.1997.
\]

ANSWER 8: C

Problem 9:

We need to add

\[
90+91+92+93+94+95+96+97+98+99.
\]

There are 10 numbers.

The average of the first and last number is

\[
\frac{90+99}{2}=\frac{189}{2}=94.5.
\]

So the sum is

\[
10 \times 94.5 = 945.
\]

Alternatively, pair the numbers:

\[
90+99=189
\]

\[
91+98=189
\]

\[
92+97=189
\]

\[
93+96=189
\]

\[
94+95=189
\]

There are 5 pairs, so

\[
5 \times 189 = 945.
\]

ANSWER 9: B

Problem 10:

We compare the number of lines of symmetry for each figure.

- Equilateral triangle: 3 lines of symmetry.
- Non-square rhombus: 2 lines of symmetry.
- Non-square rectangle: 2 lines of symmetry.
- Isosceles trapezoid: 1 line of symmetry.
- Square: 4 lines of symmetry.

The square has the greatest number of lines of symmetry.

ANSWER 10: E

Problem 11:

We are told:

- Neither Jo nor Bo has as much money as Flo, so Flo has more than Jo and Bo.
- Both Bo and Coe have more than Moe.
- Jo has more than Moe, but less than Bo.

From this:

\[
Bo > Moe
\]

\[
Coe > Moe
\]

\[
Jo > Moe
\]

Also, since Flo has more than Jo,

\[
Flo > Jo > Moe.
\]

So Flo also has more than Moe.

Therefore, everyone has more money than Moe.

So Moe has the least amount of money.

ANSWER 11: E

Problem 12:

We need to find the pair whose product is NOT equal to 36.

Check each pair:

A. \(\{-4,-9\}\)

\[
(-4)(-9)=36
\]

So A works.

B. \(\{-3,-12\}\)

\[
(-3)(-12)=36
\]

So B works.

C. \(\left\{\frac12,-72\right\}\)

\[
\frac12 \cdot (-72)=-36
\]

This is not 36.

D. \(\{1,36\}\)

\[
1 \cdot 36=36
\]

So D works.

E. \(\left\{\frac32,24\right\}\)

\[
\frac32 \cdot 24=36
\]

So E works.

The only pair whose product is not 36 is choice C.

ANSWER 12: C

**Problem 1:**
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

**Step-by-step working:**
1. Let the side length of the square be $s$ and the radius of the circle be $r$.
2. The area of the square is $s^2$.
3. The area of the circle is $\pi r^2$.
4. Since their areas are equal, we can set up the equation: $s^2 = \pi r^2$.
5. To find the ratio of the side length to the radius ($s/r$), we first divide both sides by $r^2$: $\frac{s^2}{r^2} = \pi$.
6. Taking the square root of both sides gives us: $\frac{s}{r} = \sqrt{\pi}$.
7. This matches choice B.

ANSWER 1: B

**Problem 2:**
How many digits are in the product $4^5 \cdot 5^{10}$?

**Step-by-step working:**
1. We want to find the number of digits in the number $4^5 \cdot 5^{10}$.
2. First, express the base $4$ as a power of $2$: $4 = 2^2$.
3. Substitute this into the expression: $(2^2)^5 \cdot 5^{10}$.
4. Use the power of a power property $(a^m)^n = a^{m \cdot n}$: $2^{10} \cdot 5^{10}$.
5. Since the exponents are the same, we can combine the bases: $(2 \cdot 5)^{10} = 10^{10}$.
6. The number $10^{10}$ is written as a $1$ followed by $10$ zeros.
7. Therefore, it has $1 + 10 = 11$ digits.

ANSWER 2: D

**Problem 3:**
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

**Step-by-step working:**
1. Judi's portion of the bill was covered by her $7$ friends.
2. Each of the $7$ friends paid an extra $\$2.50$.
3. The total amount they paid to cover Judi's share is $7 \times \$2.50 = \$17.50$.
4. This means that an equal share of the bill for one person is $\$17.50$.
5. Since there are $8$ friends in total, the total bill is $8 \times \$17.50$.
6. $8 \times \$17.50 = \$140.00$.

ANSWER 3: C

**Problem 4:**
Carrie has a rectangular garden that measures 6 feet by 8 feet. She plants the entire garden with strawberry plants. Carrie is able to plant 4 strawberry plants per square foot, and she harvests an average of 10 strawberries per plant. How many strawberries can she expect to harvest?

**Step-by-step working:**
1. First, find the area of the rectangular garden: $6 \text{ feet} \times 8 \text{ feet} = 48 \text{ square feet}$.
2. Next, calculate the total number of strawberry plants. She plants $4$ plants per square foot: $48 \times 4 = 192 \text{ plants}$.
3. Finally, calculate the total expected strawberry harvest. Each plant yields an average of $10$ strawberries: $192 \times 10 = 1920 \text{ strawberries}$.

ANSWER 4: D

**Problem 5:**
Which of the following is the largest?

**Step-by-step working:**
1. The choices given are the fractions: $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$.
2. To compare them easily, find a common denominator. The least common multiple (LCM) of $3, 4, 8, 12,$ and $24$ is $24$.
3. Convert each fraction:
   - A. $\frac{1}{3} = \frac{8}{24}$
   - B. $\frac{1}{4} = \frac{6}{24}$
   - C. $\frac{3}{8} = \frac{9}{24}$
   - D. $\frac{5}{12} = \frac{10}{24}$
   - E. $\frac{7}{24} = \frac{7}{24}$
4. Comparing the numerators ($8, 6, 9, 10, 7$), the largest is $10$.
5. Therefore, $\frac{10}{24}$ (which is $\frac{5}{12}$) is the largest fraction.

ANSWER 5: D

**Problem 6:**
The length of a rectangle is increased by 10% and the width is decreased by 10%. What percent of the old area is the new area?

**Step-by-step working:**
1. Let the original length be $L$ and the original width be $W$.
2. The old area is $A_{\text{old}} = L \times W$.
3. The new length is increased by $10\%$, making it $1.10L$.
4. The new width is decreased by $10\%$, making it $0.90W$.
5. The new area is $A_{\text{new}} = (1.10L) \times (0.90W) = 0.99(L \times W)$.
6. This means the new area is $0.99$ times the old area, which corresponds to $99\%$.

ANSWER 6: B

**Problem 7:**
$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}} =$

**Step-by-step working:**
1. First, evaluate the numerator by adding the fractions: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8}$.
2. Simplify the fraction: $\frac{10}{8} = \frac{5}{4}$.
3. Now, divide this result by the denominator $\frac{4}{5}$: $\frac{5}{4} \div \frac{4}{5}$.
4. Dividing by a fraction is the same as multiplying by its reciprocal: $\frac{5}{4} \times \frac{5}{4}$.
5. Multiply the numerators and the denominators: $\frac{5 \times 5}{4 \times 4} = \frac{25}{16}$.

ANSWER 7: B

**Problem 8:**
$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000} =$

**Step-by-step working:**
1. Convert each fraction into its decimal equivalent:
   - $\frac{1}{10} = 0.1$
   - $\frac{9}{100} = 0.09$
   - $\frac{9}{1000} = 0.009$
   - $\frac{7}{10000} = 0.0007$
2. Add the decimals together:
   $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$.

ANSWER 8: C

**Problem 9:**
$90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =$

**Step-by-step working:**
1. We are adding an arithmetic sequence of $10$ consecutive integers from $90$ to $99$.
2. We can split the sum into the tens and the units:
   - The tens contribute $90 \times 10 = 900$.
   - The units contribute $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$.
3. The sum of the digits from $0$ to $9$ is $\frac{9 \times 10}{2} = 45$.
4. Adding these together gives $900 + 45 = 945$.
*(Alternatively, using the formula for the sum of an arithmetic sequence: $S = \frac{n(a_1 + a_n)}{2} = \frac{10(90 + 99)}{2} = 5 \times 189 = 945$.)*

ANSWER 9: B

**Problem 10:**
Which of the following figures has the greatest number of lines of symmetry?

**Step-by-step working:**
1. Let's count the lines of symmetry for each given figure:
   - A. Equilateral triangle: $3$ lines of symmetry (from each vertex to the midpoint of the opposite side).
   - B. Non-square rhombus: $2$ lines of symmetry (along its diagonals).
   - C. Non-square rectangle: $2$ lines of symmetry (connecting the midpoints of opposite sides).
   - D. Isosceles trapezoid: $1$ line of sy

Problem 1:
The problem asks for the ratio of the side length of a square to the radius of a circle when both have the same area. Let's denote the side length of the square as $s$ and the radius of the circle as $r$. The area of the square is $s^2$ and the area of the circle is $\pi r^2$. Since they have the same area, we have $s^2 = \pi r^2$. To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$. Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$. Let's examine the choices: A. $\sqrt{\pi}^2 = \pi$, B. $\sqrt{\pi}$, C. $\pi$, D. $2\pi$, E. $\pi^2$. The correct ratio is given by choice B.

ANSWER 1: B

Problem 2:
To find the number of digits in $4^5 \cdot 5^{10}$, let's simplify the expression. $4^5 = (2^2)^5 = 2^{10}$, so $4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$. $10^{10}$ is a 1 followed by 10 zeros, making it an 11-digit number. Let's check the choices: A. 8, B. 9, C. 10, D. 11, E. 12. The correct number of digits is given by choice D.

ANSWER 2: D

Problem 3:
The problem states that 8 friends agreed to share a bill equally, but because Judi forgot her money, each of her 7 friends paid an extra $2.50. Let's denote the total bill as $B$. The amount each friend was originally supposed to pay is $\frac{B}{8}$. When Judi's portion is covered by her 7 friends, each pays $\frac{B}{8} + $2.50$. The total amount paid by the 7 friends is $7(\frac{B}{8} + 2.50)$, which equals $B$. So, we have $7(\frac{B}{8} + 2.50) = B$. Simplifying this gives $\frac{7B}{8} + 17.50 = B$. Subtracting $\frac{7B}{8}$ from both sides gives $17.50 = \frac{B}{8}$. Multiplying both sides by 8 gives $B = 140$. Let's check the choices: A. $120, B. $128, C. $140, D. $144, E. $160. The total bill matches choice C.

ANSWER 3: C

Problem 4:
Carrie has a garden measuring 6 feet by 8 feet, so its area is $6 \times 8 = 48$ square feet. She plants 4 strawberry plants per square foot, so she plants $48 \times 4 = 192$ plants. With an average of 10 strawberries per plant, she can expect to harvest $192 \times 10 = 1920$ strawberries. Let's check the choices: A. 560, B. 960, C. 1120, D. 1920, E. 3840. The expected harvest matches choice D.

