# 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: $$x=\frac{45}{1.50}=30$$ ## Best SSAT Upper Level Math Prep Resource for 2022 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 $$\frac{4}{6}×36=\frac{144}{6}=24$$ ## The Best Books to Ace the SSAT Upper Level MathTest ## Related to This Article ### More math articles ### What people say about "Full-Length SSAT Upper Level Practice Test-Answers and Explanations - Effortless Math: We Help Students Learn to LOVE Mathematics"? No one replied yet. X 52% OFF Limited time only! Save Over 52% SAVE$40

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