Did you take the ISEE Middle Level Math Practice Test? If so, then it’s time to review your results to see where you went wrong and what areas you need to improve.

## ISEE Middle Level Math Practice Test Answers and Explanations

**ISEE Middle Level Practice Test**

**Quantitative Reasoning**

1- **Choice D is correct**

\(15\%\) off equals \($18\). Let x be the original price of the table. Then:

\(15\%\) of \(x=18→0.15x=18→x=\frac{18}{0.15}=120\)

2- **Choice B is correct**

\(981,364,454×\frac{1}{10,000,000}=98.1364454\)

3- **Choice C is correct**

\(\frac{1}{8}≅0.125, \frac{1}{3}≅0.33, \frac{1}{2}=0.5, \frac{3}{5}=0.6\)

4- **Choice C is correct**

Method 1: \(5×5×5×5=625\), Method 2: \(8^x=512\)

Let’s review the choices provided:

A. \(2\) →\(5^x=625→5^2=25\)

B. \(3\) →\(5^x=625→5^3=125\)

C. \(4\) →\(5^x=625→5^4=625\)

D. \(5\) →\(5^x=625→5^5=3125\)

Choice C is correct.

5- **Choice C is correct**

\(2g=2×(-2x-2y)=-4x-4y\)

\( f-2g=4x-y-(-4x-4y)=8x+3y\)

6- **Choice A is correct**

\(\frac{-45 × 0.8}{4}=-\frac{45×\frac{8}{10}}{4}=-\frac{\frac{360}{10}}{4}=-\frac{360}{40}=-9\)

7- **Choice A is correct**

Use the formula for Percent of Change: \(\frac{New \space Value\space-\space Old \space Value}{Old \space Value}× 100 \%\)

\(\frac{43.2-54}{54}× 100\% = –20\%\) (negative sign here means that the new price is less than old price)

8- **Choice A is correct**

Supplementary angles sum up to 180 degrees. x and 33 degrees are supplementary angles. Then: \(x=180^\circ -33^\circ=147^\circ\)

9- **Choice C is correct**

Use the formula of areas of circles.

Area of a circe \(= πr^2 ⇒ 36 π = r^2 ⇒ 36 = r^2 ⇒ r = 6\)

Radius of the circle is 6. Now, use the circumference formula:

Circumference \(= 2πr = 2π(6) = 12π\)

10- **Choice D is correct**

\(2.21=\frac{221}{100} \) and \(6.5=\frac{65}{10} →2.21×6.5=\frac{221}{100}×\frac{65}{10}=\frac{14365}{1000}=14.365≅14.4\)

11- **Choice C is correct**

If the score of Mia was 80, therefore the score of Ava is 40. Since, the score of Emma was half as that of Ava, therefore, the score of Emma is 20.

12- **Choice A is correct**

The perimeter of the trapezoid is 60.

Therefore, the missing side (height) is \(= 60 – 10 – 20 – 12 = 18\)

Area of the trapezoid: A \(= \frac{1}{2} h (b_1 + b_2) = \frac{1}{2}(18) (10 + 12) = 198\)

13- **Choice C is correct**

Add the first 4 numbers. \(50 + 48 + 42 + 52 = 192\)

To find the distance traveled in the next 4 hours, multiply the average by number of hours.

Distance = Average \(×\) Rate\( = 50 × 4 = 200\), Add both numbers.\( 200 + 192 = 392\)

14- **Choice D is correct**

Mean\(=\frac{12+22+34+46+52+68+72+86}{8}=\frac{392}{8}=49\)

15- **Choice D is correct**

Let \(x\) be the number. Write the equation and solve for \(x\).

\(\frac{2}{3}×36=\frac{4}{9}\). \(x ⇒ \frac{2×36}{3}= \frac{4x}{9}\), use cross multiplication to solve for \(x\).

\(9×72=4x×3 ⇒648=12x ⇒ x=54\)

**16-Choice B is correct**

Simplify: \(-2(2x+3)=3(4-2x), -4x-6=12-6x\), Add \(4x\) from both sides: \(-6=-2x+12\), subtract 12 to both sides: \(-18 = -2x, 9 = x\)

**17**–**Choice B is correct **

\(\frac{3}{4}\) of \(240=\frac{3}{4}×240=180\), \(\frac{1}{3}\) of \(180=\frac{1}{3}×180=60\)

**18**–**Choice C is correct**

The ratio of boy to girls is 6:8. Therefore, there are 6 boys out of 14 students. To find the answer, first divide the total number of students by 14, then multiply the result by 6.