ANSWER 4: D

Problem 5:
To compare the given fractions, let's convert them to equivalent decimals or find a common denominator. The fractions are $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$. Converting them to equivalent fractions with a common denominator (24) gives $\frac{8}{24}$, $\frac{6}{24}$, $\frac{9}{24}$, $\frac{10}{24}$, and $\frac{7}{24}$. Comparing these, $\frac{10}{24}$ is the largest. So, $\frac{5}{12}$ is the largest fraction. Let's check the choices: A. $\frac{1}{3}$, B. $\frac{1}{4}$, C. $\frac{3}{8}$, D. $\frac{5}{12}$, E. $\frac{7}{24}$. The largest fraction is given by choice D.

ANSWER 5: D

Problem 6:
Let's denote the original length and width of the rectangle as $L$ and $W$, respectively. The original area is $L \times W$. After increasing the length by 10% and decreasing the width by 10%, the new length is $1.1L$ and the new width is $0.9W$. The new area is $1.1L \times 0.9W = 0.99LW$. To find what percent of the old area the new area is, we calculate $\frac{0.99LW}{LW} \times 100\% = 99\%$. Let's check the choices: A. 90, B. 99, C. 100, D. 101, E. 110. The new area is 99% of the old area, matching choice B.

ANSWER 6: B

Problem 7:
To evaluate $\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$, first simplify the numerator: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$. Then, divide by $\frac{4}{5}$, which is equivalent to multiplying by $\frac{5}{4}$. So, $\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$. Let's check the choices: A. 1, B. $\frac{25}{16}$, C. 2, D. $\frac{43}{20}$, E. $\frac{47}{16}$. The result matches choice B.

ANSWER 7: B

Problem 8:
To evaluate $\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$, convert each fraction to a decimal: $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$. Let's check the choices: A. 0.0026, B. 0.0197, C. 0.1997, D. 0.26, E. 1.997. The sum matches choice C.

ANSWER 8: C

Problem 9:
The sum $90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99$ is an arithmetic series with the first term $a = 90$, the last term $l = 99$, and $n = 10$ terms. The sum of an arithmetic series is given by $S = \frac{n}{2}(a + l)$. So, $S = \frac{10}{2}(90 + 99) = 5 \times 189 = 945$. Let's check the choices: A. 845, B. 945, C. 1005, D. 1025, E. 1045. The sum matches choice B.

ANSWER 9: B

Problem 10:
Let's analyze the number of lines of symmetry for each figure: A. An equilateral triangle has 3 lines of symmetry. B. A non-square rhombus has 2 lines of symmetry. C. A non-square rectangle has 2 lines of symmetry. D. An isosceles trapezoid has 1 line of symmetry. E. A square has 4 lines of symmetry. The figure with the greatest number of lines of symmetry is the square. Let's check the choices: the square is choice E.

ANSWER 10: E

Problem 11:
From the given information: Neither Jo nor Bo has as much money as Flo, so $Jo < Flo$ and $Bo < Flo$. Both Bo and Coe have more than Moe, so $Moe < Bo$ and $Moe < Coe$. Jo has more than Moe but less than Bo, so $Moe < Jo < Bo$. Combining these, we have $Moe < Jo < Bo < Flo$ and $Moe < Coe$. Since we don't have a direct comparison between Coe and the others except Moe, and we know $Bo < Flo$, the order for the least amount of money is clear for Moe. Let's check the choices: A. Bo, B. Coe, C. Flo, D. Jo, E. Moe. Moe has the least amount of money, matching choice E.

ANSWER 11: E

Problem 12:
Let's check each pair to see if their product equals 36: A. $(-4) \times (-9) = 36$, B. $(-3) \times (-12) = 36$, C. $\frac{1}{2} \times (-72) = -36$, D. $1 \times 36 = 36$, E. $\frac{3}{2} \times 24 = 36$. The pair that does NOT have a product equal to 36 is C. $\{\frac{1}{2}, -72\}$.

ANSWER 12: C

Problem 1:

We are asked for the ratio

\[
\frac{\text{side length of square}}{\text{radius of circle}}.
\]

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The square has area

\[
s^2.
\]

The circle has area

\[
\pi r^2.
\]

They have the same area, so

\[
s^2=\pi r^2.
\]

Taking the square root of both sides,

\[
s=r\sqrt{\pi}.
\]

So

\[
\frac{s}{r}=\sqrt{\pi}.
\]

The choices involving \(\pi\), \(2\pi\), or fractions of \(\pi\) are too large or have the wrong form. The correct ratio is \(\sqrt{\pi}\).

ANSWER 1: B

Problem 2:

We need the number of digits in

\[
4^5 \cdot 5^{10}.
\]

Rewrite \(4^5\):

\[
4^5=(2^2)^5=2^{10}.
\]

So

\[
4^5 \cdot 5^{10}=2^{10}\cdot 5^{10}.
\]

Since \(2^{10}\cdot 5^{10}=(2\cdot 5)^{10}\),

\[
2^{10}\cdot 5^{10}=10^{10}.
\]

Now,

\[
10^{10}=10,000,000,000,
\]

which is a 1 followed by 10 zeros.

So it has

\[
11
\]

digits.

ANSWER 2: D

Problem 3:

Eight friends were supposed to share the bill equally. Judi forgot her money, so the other 7 friends each paid an extra \(\$2.50\).

The total extra money paid by the 7 friends was

\[
7 \times 2.50 = 17.50.
\]

This extra money covered Judi’s share of the bill.

So Judi’s share was \(\$17.50\). Since there were 8 equal shares, the total bill was

\[
8 \times 17.50 = 140.
\]

Thus the total bill was \(\$140\).

ANSWER 3: C

Problem 4:

Carrie’s garden is a rectangle measuring 6 feet by 8 feet.

First find the area:

\[
6 \times 8 = 48
\]

square feet.

She plants 4 strawberry plants per square foot, so the number of plants is

\[
48 \times 4 = 192.
\]

Each plant produces an average of 10 strawberries, so the total expected harvest is

\[
192 \times 10 = 1920.
\]

So Carrie can expect to harvest 1920 strawberries.

ANSWER 4: D

Problem 5:

We need to find the largest fraction among:

\[
\frac13,\frac14,\frac38,\frac5{12},\frac7{24}.
\]

Use a common denominator of 24:

\[
\frac13=\frac8{24}
\]

\[
\frac14=\frac6{24}
\]

\[
\frac38=\frac9{24}
\]

\[
\frac5{12}=\frac{10}{24}
\]

\[
\frac7{24}=\frac7{24}
\]

Now compare:

\[
\frac6{24},\frac7{24},\frac8{24},\frac9{24},\frac{10}{24}.
\]

The largest is

\[
\frac{10}{24}=\frac5{12}.
\]

ANSWER 5: D

Problem 6:

Let the original length be \(L\), and let the original width be \(W\).

The original area is

\[
LW.
\]

The length is increased by 10%, so the new length is

\[
1.10L.
\]

The width is decreased by 10%, so the new width is

\[
0.90W.
\]

The new area is

\[
(1.10L)(0.90W)=0.99LW.
\]

So the new area is \(99\%\) of the old area.

A common mistake is to think the \(+10\%\) and \(-10\%\) cancel, but they are applied to different dimensions, so the area changes by the product \(1.10 \times 0.90=0.99\).

ANSWER 6: B

Problem 7:

The expression is

\[
\left(\frac38+\frac78\right)\div \frac45.
\]

First add the fractions in parentheses:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45.
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54 = \frac{25}{16}.
\]

So the value is

\[
\frac{25}{16}.
\]

ANSWER 7: B

Problem 8:

We need to compute

\[
\frac1{10}+\frac9{100}+\frac9{1000}+\frac7{10000}.
\]

Convert each fraction to a decimal:

\[
\frac1{10}=0.1
\]

\[
\frac9{100}=0.09
\]

\[
\frac9{1000}=0.009
\]

\[
\frac7{10000}=0.0007
\]

Now add:

\[
0.1+0.09=0.19
\]

\[
0.19+0.009=0.199
\]

\[
0.199+0.0007=0.1997.
\]

So the answer is

\[
0.1997.
\]

ANSWER 8: C

Problem 9:

We need to add

\[
90+91+92+93+94+95+96+97+98+99.
\]

There are 10 numbers.

The average of the first and last number is

\[
\frac{90+99}{2}=\frac{189}{2}=94.5.
\]

So the sum is

\[
10 \times 94.5 = 945.
\]

Alternatively, pair the numbers:

\[
90+99=189
\]

\[
91+98=189
\]

\[
92+97=189
\]

\[
93+96=189
\]

\[
94+95=189
\]

There are 5 pairs, so

\[
5 \times 189 = 945.
\]

ANSWER 9: B

Problem 10:

We compare the number of lines of symmetry for each figure.

- Equilateral triangle: 3 lines of symmetry.
- Non-square rhombus: 2 lines of symmetry.
- Non-square rectangle: 2 lines of symmetry.
- Isosceles trapezoid: 1 line of symmetry.
- Square: 4 lines of symmetry.

The square has the greatest number of lines of symmetry.

ANSWER 10: E

Problem 11:

We are told:

- Neither Jo nor Bo has as much money as Flo, so Flo has more than Jo and Bo.
- Both Bo and Coe have more than Moe.
- Jo has more than Moe, but less than Bo.

From this:

\[
Bo > Moe
\]

\[
Coe > Moe
\]

\[
Jo > Moe
\]

Also, since Flo has more than Jo,

\[
Flo > Jo > Moe.
\]

So Flo also has more than Moe.

Therefore, everyone has more money than Moe.

So Moe has the least amount of money.

ANSWER 11: E

Problem 12:

We need to find the pair whose product is NOT equal to 36.

Check each pair:

A. \(\{-4,-9\}\)

\[
(-4)(-9)=36
\]

So A works.

B. \(\{-3,-12\}\)

\[
(-3)(-12)=36
\]

So B works.

C. \(\left\{\frac12,-72\right\}\)

\[
\frac12 \cdot (-72)=-36
\]

This is not 36.

D. \(\{1,36\}\)

\[
1 \cdot 36=36
\]

So D works.

E. \(\left\{\frac32,24\right\}\)

\[
\frac32 \cdot 24=36
\]

So E works.

The only pair whose product is not 36 is choice C.

