Did you take the SSAT Upper 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.

## SSAT Upper Level Mathematical Reasoning Practice Test Answers and Explanations

1- **Choice E is correct**

The area of the floor is: 9 cm \(×\) 36 cm \(=\) 324 cm\(^2\), The number of tiles needed \(= 324 ÷ 6 = 54\)

2- **Choice E is correct**

Number of packs equal to: \(31/4≅7.75\), Therefore, the school must purchase 7 packs.

3- **Choice D is correct**

\(\frac{35}{A}+2=7→\frac{35}{A}=7-2=5, →35=5A→A=\frac{35}{5}=7, 35+A=35+7=42\)

4- **Choice C is correct**

Number of rotates in 12 second equals to: \(\frac{400×15}{6}=1125\)

5- **Choice E is correct**

A. Number of books sold in April is: 610

Number of books sold in July is: 845→ \(\frac{610}{845}≅0.721\)

B. number of books sold in July is: 845

Half the number of books sold in May is: \(\frac{1560}{2}=780→780<845 \)

C. number of books sold in June is: 210

Half the number of books sold in April is: \(\frac{610}{2}=305→210<305\)

D. \(610+210=820<845 \)

E. \(610<845\)

6- **Choice D is correct**

The amount of money that Jack earns for one hour: \(\frac{$756}{54}=$14\)

Number of additional hours that he needs to work in order to make enough money is:

\(\frac{$924-$756}{1.5×$14}=8\)

Number of total hours is: \(54+8=62\)

7- **Choice C is correct**

The digit in tens place is 8. The digit in the thousandths place is 3. Therefore; \(8+3=11\)

8- **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 80, then: \(1.6×x=80→x=80÷1.6=50, 70 \%\) of 50 is: \(0.7×50=35\)

9- **Choice E is correct**

\(18.236÷0.004=\frac{\frac{18,236}{1,000}}{\frac{4}{1,000}}=\frac{18,236}{4}=4559\)

10- **Choice E is correct**

Average \(=\frac{sum of terms }{number of terms}\)

The sum of the weight of all girls is: \(12 × 55 = 660\) kg

The sum of the weight of all boys is: \(21 × 59 = 1,239\) kg

The sum of the weight of all students is: \(660 + 1,239 = 1,899\) kg

The average weight of the 50 students: \(\frac{1899}{33}=57.54\)

11- **Choice D is correct**

\(3≤x<6→\) Multiply all sides of the inequality by 3. Then:

\(3×3≤3×x<3×6→9≤3x<18\)

Add 5 to all sides. Then: \(→9+5≤3x+5<18+5→ 14≤3x+5<23 \)

Minimum value of \(3x+5 is 14\)

12- **Choice E is correct**

\(6∎13=\sqrt{6^2+13}= \sqrt{36+13}=\sqrt{49}=7\)

13- **Choice C is correct**

\(\frac{2 \frac{2}{3}+\frac{1}{6}}{3 \frac{3}{4}-\frac{11}{4}}=\frac{\frac{8}{3}+\frac{1}{6}}{\frac{15}{4}-\frac{11}{4}}=\frac{\frac{16+1}{6}}{\frac{15-11}{4}}=\frac{\frac{17}{6}}{\frac{4}{4}}=\frac{17×4}{6×4}=\frac{17}{6}≅2.83\)

14- **Choice E is correct**

Let \(x\) be the capacity of one tank. Then, \(\frac{1}{8} x=350→x=\frac{350×8}{1}=2800\) Liters

The amount of water in two tanks is equal to: \(2×2800=5600\) Liters

15- **Choice E is correct**

Let’s review the choices provided.

A. 4. In 4 years, David will be 62 and Ava will be 14. 62 is not 3 times 14.

B. 6. In 6 years, David will be 64 and Ava will be 16. 62 is not 3 times 16!

C. 8. In 8 years, David will be 66 and Ava will be 18. 66 is not 3 times 18.

D. 10. In 10 years, David will be 68 and Ava will be 20. 68 is not 3 times 20.

E. 14. In 14 years, David will be 72 and Ava will be 24. 72 is 3 times 24.

16- **Choice B is correct**

Let \(x\) be the cost of one-kilogram orange, then:

\(4x+(4×4.6)=26.4→3x+18.4=26.4→4x=26.4-18.4→4x=8→x=\frac{8}{4}=$2\)

17- **Choice B is correct**

All angles in a triangle sum up to 180 degrees. Then: \(x=40+105=145\)

18- **Choice A is correct**

\(9.5÷0.25=\frac{9.5}{0.25}=\frac{\frac{95}{10}}{\frac{25}{100}}=\frac{95×100}{25×10}=\frac{95}{25}×\frac{100}{10}=3.8×10=38\)

19- **Choice D is correct**

Let b be the amount of time Alec can do the job, then,

\(\frac{1}{a}+\frac{1}{b}=\frac{1}{140}→\frac{1}{420}+\frac{1}{b}=\frac{1}{140}→\frac{1}{b}=\frac{1}{140}-\frac{1}{420}=\frac{2}{420}=\frac{1}{210}\),

Then: \(b=210\) minutes

20- **Choice E is correct**

Choices A,B, C and D are incorrect because \(60\%\) of each of the numbers is non-whole number.