\( 70 ÷ 14 = 5 ⇒ 5× 6 = 30\)

There are 30 boys and 40 (\(70 – 30\)) girls. So, 10 more boys should be enrolled to make the ratio 1:1

**19**–**Choice D is correct**

Let \(x\) be the original price. If the price of a laptop is decreased by \(20\%\) to $420, then: \(0.8\%\) of \(x=420⇒ 0.8x=420 ⇒ x=420÷0.8=525\)

**20-Choice C is correct **Let \(x\) be the sales profit. Then, \(2\%\) of sales profit is \(0.03x\). Employee’s revenue: \(0.03x+6,500\)

**21-Choice C is correct**

Petrol of car A in 300km=\(\frac{7×300}{100}=21\), Petrol of car B in 300km = \(\frac{9×300}{100}=27, 27-21=6\)

**22-Choice B is correct**

\(x=105+35=140\)

**23-Choice D is correct **

The diagonal of the square is 10. Let \(x\) be the side.

Use Pythagorean Theorem:\(a^2 + b^2 = c^2\)

\(x^2 + x^2 = 102 ⇒ 2x^2 = 102 ⇒ 2x^2 = 100 ⇒x^2 = 50 ⇒x= \sqrt{50}\)

The area of the square is:

\(\sqrt{50} × \sqrt{50} = 50\)

**24-Choice A is correct**

\(-42-(-73)=-42+73=73-42=31\)

**25-Choice D is correct**

Choice B is correct Simplify: \(-2(2x+3)=3(4-2x), -4x-6=12-6x\), Add \(4x\) from both sides: \(-6=-2x+12\), subtract 12 to both sides: \(-18 = -2x, 9 = x\)

Let’s review the choices:

A. \(4/5>0.8\) This is not a correct statement. Because \(\frac{4}{5}=0.8\) and it’s less than 0.8.

B. \(12\%=\frac{1}{5}\) This is not a correct statement. Because \(12\% = 0.12\) and \(\frac{1}{5}=0.2\)

C. \(4<\frac{6}{2}\) This is not a correct statement. Because \(\frac{6}{2}=3\) and it’s less than 4.

D. \(\frac{8}{9}>0.8\) This is a correct statement.

\(\frac{8}{9}=0.88→0.8<\frac{8}{9}\)

## The Absolute Best Book** to Ace the ISEE Middle Level** **Math** Test

**26-Choice A is correct**

Column A: Use order of operation to calculate the result. \(5+5×7-4=5+35-4=36\)

Column B: \(4+6×6-6→4+36-6=34\)

**27-Choice A is correct**

Column A: The value of \(x\) when \(y=9\) : \(y=3x+18→9=3x+18→3x=-9→x=-3\)

Column B: \(-4\), \(-3\) is greater than \(-4\).

**28-Choice C is correct**

Column A: Simplify. \(\sqrt{25}+\sqrt{25}=5+5=10, 10\) is equal to \(\sqrt{100}(\sqrt{100}=10)\)

**29-Choice D is correct****Column A:** Based on information provided, we cannot find the average age of Joe and Michelle or average age of Michelle and Nicole.

**30-Choice A is correct**

Column A: Simplify. \(\sqrt{144-12}=\sqrt{23}\)

Column B: \(\sqrt{144}-\sqrt{121}=12-11=1\) , \(\sqrt{23}\) is bigger than 1.

**31-Choice A is correct**

Volume of a right cylinder \(= πr^2 h→80π=πr^2 h=π(5)^2 h→h=3.2\)

The height of the cylinder is 3.2 inches which is bigger than 3 inches.

**32-Choice D is correct**

Choose different values for x and find the value of quantity A.

\(x=1\), then: Quantity A: \(\frac{2}{X}+x= \frac{2}{1}+1=3\), Quantity B is greater

\(x=0.1\), then: Quantity A: \(\frac{2}{X}+x= \frac{2}{0.1}+1=20+1=21\), Quantity A is greater The relationship cannot be determined from the information given.

**33-Choice B is correct**

Simply change the fractions to decimals. \(\frac{2}{5}=0.40, \frac{5}{9}=0.55…, \frac{5}{8}=0.625,\) Quantity B is greater

**34-Choice A is correct**

Simplify quantity B. Quantity B: \((\frac{x}{3})^3=\frac{x^3}{3^3 }\)

Since, the two quantities have the same numerator (\(x^3\)) and the denominator in quantity B is bigger \((3^3>3)\), then the quantity A is greater.

**35-Choice A is correct**

Quantity A is: \(\frac{5+6+x}{3}=6→x=7\), Quantity B is: \(\frac{(7+(7+3)+(7+1)-7)}{4}=4.5\)

**36-Choice C is correct**

Choose different values for a and b and find the values of quantity A and quantity B.