ANSWER 12: C

**Problem 1:**
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

**Step-by-step working:**
1. Let the side length of the square be $s$ and the radius of the circle be $r$.
2. The area of the square is $s^2$.
3. The area of the circle is $\pi r^2$.
4. Since their areas are equal, we can set up the equation: $s^2 = \pi r^2$.
5. To find the ratio of the side length to the radius ($s/r$), we first divide both sides by $r^2$: $\frac{s^2}{r^2} = \pi$.
6. Taking the square root of both sides gives us: $\frac{s}{r} = \sqrt{\pi}$.
7. This matches choice B.

ANSWER 1: B

**Problem 2:**
How many digits are in the product $4^5 \cdot 5^{10}$?

**Step-by-step working:**
1. We want to find the number of digits in the number $4^5 \cdot 5^{10}$.
2. First, express the base $4$ as a power of $2$: $4 = 2^2$.
3. Substitute this into the expression: $(2^2)^5 \cdot 5^{10}$.
4. Use the power of a power property $(a^m)^n = a^{m \cdot n}$: $2^{10} \cdot 5^{10}$.
5. Since the exponents are the same, we can combine the bases: $(2 \cdot 5)^{10} = 10^{10}$.
6. The number $10^{10}$ is written as a $1$ followed by $10$ zeros.
7. Therefore, it has $1 + 10 = 11$ digits.

ANSWER 2: D

**Problem 3:**
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

**Step-by-step working:**
1. Judi's portion of the bill was covered by her $7$ friends.
2. Each of the $7$ friends paid an extra $\$2.50$.
3. The total amount they paid to cover Judi's share is $7 \times \$2.50 = \$17.50$.
4. This means that an equal share of the bill for one person is $\$17.50$.
5. Since there are $8$ friends in total, the total bill is $8 \times \$17.50$.
6. $8 \times \$17.50 = \$140.00$.

ANSWER 3: C

**Problem 4:**
Carrie has a rectangular garden that measures 6 feet by 8 feet. She plants the entire garden with strawberry plants. Carrie is able to plant 4 strawberry plants per square foot, and she harvests an average of 10 strawberries per plant. How many strawberries can she expect to harvest?

**Step-by-step working:**
1. First, find the area of the rectangular garden: $6 \text{ feet} \times 8 \text{ feet} = 48 \text{ square feet}$.
2. Next, calculate the total number of strawberry plants. She plants $4$ plants per square foot: $48 \times 4 = 192 \text{ plants}$.
3. Finally, calculate the total expected strawberry harvest. Each plant yields an average of $10$ strawberries: $192 \times 10 = 1920 \text{ strawberries}$.

ANSWER 4: D

**Problem 5:**
Which of the following is the largest?

**Step-by-step working:**
1. The choices given are the fractions: $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$.
2. To compare them easily, find a common denominator. The least common multiple (LCM) of $3, 4, 8, 12,$ and $24$ is $24$.
3. Convert each fraction:
   - A. $\frac{1}{3} = \frac{8}{24}$
   - B. $\frac{1}{4} = \frac{6}{24}$
   - C. $\frac{3}{8} = \frac{9}{24}$
   - D. $\frac{5}{12} = \frac{10}{24}$
   - E. $\frac{7}{24} = \frac{7}{24}$
4. Comparing the numerators ($8, 6, 9, 10, 7$), the largest is $10$.
5. Therefore, $\frac{10}{24}$ (which is $\frac{5}{12}$) is the largest fraction.

ANSWER 5: D

**Problem 6:**
The length of a rectangle is increased by 10% and the width is decreased by 10%. What percent of the old area is the new area?

**Step-by-step working:**
1. Let the original length be $L$ and the original width be $W$.
2. The old area is $A_{\text{old}} = L \times W$.
3. The new length is increased by $10\%$, making it $1.10L$.
4. The new width is decreased by $10\%$, making it $0.90W$.
5. The new area is $A_{\text{new}} = (1.10L) \times (0.90W) = 0.99(L \times W)$.
6. This means the new area is $0.99$ times the old area, which corresponds to $99\%$.

ANSWER 6: B

**Problem 7:**
$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}} =$

**Step-by-step working:**
1. First, evaluate the numerator by adding the fractions: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8}$.
2. Simplify the fraction: $\frac{10}{8} = \frac{5}{4}$.
3. Now, divide this result by the denominator $\frac{4}{5}$: $\frac{5}{4} \div \frac{4}{5}$.
4. Dividing by a fraction is the same as multiplying by its reciprocal: $\frac{5}{4} \times \frac{5}{4}$.
5. Multiply the numerators and the denominators: $\frac{5 \times 5}{4 \times 4} = \frac{25}{16}$.

ANSWER 7: B

**Problem 8:**
$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000} =$

**Step-by-step working:**
1. Convert each fraction into its decimal equivalent:
   - $\frac{1}{10} = 0.1$
   - $\frac{9}{100} = 0.09$
   - $\frac{9}{1000} = 0.009$
   - $\frac{7}{10000} = 0.0007$
2. Add the decimals together:
   $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$.

ANSWER 8: C

**Problem 9:**
$90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =$

**Step-by-step working:**
1. We are adding an arithmetic sequence of $10$ consecutive integers from $90$ to $99$.
2. We can split the sum into the tens and the units:
   - The tens contribute $90 \times 10 = 900$.
   - The units contribute $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$.
3. The sum of the digits from $0$ to $9$ is $\frac{9 \times 10}{2} = 45$.
4. Adding these together gives $900 + 45 = 945$.
*(Alternatively, using the formula for the sum of an arithmetic sequence: $S = \frac{n(a_1 + a_n)}{2} = \frac{10(90 + 99)}{2} = 5 \times 189 = 945$.)*

ANSWER 9: B

**Problem 10:**
Which of the following figures has the greatest number of lines of symmetry?

**Step-by-step working:**
1. Let's count the lines of symmetry for each given figure:
   - A. Equilateral triangle: $3$ lines of symmetry (from each vertex to the midpoint of the opposite side).
   - B. Non-square rhombus: $2$ lines of symmetry (along its diagonals).
   - C. Non-square rectangle: $2$ lines of symmetry (connecting the midpoints of opposite sides).
   - D. Isosceles trapezoid: $1$ line of sy

Problem 1:
The problem asks for the ratio of the side length of a square to the radius of a circle when both have the same area. Let's denote the side length of the square as $s$ and the radius of the circle as $r$. The area of the square is $s^2$ and the area of the circle is $\pi r^2$. Since they have the same area, we have $s^2 = \pi r^2$. To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$. Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$. Let's examine the choices: A. $\sqrt{\pi}^2 = \pi$, B. $\sqrt{\pi}$, C. $\pi$, D. $2\pi$, E. $\pi^2$. The correct ratio is given by choice B.

ANSWER 1: B

Problem 2:
To find the number of digits in $4^5 \cdot 5^{10}$, let's simplify the expression. $4^5 = (2^2)^5 = 2^{10}$, so $4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$. $10^{10}$ is a 1 followed by 10 zeros, making it an 11-digit number. Let's check the choices: A. 8, B. 9, C. 10, D. 11, E. 12. The correct number of digits is given by choice D.

ANSWER 2: D

Problem 3:
The problem states that 8 friends agreed to share a bill equally, but because Judi forgot her money, each of her 7 friends paid an extra $2.50. Let's denote the total bill as $B$. The amount each friend was originally supposed to pay is $\frac{B}{8}$. When Judi's portion is covered by her 7 friends, each pays $\frac{B}{8} + $2.50$. The total amount paid by the 7 friends is $7(\frac{B}{8} + 2.50)$, which equals $B$. So, we have $7(\frac{B}{8} + 2.50) = B$. Simplifying this gives $\frac{7B}{8} + 17.50 = B$. Subtracting $\frac{7B}{8}$ from both sides gives $17.50 = \frac{B}{8}$. Multiplying both sides by 8 gives $B = 140$. Let's check the choices: A. $120, B. $128, C. $140, D. $144, E. $160. The total bill matches choice C.

ANSWER 3: C

Problem 4:
Carrie has a garden measuring 6 feet by 8 feet, so its area is $6 \times 8 = 48$ square feet. She plants 4 strawberry plants per square foot, so she plants $48 \times 4 = 192$ plants. With an average of 10 strawberries per plant, she can expect to harvest $192 \times 10 = 1920$ strawberries. Let's check the choices: A. 560, B. 960, C. 1120, D. 1920, E. 3840. The expected harvest matches choice D.

ANSWER 4: D

Problem 5:
To compare the given fractions, let's convert them to equivalent decimals or find a common denominator. The fractions are $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$. Converting them to equivalent fractions with a common denominator (24) gives $\frac{8}{24}$, $\frac{6}{24}$, $\frac{9}{24}$, $\frac{10}{24}$, and $\frac{7}{24}$. Comparing these, $\frac{10}{24}$ is the largest. So, $\frac{5}{12}$ is the largest fraction. Let's check the choices: A. $\frac{1}{3}$, B. $\frac{1}{4}$, C. $\frac{3}{8}$, D. $\frac{5}{12}$, E. $\frac{7}{24}$. The largest fraction is given by choice D.

ANSWER 5: D

Problem 6:
Let's denote the original length and width of the rectangle as $L$ and $W$, respectively. The original area is $L \times W$. After increasing the length by 10% and decreasing the width by 10%, the new length is $1.1L$ and the new width is $0.9W$. The new area is $1.1L \times 0.9W = 0.99LW$. To find what percent of the old area the new area is, we calculate $\frac{0.99LW}{LW} \times 100\% = 99\%$. Let's check the choices: A. 90, B. 99, C. 100, D. 101, E. 110. The new area is 99% of the old area, matching choice B.

ANSWER 6: B

Problem 7:
To evaluate $\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$, first simplify the numerator: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$. Then, divide by $\frac{4}{5}$, which is equivalent to multiplying by $\frac{5}{4}$. So, $\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$. Let's check the choices: A. 1, B. $\frac{25}{16}$, C. 2, D. $\frac{43}{20}$, E. $\frac{47}{16}$. The result matches choice B.

ANSWER 7: B

Problem 8:
To evaluate $\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$, convert each fraction to a decimal: $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$. Let's check the choices: A. 0.0026, B. 0.0197, C. 0.1997, D. 0.26, E. 1.997. The sum matches choice C.