A. \(11, 60\%\) of \(11 = 0.60×11=6.6 \)

B. \(16, 60\%\) of \(16=0.60×16=9.6 \)

C. \(19, 60\%\) of \(19=0.60×19=11.4 \)

D. \(23, 60\%\) of\( 23=0.60×23=13.8 \)

E. \(35, 60\%\) of \(35=0.60×35=21\)

## The Absolute Best Book** to Ace the SSAT Upper Level Math** Test

21- **Choice B is correct**

The equation of a line in slope intercept form is: \(y=mx+b\)

Solve for \(y. 8x-4y=24 ⇒ -4y=24-8x ⇒ y=(24-8x)÷(-4) ⇒ y=2x-6\), The slope is 2. The slope of the line perpendicular to this line is:

\(m_1 × m_2 = -1 ⇒ 2 × m_2 = -1 ⇒ m_2 = -\frac{1}{2}\)

22- **Choice D is correct**

Use the information provided in the question to draw the shape.

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

\(7^2+ 24^2 = c^2⇒\)\(49+ 576 =\)\(c^2 ⇒\)\( 625 = c^2 ⇒ c = 25\)

23- **Choice D is correct**

The amount of money for \(x\) bookshelf is: \(200x\)

Then, the total cost of all bookshelves is equal to: \(200x+600\)

The total cost, in dollar, per bookshelf is: \(\frac{Total cost}{number of items}=\frac{200x+600}{x}\)

24- **Choice B is correct**

The smallest number is \(-9\). To find the largest possible value of one of the other three integers, we need to choose the smallest possible integers for two of them. Let x be the largest number. Then: \(-40=(-9)+(-8)+(-7)+x→-40=-24+x, →x=-40+24=-16\)

25- **Choice D is correct**

If the length of the box is 36, then the width of the box is one third of it, 12, and the height of the box is 4 (one third of the width). The volume of the box is:

V = (length)(width)(height) = (36) (12) (4) = 1,008

26- **Choice C is correct**

\(\frac{87,436}{287}≅304.655≅305\)

27- **Choice D is correct**

Let’s review the choices provided:

A. \(x=2→\) The perimeter of the figure is: \(3+6+3+2+2=16≠30 \)

B. \( x=3→\) The perimeter of the figure is:\( 3+6+3+3+3=18≠30 \)

C. \(x=6→\) The perimeter of the figure is: \(3+6+3+6+6=24≠30 \)

D. \(x=9→\) The perimeter of the figure is: \(3+6+3+9+9=30=30 \)

E. \(x=12→\) The perimeter of the figure is: \(3+6+3+12+12=36≠30\)

28- **Choice C is correct**

\(\frac{14+22}{2}=\frac{36}{2}=18\) then \(18-14=4\)

29- **Choice E is correct**

Alex’s mark is k less than Jason’s mark. Then, from the choices provided Alex’s mark can only be \(20-k\).

30- **Choice E is correct**

Let’s review the options provided:

A. \(16×\frac{1}{2}=\frac{16}{2}=8=8 \)

B. \(40×\frac{1}{5}=\frac{40}{5}=8=8 \)

C. \(2×\frac{8}{2}=\frac{16}{2}=8=8 \)

D. \(6×\frac{8}{6}=\frac{48}{6}=8=8 \)

E. \(8×\frac{1}{8}=\frac{8}{8}=1≠8\)

31- **Choice E is correct**

Let \(x\) be the original price. If the price of the sofa is decreased by \(25\%\) to $420, then: \(50 \%\) of \(x=405 ⇒ 0.5x=405 ⇒ x=405÷0.5=810\)

32- **Choice E is correct**

\(860-6 \frac{8}{13}=(859-6)+(\frac{13}{13}-\frac{8}{13})=853 \frac{5}{13}\)

33- **Choice E is correct**

Number of times that the driver rests \(=\frac{32}{8}=4\)

Driver’s rest time = 1 hour and 5 minutes = 65 minutes

Then, \(4×65\) minutes = 260 minutes,

1 hour = 60 minutes → 260 minutes = 4 hour and 20 minutes

34- **Choice A is correct**

The Area that one liter of paint is required: 96cm \(×\) 100cm = 9,600cm\(^2\)

Remember: 1 m\(^2\) = 10,000 cm\(^2\) (\(100 × 100 = 10,000\)), then, 9,600cm\(^2\) = 0.96 m\(^2 \)

Number of liters of paint we need: \(\frac{48}{0.96}=50\) liters

35- **Choice B is correct**

Find the difference of each pairs of numbers: 3, 5, 9, 17, 33, ___, 129

The difference of 3 and 5 is 2, 5 and 9 is 4, 9 and 17 is 8, 17and 33 is 16, and next number should be 65. The number is \(33 + 32 = 65\)