\(a=4\) and \(b=3\), then: Quantity A: \(|4 – 3|=|1|=1\), Quantity B: \(|3 – 4|=|1|=1\)

The two quantities are equal. a=3 and b=-3, then: Quantity A: \(|3+3|=|6|=6\)

Quantity B: \(|- 3 – (3)|=|-6|=6\), The two quantities are equal.

Any other values of a and b provide the same answer.

**37-Choice C is correct**

\(4x^3-80=176→4x^3=176+80=256→x^3=\frac{256}{4}=64→x=\sqrt[3]{64}=\sqrt[3]{4^3}=4\)

\(3-4y=-13→-4y=-13-3=-16→y=\frac{-16}{-4}=4\)

**38-Choice B is correct**

Number of pencils are blue\(=75-33=42\), Percent of blue pencils is: \(\frac{42}{75}×100=56\%\)

**39-Choice A is correct**

\(\frac{4}{6}×120=80\)

**40-Choice C is correct**

\(2(\frac{3}{5}-\frac{6}{10})+2=2×(\frac{6-6}{10})+2=0+2=2\)

**41-Choice B is correct**

\(12\%\) of \(160=\frac{12}{100}×160=19.2\), Let \(x\) be the number then, \(x=19.2+15=34.2\)

**42-Choice C is correct**

Let \(x\) be the number. Write the equation and solve for \(x\). \((21 – 2x) ÷ x = 5\)

Multiply both sides by \(x\). \((21 – 2x) = 5x\), then add x both sides. \(21 = 7x\), now divide both sides by 7. \(x = 3\)

**43-Choice A is correct**

\(25\%\) of \(48\) is: \(\frac{25}{100}×48=\frac{1200}{100}=12\), Let \(x\) be the number then: \(x=12-3=9\)

**44-Choice A is correct**

The perimeter of rectangle is: \(2×(6+5)=2×11=22\)

The perimeter of circle is: \(2πr=2×3×\frac{12}{2}=36\), Difference in perimeter is: \(36-22=14\)

**45-Choice A is correct**

Use this formula: Percent of Change = \(\frac{New \space Value-Old \space Value}{Old \space Valu}× 100 \%\)

\(\frac{21000-25000}{25000} × 100 \% = 16 \% \space and \space \frac{17640-21000}{21000} × 100 \% = 16 \%\)

**46-Choice D is correct**

\((2x-3)^3=125→2x-3=\sqrt[3]{125}=\sqrt[3]{5^3} =5→2x=5+3→x=\frac{8}{2}=4\)

**47-Choice B is correct**

\((\frac{10}{4}×24)+(\frac{5}{2}×10)=(60)+(25)=85\)

**48-Choice D is correct**

If \(\frac{x}{2}=24\), then \(x=48\), \(\frac{5x}{8}=\frac{5×48}{8}=\frac{240}{8}=30\)

**49-Choice C is correct**

\(\frac{3}{4}=0.75, \frac{7}{9}=0.77, 85%=0.85\)

**50-Choice B is correct**

Area\(=πr^2=π×(\frac{22}{2})^2=121π=121×3.14≅379.94=380\)

**51-Choice A is correct**

First, find the number. Let \(x\) be the number. Write the equation and solve for \(x\).

\(160 \%\) of a number is \(128\), then: \(1.6×x=128 ⇒ x=128÷1.6=80\),

\(60 \% \space \)of\( \space 80 \space is: 0.6 × 80 = 48\)

**52-Choice D is correct**

\(\frac{3}{5}×45=\frac{135}{5}=27\)

**53-Choice A is correct**

All angles in a triangle sum up to 180 degrees. Then:

\(2α+90^\circ=180^\circ→2α=90→α=45^\circ\)

**54-Choice B is correct**

The sum of angles in rectangle is 360.

**55-Choice B is correct**

Find the difference of each pairs of numbers: 5,8,13,21,34,*_*,89

The sum of \(5,8\) is \(13\). The sum of \(8,13\) is \(21\). The sum of \(13,21\) is \(34\). The sum of \(34\) and x is \(89\).Thus \(x=89-34=55\)

**56-Choice D is correct**

\(\frac{4}{6}×96=\frac{384}{6}=64\)

**57-Choice D is correct**

The capacity of a red box is \(25\%\) greater than a blue box. Let \(x\) be the capacity of the blue box. Then: \(x+25\% \space of \space x=40→1.25x=40→x=\frac{40}{1.25}=32\)

**58-Choice C is correct**

The question is this: 1.65 is what percent of 1.2?

Use percent formula: Part = \(\frac{percent}{100}× whole, 1.65= \frac{percent}{100}× 1.2 ⇒ 1.65 = \frac{percent ×1.2}{100} ⇒165 = percent ×1.2 ⇒ percent = \frac{165}{1.2} =137.5\)

**59-Choice A is correct**

60 minutes = 1Hours→\(\frac{315}{60}=5.25\) Hours

**60-Choice A is correct**

Let \(x\) be the number of shoes the team can purchase. Therefore, the team can purchase \(120 x\).