ANSWER 8: C

Problem 9:
The sum $90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99$ is an arithmetic series with the first term $a = 90$, the last term $l = 99$, and $n = 10$ terms. The sum of an arithmetic series is given by $S = \frac{n}{2}(a + l)$. So, $S = \frac{10}{2}(90 + 99) = 5 \times 189 = 945$. Let's check the choices: A. 845, B. 945, C. 1005, D. 1025, E. 1045. The sum matches choice B.

ANSWER 9: B

Problem 10:
Let's analyze the number of lines of symmetry for each figure: A. An equilateral triangle has 3 lines of symmetry. B. A non-square rhombus has 2 lines of symmetry. C. A non-square rectangle has 2 lines of symmetry. D. An isosceles trapezoid has 1 line of symmetry. E. A square has 4 lines of symmetry. The figure with the greatest number of lines of symmetry is the square. Let's check the choices: the square is choice E.

ANSWER 10: E

Problem 11:
From the given information: Neither Jo nor Bo has as much money as Flo, so $Jo < Flo$ and $Bo < Flo$. Both Bo and Coe have more than Moe, so $Moe < Bo$ and $Moe < Coe$. Jo has more than Moe but less than Bo, so $Moe < Jo < Bo$. Combining these, we have $Moe < Jo < Bo < Flo$ and $Moe < Coe$. Since we don't have a direct comparison between Coe and the others except Moe, and we know $Bo < Flo$, the order for the least amount of money is clear for Moe. Let's check the choices: A. Bo, B. Coe, C. Flo, D. Jo, E. Moe. Moe has the least amount of money, matching choice E.

ANSWER 11: E

Problem 12:
Let's check each pair to see if their product equals 36: A. $(-4) \times (-9) = 36$, B. $(-3) \times (-12) = 36$, C. $\frac{1}{2} \times (-72) = -36$, D. $1 \times 36 = 36$, E. $\frac{3}{2} \times 24 = 36$. The pair that does NOT have a product equal to 36 is C. $\{\frac{1}{2}, -72\}$.

ANSWER 12: C

Problem 1:

We are asked for the ratio

\[
\frac{\text{side length of square}}{\text{radius of circle}}.
\]

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The square has area

\[
s^2.
\]

The circle has area

\[
\pi r^2.
\]

They have the same area, so

\[
s^2=\pi r^2.
\]

Taking the square root of both sides,

\[
s=r\sqrt{\pi}.
\]

So

\[
\frac{s}{r}=\sqrt{\pi}.
\]

The choices involving \(\pi\), \(2\pi\), or fractions of \(\pi\) are too large or have the wrong form. The correct ratio is \(\sqrt{\pi}\).

ANSWER 1: B

Problem 2:

We need the number of digits in

\[
4^5 \cdot 5^{10}.
\]

Rewrite \(4^5\):

\[
4^5=(2^2)^5=2^{10}.
\]

So

\[
4^5 \cdot 5^{10}=2^{10}\cdot 5^{10}.
\]

Since \(2^{10}\cdot 5^{10}=(2\cdot 5)^{10}\),

\[
2^{10}\cdot 5^{10}=10^{10}.
\]

Now,

\[
10^{10}=10,000,000,000,
\]

which is a 1 followed by 10 zeros.

So it has

\[
11
\]

digits.

ANSWER 2: D

Problem 3:

Eight friends were supposed to share the bill equally. Judi forgot her money, so the other 7 friends each paid an extra \(\$2.50\).

The total extra money paid by the 7 friends was

\[
7 \times 2.50 = 17.50.
\]

This extra money covered Judi’s share of the bill.

So Judi’s share was \(\$17.50\). Since there were 8 equal shares, the total bill was

\[
8 \times 17.50 = 140.
\]

Thus the total bill was \(\$140\).

ANSWER 3: C

Problem 4:

Carrie’s garden is a rectangle measuring 6 feet by 8 feet.

First find the area:

\[
6 \times 8 = 48
\]

square feet.

She plants 4 strawberry plants per square foot, so the number of plants is

\[
48 \times 4 = 192.
\]

Each plant produces an average of 10 strawberries, so the total expected harvest is

\[
192 \times 10 = 1920.
\]

So Carrie can expect to harvest 1920 strawberries.

ANSWER 4: D

Problem 5:

We need to find the largest fraction among:

\[
\frac13,\frac14,\frac38,\frac5{12},\frac7{24}.
\]

Use a common denominator of 24:

\[
\frac13=\frac8{24}
\]

\[
\frac14=\frac6{24}
\]

\[
\frac38=\frac9{24}
\]

\[
\frac5{12}=\frac{10}{24}
\]

\[
\frac7{24}=\frac7{24}
\]

Now compare:

\[
\frac6{24},\frac7{24},\frac8{24},\frac9{24},\frac{10}{24}.
\]

The largest is

\[
\frac{10}{24}=\frac5{12}.
\]

ANSWER 5: D

Problem 6:

Let the original length be \(L\), and let the original width be \(W\).

The original area is

\[
LW.
\]

The length is increased by 10%, so the new length is

\[
1.10L.
\]

The width is decreased by 10%, so the new width is

\[
0.90W.
\]

The new area is

\[
(1.10L)(0.90W)=0.99LW.
\]

So the new area is \(99\%\) of the old area.

A common mistake is to think the \(+10\%\) and \(-10\%\) cancel, but they are applied to different dimensions, so the area changes by the product \(1.10 \times 0.90=0.99\).

ANSWER 6: B

Problem 7:

The expression is

\[
\left(\frac38+\frac78\right)\div \frac45.
\]

First add the fractions in parentheses:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45.
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54 = \frac{25}{16}.
\]

So the value is

\[
\frac{25}{16}.
\]

ANSWER 7: B

Problem 8:

We need to compute

\[
\frac1{10}+\frac9{100}+\frac9{1000}+\frac7{10000}.
\]

Convert each fraction to a decimal:

\[
\frac1{10}=0.1
\]

\[
\frac9{100}=0.09
\]

\[
\frac9{1000}=0.009
\]

\[
\frac7{10000}=0.0007
\]

Now add:

\[
0.1+0.09=0.19
\]

\[
0.19+0.009=0.199
\]

\[
0.199+0.0007=0.1997.
\]

So the answer is

\[
0.1997.
\]

ANSWER 8: C

Problem 9:

We need to add

\[
90+91+92+93+94+95+96+97+98+99.
\]

There are 10 numbers.

The average of the first and last number is

\[
\frac{90+99}{2}=\frac{189}{2}=94.5.
\]

So the sum is

\[
10 \times 94.5 = 945.
\]

Alternatively, pair the numbers:

\[
90+99=189
\]

\[
91+98=189
\]

\[
92+97=189
\]

\[
93+96=189
\]

\[
94+95=189
\]

There are 5 pairs, so

\[
5 \times 189 = 945.
\]

ANSWER 9: B

Problem 10:

We compare the number of lines of symmetry for each figure.

- Equilateral triangle: 3 lines of symmetry.
- Non-square rhombus: 2 lines of symmetry.
- Non-square rectangle: 2 lines of symmetry.
- Isosceles trapezoid: 1 line of symmetry.
- Square: 4 lines of symmetry.

The square has the greatest number of lines of symmetry.

ANSWER 10: E

Problem 11:

We are told:

- Neither Jo nor Bo has as much money as Flo, so Flo has more than Jo and Bo.
- Both Bo and Coe have more than Moe.
- Jo has more than Moe, but less than Bo.

From this:

\[
Bo > Moe
\]

\[
Coe > Moe
\]

\[
Jo > Moe
\]

Also, since Flo has more than Jo,

\[
Flo > Jo > Moe.
\]

So Flo also has more than Moe.

Therefore, everyone has more money than Moe.

So Moe has the least amount of money.

ANSWER 11: E

Problem 12:

We need to find the pair whose product is NOT equal to 36.

Check each pair:

A. \(\{-4,-9\}\)

\[
(-4)(-9)=36
\]

So A works.

B. \(\{-3,-12\}\)

\[
(-3)(-12)=36
\]

So B works.

C. \(\left\{\frac12,-72\right\}\)

\[
\frac12 \cdot (-72)=-36
\]

This is not 36.

D. \(\{1,36\}\)

\[
1 \cdot 36=36
\]

So D works.

E. \(\left\{\frac32,24\right\}\)

\[
\frac32 \cdot 24=36
\]

So E works.

The only pair whose product is not 36 is choice C.

ANSWER 12: C

**Problem 1:**
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

**Step-by-step working:**
1. Let the side length of the square be $s$ and the radius of the circle be $r$.
2. The area of the square is $s^2$.
3. The area of the circle is $\pi r^2$.
4. Since their areas are equal, we can set up the equation: $s^2 = \pi r^2$.
5. To find the ratio of the side length to the radius ($s/r$), we first divide both sides by $r^2$: $\frac{s^2}{r^2} = \pi$.
6. Taking the square root of both sides gives us: $\frac{s}{r} = \sqrt{\pi}$.
7. This matches choice B.

ANSWER 1: B

**Problem 2:**
How many digits are in the product $4^5 \cdot 5^{10}$?

**Step-by-step working:**
1. We want to find the number of digits in the number $4^5 \cdot 5^{10}$.
2. First, express the base $4$ as a power of $2$: $4 = 2^2$.
3. Substitute this into the expression: $(2^2)^5 \cdot 5^{10}$.
4. Use the power of a power property $(a^m)^n = a^{m \cdot n}$: $2^{10} \cdot 5^{10}$.
5. Since the exponents are the same, we can combine the bases: $(2 \cdot 5)^{10} = 10^{10}$.
6. The number $10^{10}$ is written as a $1$ followed by $10$ zeros.
7. Therefore, it has $1 + 10 = 11$ digits.

ANSWER 2: D

**Problem 3:**
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

**Step-by-step working:**
1. Judi's portion of the bill was covered by her $7$ friends.
2. Each of the $7$ friends paid an extra $\$2.50$.
3. The total amount they paid to cover Judi's share is $7 \times \$2.50 = \$17.50$.
4. This means that an equal share of the bill for one person is $\$17.50$.
5. Since there are $8$ friends in total, the total bill is $8 \times \$17.50$.
6. $8 \times \$17.50 = \$140.00$.

ANSWER 3: C

**Problem 4:**
Carrie has a rectangular garden that measures 6 feet by 8 feet. She plants the entire garden with strawberry plants. Carrie is able to plant 4 strawberry plants per square foot, and she harvests an average of 10 strawberries per plant. How many strawberries can she expect to harvest?

