# Full-Length SSAT Upper Level Practice Test-Answers and Explanations

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**

The number of packs is 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

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

B. the 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: \(\frac{$756}{54}=$14\)

Several additional hours that he needs to work to make enough money is:

\(\frac{$924-$756}{1.5×$14}=8\)

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

7- **Choice C is correct**

The digit in tens place is 8. The digit in the thousandth 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 \(35=0.60×35=21\)

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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 pair 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 the 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 \(30\%\) is 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:

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

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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 triangles than \(4x=200→x=50\)

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**

\(\frac{4}{6}×36=\frac{144}{6}=24\)

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