The team had \($32,000\) and spent \($20,000\). Now the team can spend on new shoes \($12000\) at most. Now, write the inequality: \(140x+20,000≤32,000\)

## Best *ISEE Middle Level* *Math* Prep Resource for 2020

*ISEE Middle Level*

*Math*

**61-Choice D is correct**

78 is not prime number, it is divisible by 3 and13.

**62-Choice A is correct**

The width of a rectangle is \(5x\) and its length is \(7x\). Therefore, the perimeter of the rectangle is 24x. Perimeter of a rectangle=2(width+length)\(=2(5x+7x)=2(12x)=24x\)

The perimeter of the rectangle is 108. Then: \(24x=108→x=4.5\)

**63-Choice D is correct**

$18 is what percent of $63? \(18 ÷ 60 = 0.3 = 30\%\)

**64-Choice B is correct**

Let x be one-kilogram orange cost, then: \(2x+(5×5.2)=34→2x+26=34→x=34-26→2x=8→x=\frac{8}{2}=$4\)

**65-Choice D is correct**

\((((-18)+44)×(-4))+80=(26×(-4)+80=(-104)+80=24\)

**66-Choice C is correct**

The area of the square is 81 inches. Therefore, the side of the square is square root of the area. \(\sqrt{81}=9\) inches, Four times the side of the square is the perimeter: \(4 × 9 = 36\) inches

**67-Choice C is correct**

The distance between Jason and Joe is 12 miles. Jason running at 6.4 miles per hour and Joe is running at the speed of 8 miles per hour. Therefore, every hour the distance is 1.6 miles less. \(12 ÷ 1.6 = 7.5\)

**68-Choice A is correct**

The ratio of lions to tigers is 6 to 4 at the zoo. Therefore, total number of lions and tigers must be divisible by 10. \(6+4=10\), From the numbers provided, only 64 is not divisible by 10.

**69-Choice B is correct**

The percent of girls playing tennis is: \(30 \% × 50 \% = 0.3 × 0.5 = 0.15 = 15 \%\)

**70-Choice A is correct**

There are triple as many girls as boys. Let x be the number of girls in the class. Then:

\(x+3x=56→4x=56→x=14\)

**71-Choice D is correct**

Use PEMDAS (order of operation):

\([(-3)×(-12)+4]-(-2)+[3×6]÷6=[36+4]-(-2)+18÷6=40+2+3=45\)

**72-Choice C is correct**

\(1024=4^x →4^x=4^5→x=5\)

**73-Choice A is correct**

Number of rotates in 15 second\(=\frac{360×15}{12}=450\)

**74-Choice C is correct**

\(6x+8=2x-12→4x+8=-12→4x=-20→x=\frac{-20}{4}=-5\)

**75-Choice B is correct**

Use formula of rectangle prism volume.

V = (length) (width) (height) ⇒ 1260 = (18) (14) (height) ⇒ height = \(1260 ÷ 252 = 5\)

**76-Choice B is correct**

Let x be the original price. If the price of the sofa is decreased by \(20\%\) to $380, then: \(80 \% \space\) of\( \space x=380 ⇒ 0.80 x=380 ⇒ x=380÷0.80=475\)

**77-Choice A is correct**

The area of trapezoid is: \((\frac{(10+12)}{2})×8=88\)

**78-Choice D is correct**

Write the equation and solve for B:

\(0.20A=0.40B \), divide both sides by \(0.40\), then you will have \(\frac{0.20}{0.40} A=B\) , therefore: \(B=\frac{1}{2} A\), and B is \(0.5\) times of A or it’s \(50\%\) of A.

**79-Choice B is correct **

\(2x-(12-x)=x+20→2x-12+x=x+20→2x=32→x=16\)

80-**Choice B is correct**

15-11.56=$3.44

81-**Choice D is correct**

\(12.56÷0.08=\frac{\frac{1256}{100}}{\frac{8}{100}}=\frac{1256}{8}=157\)

**82-Choice B is correct **

The probability of choosing a Hearts or 10 is(one of 10 is heart) \(\frac{13+(4-1)}{52}=\frac{16}{52}=\frac{4}{13}\)

**83**–**Choice C is correct**

\(\frac{1}{3}+\frac{\frac{8}{10}}{\frac{6}{12}}=\frac{1}{3}+\frac{812}{106}=\frac{1}{3}+\frac{96}{60}=\frac{20+96}{60}=\frac{116}{60}=1\frac{56}{60}=1\frac{14}{15}\)

**84-Choice A is correct**

\(\frac{8×14}{90}=\frac{112}{90}=1.24≅1.2\)