**Step-by-step working:**
1. First, find the area of the rectangular garden: $6 \text{ feet} \times 8 \text{ feet} = 48 \text{ square feet}$.
2. Next, calculate the total number of strawberry plants. She plants $4$ plants per square foot: $48 \times 4 = 192 \text{ plants}$.
3. Finally, calculate the total expected strawberry harvest. Each plant yields an average of $10$ strawberries: $192 \times 10 = 1920 \text{ strawberries}$.

ANSWER 4: D

**Problem 5:**
Which of the following is the largest?

**Step-by-step working:**
1. The choices given are the fractions: $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$.
2. To compare them easily, find a common denominator. The least common multiple (LCM) of $3, 4, 8, 12,$ and $24$ is $24$.
3. Convert each fraction:
   - A. $\frac{1}{3} = \frac{8}{24}$
   - B. $\frac{1}{4} = \frac{6}{24}$
   - C. $\frac{3}{8} = \frac{9}{24}$
   - D. $\frac{5}{12} = \frac{10}{24}$
   - E. $\frac{7}{24} = \frac{7}{24}$
4. Comparing the numerators ($8, 6, 9, 10, 7$), the largest is $10$.
5. Therefore, $\frac{10}{24}$ (which is $\frac{5}{12}$) is the largest fraction.

ANSWER 5: D

**Problem 6:**
The length of a rectangle is increased by 10% and the width is decreased by 10%. What percent of the old area is the new area?

**Step-by-step working:**
1. Let the original length be $L$ and the original width be $W$.
2. The old area is $A_{\text{old}} = L \times W$.
3. The new length is increased by $10\%$, making it $1.10L$.
4. The new width is decreased by $10\%$, making it $0.90W$.
5. The new area is $A_{\text{new}} = (1.10L) \times (0.90W) = 0.99(L \times W)$.
6. This means the new area is $0.99$ times the old area, which corresponds to $99\%$.

ANSWER 6: B

**Problem 7:**
$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}} =$

**Step-by-step working:**
1. First, evaluate the numerator by adding the fractions: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8}$.
2. Simplify the fraction: $\frac{10}{8} = \frac{5}{4}$.
3. Now, divide this result by the denominator $\frac{4}{5}$: $\frac{5}{4} \div \frac{4}{5}$.
4. Dividing by a fraction is the same as multiplying by its reciprocal: $\frac{5}{4} \times \frac{5}{4}$.
5. Multiply the numerators and the denominators: $\frac{5 \times 5}{4 \times 4} = \frac{25}{16}$.

ANSWER 7: B

**Problem 8:**
$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000} =$

**Step-by-step working:**
1. Convert each fraction into its decimal equivalent:
   - $\frac{1}{10} = 0.1$
   - $\frac{9}{100} = 0.09$
   - $\frac{9}{1000} = 0.009$
   - $\frac{7}{10000} = 0.0007$
2. Add the decimals together:
   $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$.

ANSWER 8: C

**Problem 9:**
$90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =$

**Step-by-step working:**
1. We are adding an arithmetic sequence of $10$ consecutive integers from $90$ to $99$.
2. We can split the sum into the tens and the units:
   - The tens contribute $90 \times 10 = 900$.
   - The units contribute $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$.
3. The sum of the digits from $0$ to $9$ is $\frac{9 \times 10}{2} = 45$.
4. Adding these together gives $900 + 45 = 945$.
*(Alternatively, using the formula for the sum of an arithmetic sequence: $S = \frac{n(a_1 + a_n)}{2} = \frac{10(90 + 99)}{2} = 5 \times 189 = 945$.)*

ANSWER 9: B

**Problem 10:**
Which of the following figures has the greatest number of lines of symmetry?

**Step-by-step working:**
1. Let's count the lines of symmetry for each given figure:
   - A. Equilateral triangle: $3$ lines of symmetry (from each vertex to the midpoint of the opposite side).
   - B. Non-square rhombus: $2$ lines of symmetry (along its diagonals).
   - C. Non-square rectangle: $2$ lines of symmetry (connecting the midpoints of opposite sides).
   - D. Isosceles trapezoid: $1$ line of sy

Problem 1:
The problem asks for the ratio of the side length of a square to the radius of a circle when both have the same area. Let's denote the side length of the square as $s$ and the radius of the circle as $r$. The area of the square is $s^2$ and the area of the circle is $\pi r^2$. Since they have the same area, we have $s^2 = \pi r^2$. To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$. Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$. Let's examine the choices: A. $\sqrt{\pi}^2 = \pi$, B. $\sqrt{\pi}$, C. $\pi$, D. $2\pi$, E. $\pi^2$. The correct ratio is given by choice B.

ANSWER 1: B

Problem 2:
To find the number of digits in $4^5 \cdot 5^{10}$, let's simplify the expression. $4^5 = (2^2)^5 = 2^{10}$, so $4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$. $10^{10}$ is a 1 followed by 10 zeros, making it an 11-digit number. Let's check the choices: A. 8, B. 9, C. 10, D. 11, E. 12. The correct number of digits is given by choice D.

ANSWER 2: D

Problem 3:
The problem states that 8 friends agreed to share a bill equally, but because Judi forgot her money, each of her 7 friends paid an extra $2.50. Let's denote the total bill as $B$. The amount each friend was originally supposed to pay is $\frac{B}{8}$. When Judi's portion is covered by her 7 friends, each pays $\frac{B}{8} + $2.50$. The total amount paid by the 7 friends is $7(\frac{B}{8} + 2.50)$, which equals $B$. So, we have $7(\frac{B}{8} + 2.50) = B$. Simplifying this gives $\frac{7B}{8} + 17.50 = B$. Subtracting $\frac{7B}{8}$ from both sides gives $17.50 = \frac{B}{8}$. Multiplying both sides by 8 gives $B = 140$. Let's check the choices: A. $120, B. $128, C. $140, D. $144, E. $160. The total bill matches choice C.

ANSWER 3: C

Problem 4:
Carrie has a garden measuring 6 feet by 8 feet, so its area is $6 \times 8 = 48$ square feet. She plants 4 strawberry plants per square foot, so she plants $48 \times 4 = 192$ plants. With an average of 10 strawberries per plant, she can expect to harvest $192 \times 10 = 1920$ strawberries. Let's check the choices: A. 560, B. 960, C. 1120, D. 1920, E. 3840. The expected harvest matches choice D.

ANSWER 4: D

Problem 5:
To compare the given fractions, let's convert them to equivalent decimals or find a common denominator. The fractions are $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$. Converting them to equivalent fractions with a common denominator (24) gives $\frac{8}{24}$, $\frac{6}{24}$, $\frac{9}{24}$, $\frac{10}{24}$, and $\frac{7}{24}$. Comparing these, $\frac{10}{24}$ is the largest. So, $\frac{5}{12}$ is the largest fraction. Let's check the choices: A. $\frac{1}{3}$, B. $\frac{1}{4}$, C. $\frac{3}{8}$, D. $\frac{5}{12}$, E. $\frac{7}{24}$. The largest fraction is given by choice D.

ANSWER 5: D

Problem 6:
Let's denote the original length and width of the rectangle as $L$ and $W$, respectively. The original area is $L \times W$. After increasing the length by 10% and decreasing the width by 10%, the new length is $1.1L$ and the new width is $0.9W$. The new area is $1.1L \times 0.9W = 0.99LW$. To find what percent of the old area the new area is, we calculate $\frac{0.99LW}{LW} \times 100\% = 99\%$. Let's check the choices: A. 90, B. 99, C. 100, D. 101, E. 110. The new area is 99% of the old area, matching choice B.

ANSWER 6: B

Problem 7:
To evaluate $\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$, first simplify the numerator: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$. Then, divide by $\frac{4}{5}$, which is equivalent to multiplying by $\frac{5}{4}$. So, $\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$. Let's check the choices: A. 1, B. $\frac{25}{16}$, C. 2, D. $\frac{43}{20}$, E. $\frac{47}{16}$. The result matches choice B.

ANSWER 7: B

Problem 8:
To evaluate $\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$, convert each fraction to a decimal: $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$. Let's check the choices: A. 0.0026, B. 0.0197, C. 0.1997, D. 0.26, E. 1.997. The sum matches choice C.

ANSWER 8: C

Problem 9:
The sum $90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99$ is an arithmetic series with the first term $a = 90$, the last term $l = 99$, and $n = 10$ terms. The sum of an arithmetic series is given by $S = \frac{n}{2}(a + l)$. So, $S = \frac{10}{2}(90 + 99) = 5 \times 189 = 945$. Let's check the choices: A. 845, B. 945, C. 1005, D. 1025, E. 1045. The sum matches choice B.

ANSWER 9: B

Problem 10:
Let's analyze the number of lines of symmetry for each figure: A. An equilateral triangle has 3 lines of symmetry. B. A non-square rhombus has 2 lines of symmetry. C. A non-square rectangle has 2 lines of symmetry. D. An isosceles trapezoid has 1 line of symmetry. E. A square has 4 lines of symmetry. The figure with the greatest number of lines of symmetry is the square. Let's check the choices: the square is choice E.

ANSWER 10: E

Problem 11:
From the given information: Neither Jo nor Bo has as much money as Flo, so $Jo < Flo$ and $Bo < Flo$. Both Bo and Coe have more than Moe, so $Moe < Bo$ and $Moe < Coe$. Jo has more than Moe but less than Bo, so $Moe < Jo < Bo$. Combining these, we have $Moe < Jo < Bo < Flo$ and $Moe < Coe$. Since we don't have a direct comparison between Coe and the others except Moe, and we know $Bo < Flo$, the order for the least amount of money is clear for Moe. Let's check the choices: A. Bo, B. Coe, C. Flo, D. Jo, E. Moe. Moe has the least amount of money, matching choice E.

ANSWER 11: E

Problem 12:
Let's check each pair to see if their product equals 36: A. $(-4) \times (-9) = 36$, B. $(-3) \times (-12) = 36$, C. $\frac{1}{2} \times (-72) = -36$, D. $1 \times 36 = 36$, E. $\frac{3}{2} \times 24 = 36$. The pair that does NOT have a product equal to 36 is C. $\{\frac{1}{2}, -72\}$.

ANSWER 12: C

Problem 1:

We are asked for the ratio

\[
\frac{\text{side length of square}}{\text{radius of circle}}.
\]

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The square has area

\[
s^2.
\]

The circle has area

\[
\pi r^2.
\]

They have the same area, so

\[
s^2=\pi r^2.
\]

Taking the square root of both sides,

\[
s=r\sqrt{\pi}.
\]

So

\[
\frac{s}{r}=\sqrt{\pi}.
\]

The choices involving \(\pi\), \(2\pi\), or fractions of \(\pi\) are too large or have the wrong form. The correct ratio is \(\sqrt{\pi}\).