36- **Choice C is correct**

Number of Mathematics book: \(0.3×960=288 \)

Number of English books: \(0.15×9600=144\)

Product of number of Mathematics and number of English books: \(288×144=41,472\)

37- **Choice D is correct**

The angle α is: \(0.3×360=108^\circ\) , The angle \(β\) is: \(0.15×360=54^\circ\)

38- **Choice D is correct**

\(x+1=1+1+1→x=2 \)

\(y+8+3=7+5→y+11=12→y=1\)

Then, the perimeter is:

\(1+7+1+5+1+3+1+8+2+1=30\)

39- **Choice E is correct**

Let put some values for a and b. If \(a=12\) and \(b=3 →a×b=36→\frac{36}{4}=9→36\) is divisible by 4 then;

A. \(a+b=12+3=15\) is not divisible by 4

B. \(3a-b=(3×12)-3=36-3=33\) is not divisible by 4

If \(a=16\) and\( b=2 →a×b=32 →\frac{32}{4}=8\)

32 is divisible by 4 then;

C. \(a-3b=16-(3×2)=16-6=10\) is not divisible by 4

D. \(\frac{a}{3.5b}=\frac{16}{7}\) is not divisible by 4

E. \(4 ×16×2=128\)

128 is divisible by 4.

If you try any other numbers for a and b, you will get the same result.

40- **Choice E is correct**

The capacity of a red box is \(50\%\) bigger than the capacity of a blue box and it can hold 45 books. Therefore, we want to find a number that \(30\%\) bigger than that number is 60. Let \(x\) be that number. Then:

\(1.50×x=45\), Divide both sides of the equation by 1. 5. Then:

\(x=\frac{45}{1.50}=30\)

## Best *SSAT Upper Level* *Math *Prep Resource for 2021

41- **Choice C is correct**

Since, E is the midpoint of AB, then the area of all triangles DAE, DEF, CFE and CBE are equal.

Let \(x\) be the area of one of the triangle, then: \(4x=200→x=50\)

The area of DEC \(=2x=2(50)=100\)

42- **Choice D is correct**

Amount of available petrol in tank: \(48.3-7.96-18.21+16.47=38.6\) liters

43- **Choice E is correct**

We have two equations and three unknown variables, therefore \(x\) cannot be obtained.

44- **Choice A is correct**

\(5y+5<51→5y<51-6→5y<45→y<9\)

The only choice that is less than 8 is E.

45- **Choice C is correct**

Perimeter of figure A is: \(2πr=2π \frac{14}{2}=14π=14×3=42 \)

Area of figure B is:\( 15×7=105\), Average \(=\frac{42+105}{2}=\frac{150}{2}=75\)

46-** Choice C is correct**

The angles on a straight line add up to 180 degrees. Let’s review the choices provided:

A. \(y=15→ x+25+y+2x+y=25+25+15+2(25)+15=130≠180\)

B. \(y=25→ x+25+y+2x+y=25+25+25+2(25)+25=150≠180 \)

C. \(y=40→ x+25+y+2x+y=25+25+40+2(25)+40=180=180 \)

D. \(y=45→ x+25+y+2x+y=25+25+45+2(25)+45=190≠180 \)

E. \(y=55→ x+25+y+2x+y=25+25+55+2(25)+55=210≠180\)

47-** Choice E is correct**

Let’s review the choices provided:

A. \(x=\frac{1}{2}→ \frac{2}{7}+\frac{1}{2}=\frac{4+7}{14}=\frac{11}{14}≅0.78<3 \)

B. \(x=\frac{3}{5}→ \frac{2}{7}+\frac{3}{5}=\frac{10+21}{35}=\frac{31}{35}≅0.88<3 \)

C. \(x=\frac{4}{5}→ \frac{2}{7}+\frac{4}{5}=\frac{10+28}{35}=\frac{38}{35}≅1.085<3 \)

D. \(x=\frac{4}{3}→ \frac{2}{7}+\frac{4}{3}=\frac{6+28}{21}=\frac{34}{21}≅1.61<3 \)

E. \(x=\frac{10}{3}→ \frac{2}{7}+\frac{10}{3}=\frac{6+70}{21}=\frac{76}{21}≅3.61>3 \)

Only choice E be is correct.

48- **Choice B is correct**

Set of numbers that are not composite between 8 and 17: A= {11, 13, 17}

Probability = \(\frac{number of desired outcomes}{number of total outcomes}=\frac{ 3}{10}\)

49- **Choice C is correct**

\(432÷4=\frac{432}{4}=\frac{400+30+2}{4}=\frac{400}{4}+\frac{30}{4}+\frac{2}{4}\)

50- **Choice E is correct**

\(\frac{4}{6}×36=\frac{144}{6}=24\)