ANSWER 1: B

Problem 2:

We need the number of digits in

\[
4^5 \cdot 5^{10}.
\]

Rewrite \(4^5\):

\[
4^5=(2^2)^5=2^{10}.
\]

So

\[
4^5 \cdot 5^{10}=2^{10}\cdot 5^{10}.
\]

Since \(2^{10}\cdot 5^{10}=(2\cdot 5)^{10}\),

\[
2^{10}\cdot 5^{10}=10^{10}.
\]

Now,

\[
10^{10}=10,000,000,000,
\]

which is a 1 followed by 10 zeros.

So it has

\[
11
\]

digits.

ANSWER 2: D

Problem 3:

Eight friends were supposed to share the bill equally. Judi forgot her money, so the other 7 friends each paid an extra \(\$2.50\).

The total extra money paid by the 7 friends was

\[
7 \times 2.50 = 17.50.
\]

This extra money covered Judi’s share of the bill.

So Judi’s share was \(\$17.50\). Since there were 8 equal shares, the total bill was

\[
8 \times 17.50 = 140.
\]

Thus the total bill was \(\$140\).

ANSWER 3: C

Problem 4:

Carrie’s garden is a rectangle measuring 6 feet by 8 feet.

First find the area:

\[
6 \times 8 = 48
\]

square feet.

She plants 4 strawberry plants per square foot, so the number of plants is

\[
48 \times 4 = 192.
\]

Each plant produces an average of 10 strawberries, so the total expected harvest is

\[
192 \times 10 = 1920.
\]

So Carrie can expect to harvest 1920 strawberries.

ANSWER 4: D

Problem 5:

We need to find the largest fraction among:

\[
\frac13,\frac14,\frac38,\frac5{12},\frac7{24}.
\]

Use a common denominator of 24:

\[
\frac13=\frac8{24}
\]

\[
\frac14=\frac6{24}
\]

\[
\frac38=\frac9{24}
\]

\[
\frac5{12}=\frac{10}{24}
\]

\[
\frac7{24}=\frac7{24}
\]

Now compare:

\[
\frac6{24},\frac7{24},\frac8{24},\frac9{24},\frac{10}{24}.
\]

The largest is

\[
\frac{10}{24}=\frac5{12}.
\]

ANSWER 5: D

Problem 6:

Let the original length be \(L\), and let the original width be \(W\).

The original area is

\[
LW.
\]

The length is increased by 10%, so the new length is

\[
1.10L.
\]

The width is decreased by 10%, so the new width is

\[
0.90W.
\]

The new area is

\[
(1.10L)(0.90W)=0.99LW.
\]

So the new area is \(99\%\) of the old area.

A common mistake is to think the \(+10\%\) and \(-10\%\) cancel, but they are applied to different dimensions, so the area changes by the product \(1.10 \times 0.90=0.99\).

ANSWER 6: B

Problem 7:

The expression is

\[
\left(\frac38+\frac78\right)\div \frac45.
\]

First add the fractions in parentheses:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45.
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54 = \frac{25}{16}.
\]

So the value is

\[
\frac{25}{16}.
\]

ANSWER 7: B

Problem 8:

We need to compute

\[
\frac1{10}+\frac9{100}+\frac9{1000}+\frac7{10000}.
\]

Convert each fraction to a decimal:

\[
\frac1{10}=0.1
\]

\[
\frac9{100}=0.09
\]

\[
\frac9{1000}=0.009
\]

\[
\frac7{10000}=0.0007
\]

Now add:

\[
0.1+0.09=0.19
\]

\[
0.19+0.009=0.199
\]

\[
0.199+0.0007=0.1997.
\]

So the answer is

\[
0.1997.
\]

ANSWER 8: C

Problem 9:

We need to add

\[
90+91+92+93+94+95+96+97+98+99.
\]

There are 10 numbers.

The average of the first and last number is

\[
\frac{90+99}{2}=\frac{189}{2}=94.5.
\]

So the sum is

\[
10 \times 94.5 = 945.
\]

Alternatively, pair the numbers:

\[
90+99=189
\]

\[
91+98=189
\]

\[
92+97=189
\]

\[
93+96=189
\]

\[
94+95=189
\]

There are 5 pairs, so

\[
5 \times 189 = 945.
\]

ANSWER 9: B

Problem 10:

We compare the number of lines of symmetry for each figure.

- Equilateral triangle: 3 lines of symmetry.
- Non-square rhombus: 2 lines of symmetry.
- Non-square rectangle: 2 lines of symmetry.
- Isosceles trapezoid: 1 line of symmetry.
- Square: 4 lines of symmetry.

The square has the greatest number of lines of symmetry.

ANSWER 10: E

Problem 11:

We are told:

- Neither Jo nor Bo has as much money as Flo, so Flo has more than Jo and Bo.
- Both Bo and Coe have more than Moe.
- Jo has more than Moe, but less than Bo.

From this:

\[
Bo > Moe
\]

\[
Coe > Moe
\]

\[
Jo > Moe
\]

Also, since Flo has more than Jo,

\[
Flo > Jo > Moe.
\]

So Flo also has more than Moe.

Therefore, everyone has more money than Moe.

So Moe has the least amount of money.

ANSWER 11: E

Problem 12:

We need to find the pair whose product is NOT equal to 36.

Check each pair:

A. \(\{-4,-9\}\)

\[
(-4)(-9)=36
\]

So A works.

B. \(\{-3,-12\}\)

\[
(-3)(-12)=36
\]

So B works.

C. \(\left\{\frac12,-72\right\}\)

\[
\frac12 \cdot (-72)=-36
\]

This is not 36.

D. \(\{1,36\}\)

\[
1 \cdot 36=36
\]

So D works.

E. \(\left\{\frac32,24\right\}\)

\[
\frac32 \cdot 24=36
\]

So E works.

The only pair whose product is not 36 is choice C.

ANSWER 12: C

**Problem 1:**
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

**Step-by-step working:**
1. Let the side length of the square be $s$ and the radius of the circle be $r$.
2. The area of the square is $s^2$.
3. The area of the circle is $\pi r^2$.
4. Since their areas are equal, we can set up the equation: $s^2 = \pi r^2$.
5. To find the ratio of the side length to the radius ($s/r$), we first divide both sides by $r^2$: $\frac{s^2}{r^2} = \pi$.
6. Taking the square root of both sides gives us: $\frac{s}{r} = \sqrt{\pi}$.
7. This matches choice B.

ANSWER 1: B

**Problem 2:**
How many digits are in the product $4^5 \cdot 5^{10}$?

**Step-by-step working:**
1. We want to find the number of digits in the number $4^5 \cdot 5^{10}$.
2. First, express the base $4$ as a power of $2$: $4 = 2^2$.
3. Substitute this into the expression: $(2^2)^5 \cdot 5^{10}$.
4. Use the power of a power property $(a^m)^n = a^{m \cdot n}$: $2^{10} \cdot 5^{10}$.
5. Since the exponents are the same, we can combine the bases: $(2 \cdot 5)^{10} = 10^{10}$.
6. The number $10^{10}$ is written as a $1$ followed by $10$ zeros.
7. Therefore, it has $1 + 10 = 11$ digits.

ANSWER 2: D

**Problem 3:**
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

**Step-by-step working:**
1. Judi's portion of the bill was covered by her $7$ friends.
2. Each of the $7$ friends paid an extra $\$2.50$.
3. The total amount they paid to cover Judi's share is $7 \times \$2.50 = \$17.50$.
4. This means that an equal share of the bill for one person is $\$17.50$.
5. Since there are $8$ friends in total, the total bill is $8 \times \$17.50$.
6. $8 \times \$17.50 = \$140.00$.

ANSWER 3: C

**Problem 4:**
Carrie has a rectangular garden that measures 6 feet by 8 feet. She plants the entire garden with strawberry plants. Carrie is able to plant 4 strawberry plants per square foot, and she harvests an average of 10 strawberries per plant. How many strawberries can she expect to harvest?

**Step-by-step working:**
1. First, find the area of the rectangular garden: $6 \text{ feet} \times 8 \text{ feet} = 48 \text{ square feet}$.
2. Next, calculate the total number of strawberry plants. She plants $4$ plants per square foot: $48 \times 4 = 192 \text{ plants}$.
3. Finally, calculate the total expected strawberry harvest. Each plant yields an average of $10$ strawberries: $192 \times 10 = 1920 \text{ strawberries}$.

ANSWER 4: D

**Problem 5:**
Which of the following is the largest?

**Step-by-step working:**
1. The choices given are the fractions: $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$.
2. To compare them easily, find a common denominator. The least common multiple (LCM) of $3, 4, 8, 12,$ and $24$ is $24$.
3. Convert each fraction:
   - A. $\frac{1}{3} = \frac{8}{24}$
   - B. $\frac{1}{4} = \frac{6}{24}$
   - C. $\frac{3}{8} = \frac{9}{24}$
   - D. $\frac{5}{12} = \frac{10}{24}$
   - E. $\frac{7}{24} = \frac{7}{24}$
4. Comparing the numerators ($8, 6, 9, 10, 7$), the largest is $10$.
5. Therefore, $\frac{10}{24}$ (which is $\frac{5}{12}$) is the largest fraction.

ANSWER 5: D

**Problem 6:**
The length of a rectangle is increased by 10% and the width is decreased by 10%. What percent of the old area is the new area?

**Step-by-step working:**
1. Let the original length be $L$ and the original width be $W$.
2. The old area is $A_{\text{old}} = L \times W$.
3. The new length is increased by $10\%$, making it $1.10L$.
4. The new width is decreased by $10\%$, making it $0.90W$.
5. The new area is $A_{\text{new}} = (1.10L) \times (0.90W) = 0.99(L \times W)$.
6. This means the new area is $0.99$ times the old area, which corresponds to $99\%$.

ANSWER 6: B

**Problem 7:**
$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}} =$

**Step-by-step working:**
1. First, evaluate the numerator by adding the fractions: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8}$.
2. Simplify the fraction: $\frac{10}{8} = \frac{5}{4}$.
3. Now, divide this result by the denominator $\frac{4}{5}$: $\frac{5}{4} \div \frac{4}{5}$.
4. Dividing by a fraction is the same as multiplying by its reciprocal: $\frac{5}{4} \times \frac{5}{4}$.
5. Multiply the numerators and the denominators: $\frac{5 \times 5}{4 \times 4} = \frac{25}{16}$.

ANSWER 7: B

**Problem 8:**
$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000} =$

**Step-by-step working:**
1. Convert each fraction into its decimal equivalent:
   - $\frac{1}{10} = 0.1$
   - $\frac{9}{100} = 0.09$
   - $\frac{9}{1000} = 0.009$
   - $\frac{7}{10000} = 0.0007$
2. Add the decimals together:
   $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$.

ANSWER 8: C

**Problem 9:**
$90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =$

**Step-by-step working:**
1. We are adding an arithmetic sequence of $10$ consecutive integers from $90$ to $99$.
2. We can split the sum into the tens and the units:
   - The tens contribute $90 \times 10 = 900$.
   - The units contribute $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$.
3. The sum of the digits from $0$ to $9$ is $\frac{9 \times 10}{2} = 45$.
4. Adding these together gives $900 + 45 = 945$.
*(Alternatively, using the formula for the sum of an arithmetic sequence: $S = \frac{n(a_1 + a_n)}{2} = \frac{10(90 + 99)}{2} = 5 \times 189 = 945$.)*

ANSWER 9: B

**Problem 10:**
Which of the following figures has the greatest number of lines of symmetry?

**Step-by-step working:**
1. Let's count the lines of symmetry for each given figure:
   - A. Equilateral triangle: $3$ lines of symmetry (from each vertex to the midpoint of the opposite side).
   - B. Non-square rhombus: $2$ lines of symmetry (along its diagonals).
   - C. Non-square rectangle: $2$ lines of symmetry (connecting the midpoints of opposite sides).
   - D. Isosceles trapezoid: $1$ line of sy

Problem 1:
The problem asks for the ratio of the side length of a square to the radius of a circle when both have the same area. Let's denote the side length of the square as $s$ and the radius of the circle as $r$. The area of the square is $s^2$ and the area of the circle is $\pi r^2$. Since they have the same area, we have $s^2 = \pi r^2$. To find the ratio of $s$ to $r$, we take the square root of both sides: $s = \sqrt{\pi}r$. Thus, the ratio of $s$ to $r$ is $\sqrt{\pi}$. Let's examine the choices: A. $\sqrt{\pi}^2 = \pi$, B. $\sqrt{\pi}$, C. $\pi$, D. $2\pi$, E. $\pi^2$. The correct ratio is given by choice B.

ANSWER 1: B

Problem 2:
To find the number of digits in $4^5 \cdot 5^{10}$, let's simplify the expression. $4^5 = (2^2)^5 = 2^{10}$, so $4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$. $10^{10}$ is a 1 followed by 10 zeros, making it an 11-digit number. Let's check the choices: A. 8, B. 9, C. 10, D. 11, E. 12. The correct number of digits is given by choice D.

ANSWER 2: D

Problem 3:
The problem states that 8 friends agreed to share a bill equally, but because Judi forgot her money, each of her 7 friends paid an extra $2.50. Let's denote the total bill as $B$. The amount each friend was originally supposed to pay is $\frac{B}{8}$. When Judi's portion is covered by her 7 friends, each pays $\frac{B}{8} + $2.50$. The total amount paid by the 7 friends is $7(\frac{B}{8} + 2.50)$, which equals $B$. So, we have $7(\frac{B}{8} + 2.50) = B$. Simplifying this gives $\frac{7B}{8} + 17.50 = B$. Subtracting $\frac{7B}{8}$ from both sides gives $17.50 = \frac{B}{8}$. Multiplying both sides by 8 gives $B = 140$. Let's check the choices: A. $120, B. $128, C. $140, D. $144, E. $160. The total bill matches choice C.

ANSWER 3: C

Problem 4:
Carrie has a garden measuring 6 feet by 8 feet, so its area is $6 \times 8 = 48$ square feet. She plants 4 strawberry plants per square foot, so she plants $48 \times 4 = 192$ plants. With an average of 10 strawberries per plant, she can expect to harvest $192 \times 10 = 1920$ strawberries. Let's check the choices: A. 560, B. 960, C. 1120, D. 1920, E. 3840. The expected harvest matches choice D.

ANSWER 4: D

Problem 5:
To compare the given fractions, let's convert them to equivalent decimals or find a common denominator. The fractions are $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$. Converting them to equivalent fractions with a common denominator (24) gives $\frac{8}{24}$, $\frac{6}{24}$, $\frac{9}{24}$, $\frac{10}{24}$, and $\frac{7}{24}$. Comparing these, $\frac{10}{24}$ is the largest. So, $\frac{5}{12}$ is the largest fraction. Let's check the choices: A. $\frac{1}{3}$, B. $\frac{1}{4}$, C. $\frac{3}{8}$, D. $\frac{5}{12}$, E. $\frac{7}{24}$. The largest fraction is given by choice D.

ANSWER 5: D

Problem 6:
Let's denote the original length and width of the rectangle as $L$ and $W$, respectively. The original area is $L \times W$. After increasing the length by 10% and decreasing the width by 10%, the new length is $1.1L$ and the new width is $0.9W$. The new area is $1.1L \times 0.9W = 0.99LW$. To find what percent of the old area the new area is, we calculate $\frac{0.99LW}{LW} \times 100\% = 99\%$. Let's check the choices: A. 90, B. 99, C. 100, D. 101, E. 110. The new area is 99% of the old area, matching choice B.

ANSWER 6: B

Problem 7:
To evaluate $\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$, first simplify the numerator: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$. Then, divide by $\frac{4}{5}$, which is equivalent to multiplying by $\frac{5}{4}$. So, $\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$. Let's check the choices: A. 1, B. $\frac{25}{16}$, C. 2, D. $\frac{43}{20}$, E. $\frac{47}{16}$. The result matches choice B.

ANSWER 7: B

Problem 8:
To evaluate $\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000}$, convert each fraction to a decimal: $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$. Let's check the choices: A. 0.0026, B. 0.0197, C. 0.1997, D. 0.26, E. 1.997. The sum matches choice C.

ANSWER 8: C

Problem 9:
The sum $90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99$ is an arithmetic series with the first term $a = 90$, the last term $l = 99$, and $n = 10$ terms. The sum of an arithmetic series is given by $S = \frac{n}{2}(a + l)$. So, $S = \frac{10}{2}(90 + 99) = 5 \times 189 = 945$. Let's check the choices: A. 845, B. 945, C. 1005, D. 1025, E. 1045. The sum matches choice B.

ANSWER 9: B

Problem 10:
Let's analyze the number of lines of symmetry for each figure: A. An equilateral triangle has 3 lines of symmetry. B. A non-square rhombus has 2 lines of symmetry. C. A non-square rectangle has 2 lines of symmetry. D. An isosceles trapezoid has 1 line of symmetry. E. A square has 4 lines of symmetry. The figure with the greatest number of lines of symmetry is the square. Let's check the choices: the square is choice E.

ANSWER 10: E

Problem 11:
From the given information: Neither Jo nor Bo has as much money as Flo, so $Jo < Flo$ and $Bo < Flo$. Both Bo and Coe have more than Moe, so $Moe < Bo$ and $Moe < Coe$. Jo has more than Moe but less than Bo, so $Moe < Jo < Bo$. Combining these, we have $Moe < Jo < Bo < Flo$ and $Moe < Coe$. Since we don't have a direct comparison between Coe and the others except Moe, and we know $Bo < Flo$, the order for the least amount of money is clear for Moe. Let's check the choices: A. Bo, B. Coe, C. Flo, D. Jo, E. Moe. Moe has the least amount of money, matching choice E.

ANSWER 11: E

Problem 12:
Let's check each pair to see if their product equals 36: A. $(-4) \times (-9) = 36$, B. $(-3) \times (-12) = 36$, C. $\frac{1}{2} \times (-72) = -36$, D. $1 \times 36 = 36$, E. $\frac{3}{2} \times 24 = 36$. The pair that does NOT have a product equal to 36 is C. $\{\frac{1}{2}, -72\}$.

ANSWER 12: C

Problem 1:

We are asked for the ratio

\[
\frac{\text{side length of square}}{\text{radius of circle}}.
\]

Let the side length of the square be \(s\), and let the radius of the circle be \(r\).

The square has area

\[
s^2.
\]

The circle has area

\[
\pi r^2.
\]

They have the same area, so

\[
s^2=\pi r^2.
\]

Taking the square root of both sides,

\[
s=r\sqrt{\pi}.
\]

So

\[
\frac{s}{r}=\sqrt{\pi}.
\]

The choices involving \(\pi\), \(2\pi\), or fractions of \(\pi\) are too large or have the wrong form. The correct ratio is \(\sqrt{\pi}\).

ANSWER 1: B

Problem 2:

We need the number of digits in

\[
4^5 \cdot 5^{10}.
\]

Rewrite \(4^5\):

\[
4^5=(2^2)^5=2^{10}.
\]

So

\[
4^5 \cdot 5^{10}=2^{10}\cdot 5^{10}.
\]

Since \(2^{10}\cdot 5^{10}=(2\cdot 5)^{10}\),

\[
2^{10}\cdot 5^{10}=10^{10}.
\]

Now,

\[
10^{10}=10,000,000,000,
\]

which is a 1 followed by 10 zeros.

So it has

\[
11
\]

digits.

ANSWER 2: D

Problem 3:

Eight friends were supposed to share the bill equally. Judi forgot her money, so the other 7 friends each paid an extra \(\$2.50\).

The total extra money paid by the 7 friends was

\[
7 \times 2.50 = 17.50.
\]

This extra money covered Judi’s share of the bill.

So Judi’s share was \(\$17.50\). Since there were 8 equal shares, the total bill was

\[
8 \times 17.50 = 140.
\]

Thus the total bill was \(\$140\).

ANSWER 3: C

Problem 4:

Carrie’s garden is a rectangle measuring 6 feet by 8 feet.

First find the area:

\[
6 \times 8 = 48
\]

square feet.

She plants 4 strawberry plants per square foot, so the number of plants is

\[
48 \times 4 = 192.
\]

Each plant produces an average of 10 strawberries, so the total expected harvest is

\[
192 \times 10 = 1920.
\]

So Carrie can expect to harvest 1920 strawberries.

ANSWER 4: D

Problem 5:

We need to find the largest fraction among:

\[
\frac13,\frac14,\frac38,\frac5{12},\frac7{24}.
\]

Use a common denominator of 24:

\[
\frac13=\frac8{24}
\]

\[
\frac14=\frac6{24}
\]

\[
\frac38=\frac9{24}
\]

\[
\frac5{12}=\frac{10}{24}
\]

\[
\frac7{24}=\frac7{24}
\]

Now compare:

\[
\frac6{24},\frac7{24},\frac8{24},\frac9{24},\frac{10}{24}.
\]

The largest is

\[
\frac{10}{24}=\frac5{12}.
\]

ANSWER 5: D

Problem 6:

Let the original length be \(L\), and let the original width be \(W\).

The original area is

\[
LW.
\]

The length is increased by 10%, so the new length is

\[
1.10L.
\]

The width is decreased by 10%, so the new width is

\[
0.90W.
\]

The new area is

\[
(1.10L)(0.90W)=0.99LW.
\]

So the new area is \(99\%\) of the old area.

A common mistake is to think the \(+10\%\) and \(-10\%\) cancel, but they are applied to different dimensions, so the area changes by the product \(1.10 \times 0.90=0.99\).

ANSWER 6: B

Problem 7:

The expression is

\[
\left(\frac38+\frac78\right)\div \frac45.
\]

First add the fractions in parentheses:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45.
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54 = \frac{25}{16}.
\]

So the value is

\[
\frac{25}{16}.
\]

ANSWER 7: B

Problem 8:

We need to compute

\[
\frac1{10}+\frac9{100}+\frac9{1000}+\frac7{10000}.
\]

Convert each fraction to a decimal:

\[
\frac1{10}=0.1
\]

\[
\frac9{100}=0.09
\]

\[
\frac9{1000}=0.009
\]

\[
\frac7{10000}=0.0007
\]

Now add:

\[
0.1+0.09=0.19
\]

\[
0.19+0.009=0.199
\]

\[
0.199+0.0007=0.1997.
\]

So the answer is

\[
0.1997.
\]

ANSWER 8: C

Problem 9:

We need to add

\[
90+91+92+93+94+95+96+97+98+99.
\]

There are 10 numbers.

The average of the first and last number is

\[
\frac{90+99}{2}=\frac{189}{2}=94.5.
\]

So the sum is

\[
10 \times 94.5 = 945.
\]

Alternatively, pair the numbers:

\[
90+99=189
\]

\[
91+98=189
\]

\[
92+97=189
\]

\[
93+96=189
\]

\[
94+95=189
\]

There are 5 pairs, so

\[
5 \times 189 = 945.
\]

ANSWER 9: B

Problem 10:

We compare the number of lines of symmetry for each figure.

- Equilateral triangle: 3 lines of symmetry.
- Non-square rhombus: 2 lines of symmetry.
- Non-square rectangle: 2 lines of symmetry.
- Isosceles trapezoid: 1 line of symmetry.
- Square: 4 lines of symmetry.

The square has the greatest number of lines of symmetry.

ANSWER 10: E

Problem 11:

We are told:

- Neither Jo nor Bo has as much money as Flo, so Flo has more than Jo and Bo.
- Both Bo and Coe have more than Moe.
- Jo has more than Moe, but less than Bo.

From this:

\[
Bo > Moe
\]

\[
Coe > Moe
\]

\[
Jo > Moe
\]

Also, since Flo has more than Jo,

\[
Flo > Jo > Moe.
\]

So Flo also has more than Moe.

Therefore, everyone has more money than Moe.

So Moe has the least amount of money.

ANSWER 11: E

Problem 12:

We need to find the pair whose product is NOT equal to 36.

Check each pair:

A. \(\{-4,-9\}\)

\[
(-4)(-9)=36
\]

So A works.

B. \(\{-3,-12\}\)

\[
(-3)(-12)=36
\]

So B works.

C. \(\left\{\frac12,-72\right\}\)

\[
\frac12 \cdot (-72)=-36
\]

This is not 36.

D. \(\{1,36\}\)

\[
1 \cdot 36=36
\]

So D works.

E. \(\left\{\frac32,24\right\}\)

\[
\frac32 \cdot 24=36
\]

So E works.

The only pair whose product is not 36 is choice C.

ANSWER 12: C

**Problem 1:**
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

**Step-by-step working:**
1. Let the side length of the square be $s$ and the radius of the circle be $r$.
2. The area of the square is $s^2$.
3. The area of the circle is $\pi r^2$.
4. Since their areas are equal, we can set up the equation: $s^2 = \pi r^2$.
5. To find the ratio of the side length to the radius ($s/r$), we first divide both sides by $r^2$: $\frac{s^2}{r^2} = \pi$.
6. Taking the square root of both sides gives us: $\frac{s}{r} = \sqrt{\pi}$.
7. This matches choice B.

ANSWER 1: B

**Problem 2:**
How many digits are in the product $4^5 \cdot 5^{10}$?

**Step-by-step working:**
1. We want to find the number of digits in the number $4^5 \cdot 5^{10}$.
2. First, express the base $4$ as a power of $2$: $4 = 2^2$.
3. Substitute this into the expression: $(2^2)^5 \cdot 5^{10}$.
4. Use the power of a power property $(a^m)^n = a^{m \cdot n}$: $2^{10} \cdot 5^{10}$.
5. Since the exponents are the same, we can combine the bases: $(2 \cdot 5)^{10} = 10^{10}$.
6. The number $10^{10}$ is written as a $1$ followed by $10$ zeros.
7. Therefore, it has $1 + 10 = 11$ digits.

ANSWER 2: D

**Problem 3:**
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

**Step-by-step working:**
1. Judi's portion of the bill was covered by her $7$ friends.
2. Each of the $7$ friends paid an extra $\$2.50$.
3. The total amount they paid to cover Judi's share is $7 \times \$2.50 = \$17.50$.
4. This means that an equal share of the bill for one person is $\$17.50$.
5. Since there are $8$ friends in total, the total bill is $8 \times \$17.50$.
6. $8 \times \$17.50 = \$140.00$.

ANSWER 3: C

**Problem 4:**
Carrie has a rectangular garden that measures 6 feet by 8 feet. She plants the entire garden with strawberry plants. Carrie is able to plant 4 strawberry plants per square foot, and she harvests an average of 10 strawberries per plant. How many strawberries can she expect to harvest?

**Step-by-step working:**
1. First, find the area of the rectangular garden: $6 \text{ feet} \times 8 \text{ feet} = 48 \text{ square feet}$.
2. Next, calculate the total number of strawberry plants. She plants $4$ plants per square foot: $48 \times 4 = 192 \text{ plants}$.
3. Finally, calculate the total expected strawberry harvest. Each plant yields an average of $10$ strawberries: $192 \times 10 = 1920 \text{ strawberries}$.

ANSWER 4: D

**Problem 5:**
Which of the following is the largest?

**Step-by-step working:**
1. The choices given are the fractions: $\frac{1}{3}$, $\frac{1}{4}$, $\frac{3}{8}$, $\frac{5}{12}$, and $\frac{7}{24}$.
2. To compare them easily, find a common denominator. The least common multiple (LCM) of $3, 4, 8, 12,$ and $24$ is $24$.
3. Convert each fraction:
   - A. $\frac{1}{3} = \frac{8}{24}$
   - B. $\frac{1}{4} = \frac{6}{24}$
   - C. $\frac{3}{8} = \frac{9}{24}$
   - D. $\frac{5}{12} = \frac{10}{24}$
   - E. $\frac{7}{24} = \frac{7}{24}$
4. Comparing the numerators ($8, 6, 9, 10, 7$), the largest is $10$.
5. Therefore, $\frac{10}{24}$ (which is $\frac{5}{12}$) is the largest fraction.

ANSWER 5: D

**Problem 6:**
The length of a rectangle is increased by 10% and the width is decreased by 10%. What percent of the old area is the new area?

**Step-by-step working:**
1. Let the original length be $L$ and the original width be $W$.
2. The old area is $A_{\text{old}} = L \times W$.
3. The new length is increased by $10\%$, making it $1.10L$.
4. The new width is decreased by $10\%$, making it $0.90W$.
5. The new area is $A_{\text{new}} = (1.10L) \times (0.90W) = 0.99(L \times W)$.
6. This means the new area is $0.99$ times the old area, which corresponds to $99\%$.

ANSWER 6: B

**Problem 7:**
$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}} =$

**Step-by-step working:**
1. First, evaluate the numerator by adding the fractions: $\frac{3}{8} + \frac{7}{8} = \frac{10}{8}$.
2. Simplify the fraction: $\frac{10}{8} = \frac{5}{4}$.
3. Now, divide this result by the denominator $\frac{4}{5}$: $\frac{5}{4} \div \frac{4}{5}$.
4. Dividing by a fraction is the same as multiplying by its reciprocal: $\frac{5}{4} \times \frac{5}{4}$.
5. Multiply the numerators and the denominators: $\frac{5 \times 5}{4 \times 4} = \frac{25}{16}$.

ANSWER 7: B

**Problem 8:**
$\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000} =$

**Step-by-step working:**
1. Convert each fraction into its decimal equivalent:
   - $\frac{1}{10} = 0.1$
   - $\frac{9}{100} = 0.09$
   - $\frac{9}{1000} = 0.009$
   - $\frac{7}{10000} = 0.0007$
2. Add the decimals together:
   $0.1 + 0.09 + 0.009 + 0.0007 = 0.1997$.

ANSWER 8: C

**Problem 9:**
$90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =$

**Step-by-step working:**
1. We are adding an arithmetic sequence of $10$ consecutive integers from $90$ to $99$.
2. We can split the sum into the tens and the units:
   - The tens contribute $90 \times 10 = 900$.
   - The units contribute $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$.
3. The sum of the digits from $0$ to $9$ is $\frac{9 \times 10}{2} = 45$.
4. Adding these together gives $900 + 45 = 945$.
*(Alternatively, using the formula for the sum of an arithmetic sequence: $S = \frac{n(a_1 + a_n)}{2} = \frac{10(90 + 99)}{2} = 5 \times 189 = 945$.)*

ANSWER 9: B

**Problem 10:**
Which of the following figures has the greatest number of lines of symmetry?

**Step-by-step working:**
1. Let's count the lines of symmetry for each given figure:
   - A. Equilateral triangle: $3$ lines of symmetry (from each vertex to the midpoint of the opposite side).
   - B. Non-square rhombus: $2$ lines of symmetry (along its diagonals).
   - C. Non-square rectangle: $2$ lines of symmetry (connecting the midpoints of opposite sides).
   - D. Isosceles trapezoid: $1$ line of sy