Full-Length 8th Grade ACT Aspire Math Practice Test-Answers and Explanations

20- The answer is 20.
Five years ago, Amy was three times as old as Mike. Mike is 10 years now. Therefore, 5 years ago Mike was 5 years.
Five years ago, Amy was: \(A=3×5=15\)
Now Amy is 20 years old: \(15 + 5 = 20\)

21- Choice D is correct
\(\begin{cases}\frac{-x}{2}+\frac{y}{4}=1\\\frac{-5y}{6}+2x=4\end{cases}\)
Multiply the top equation by 4. Then,
\(\begin{cases}-2x+y=4\\\frac{-5y}{6}+2x=4\end{cases}\)
→ Add two equations.
\(\frac{1}{6}y=8→y=48\), plug in the value of y into the first equation →\(x=22\)

22- The answer is 4.8.
Two triangles ∆BAE and ∆BCD are similar. Then:
\(\frac{AE}{CD}=\frac{AB}{BC}→\frac{4}{6}=\frac{x}{12}→48-4x=6x→10x=48→x=4.8\)

23- Choice D is correct
\(\frac{2}{5}×25=\frac{50}{5}=10\)

24- Choice D is correct
The slop of line A is: \(m=\frac{y_2-y_1}{x_2-x_1}=\frac{3-2}{4-3}=1\)
Parallel lines have the same slope and only choice D \((y=x)\) has slope of 1.

25- The answer is 5.
\(x\) is directly proportional to the square of \(y\). Then:
\(x=cy^2\)
\(12=c(2)^2→12=4c→c=\frac{12}{4}=3\)
The relationship between \(x\) and \(y\) is: \(x=3y^2, x=75\), \(75=3y^2→y^2=\frac{75}{3}=25→y=5\)

26- The answer is 54.
The amount of money that jack earns for one hour: \(\frac{$616}{44}=$14\)
Number of additional hours that he work to make enough money is: \(\frac{$826-$616}{1.5×$14}=10\)
Number of total hours is: \(44+10=54\)

27- Choice C is correct
Let’s find the mean (average), mode and median of the number of cities for each type of pollution.
Number of cities for each type of pollution: 6, 3, 4, 9, 8
??????? (????) \(=\frac{sum \space of \space terms}{number \space of \space terms}= \frac{6+3+4+9+8}{5}=\frac{30}{5}=6\)
Median is the number in the middle. To find median, first list numbers in order from smallest to largest.
3, 4, 6, 8, 9
Median of the data is 6.
Mode is the number which appears most often in a set of numbers. Therefore, there is no mode in the set of numbers.
Median = Mean, then,? =?

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28- Choice A is correct
Percent of cities in the type of pollution A: \(\frac{6}{10}×100=60\%\)
Percent of cities in the type of pollution C: \(\frac{4}{10}0×100=40\%\)
Percent of cities in the type of pollution E: \(\frac{9}{10}×100=90\%\)

29- The answer is 2.
Let the number of cities should be added to type of pollution’s B be \(x\). Then: \(\frac{x+3}{8}=0.625→x+3=8×0.625→x+3=5→x=2\)

30- Choice A is correct
AB = 12 and AC = 5
\(BC=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\)
Perimeter \(=5+12+13=30\)
Area \(=\frac{5×12}{2}=5×6=30\)
In this case, the ratio of the perimeter of the triangle to its area is: \(\frac{30}{30}=1\)
If the sides AB and AC become twice longer, then:
AB = 24 and AC = 10
BC \(=\sqrt{24^2+10^2}=\sqrt{576+100}=\sqrt{676}=26\)
Perimeter \(=26+24+10=60\)
Area \(=\frac{10×24}{2}=10×12=120\)
In this case the ratio of the perimeter of the triangle to its area is: \(\frac{60}{120}=\frac{1}{2}\)

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31- The answer is 25.
The capacity of a red box is \(20\%\) bigger than the capacity of a blue box and it can hold 30 books. Therefore, we want to find a number that \(20\%\) bigger than that number is 30. Let \(x\) be that number. Then:
\(1.20×x=30\), Divide both sides of the equation by 1.2. Then:
\(x=\frac{30}{1.20}=25\)

32- Choice C is correct
The smallest number is \(-15\). To find the largest possible value of one of the other five integers, we need to choose the smallest possible integers for four of them. Let x be the largest number. Then:
\(-70=(-15)+(-14)+(-13)+(-12)+(-11)+x→-70=-65+x →x=-70+65=-5\)

33- The answer is 67.
\(α=180\circ -112\circ=68\circ\)
\(β=180\circ-135\circ=45\circ\)
\(x+α+β=180\circ→x=180\circ-68\circ-45\circ=67\circ\)

34- Choice D is correct
A. \(f(x)=x^2-5\) →if \(x=1→f(1)=(1)^2-5=1-5=-4≠5\)
B. \(f(x)=x^2-1\) →if \(x=1→f(1)=(1)^2-1=1-1=0≠5\)
C. \(f(x)=\sqrt{x+2}\)→ if \(x=1→f(1)=\sqrt{1+2}=\sqrt{3}≠5\)
D. f(x)=\sqrt{x}+4 \) →if \(x=1→f(1)=\sqrt{1}+4=5\)

35- The answer is $810.
Let \(x\) be all expenses, then \(\frac{22}{100} x=$660 →x=\frac{100×$660}{22}=$3000\)
He spent for his rent: \(\frac{27}{100}×$3000=$810\)

36- Choice C is correct
The amount of money for \(x\) bookshelf is: \(100x\)
Then, the total cost of all bookshelves is equal to: \(100x+800\)
The total cost, in dollar, per bookshelf is: \(\frac{Total \space cost}{number \space of \space items}=\frac{100x+800}{x}\)

37- The answer is 0.
\(\sqrt{x}=4→x=16\)
then; \(\sqrt{x}-7=\sqrt{16}-7=4-7=-3\) and \(\sqrt{x-7}=\sqrt{16-7}=\sqrt{9}=3\)
Then: \((\sqrt{x-7})+(\sqrt{x}-7)=3+(-3)=0\)

38- Choice B is correct
The angles on a straight line add up to 180 degrees. Then: \(x+25+y+2x+y=180\)
Then, \(3x+2y=180-25→3(35)+2y=155\)
\(→2y=155-105=50→y=25\)

39- Choice C is correct
Square root of 16 is \(\sqrt{16}=4<6\) Square root of 25 is \(\sqrt{25}=5<6 \) Square root of 37 is \(\sqrt{37}=\sqrt{36+1}>\sqrt{36}=6\)
Square root of 49 is \(\sqrt{49}=7>6\)
Since, \(\sqrt{37}<\sqrt{49}\), then the answer is C.

40- The answer is 11.
\(|-12-5|-|-8+2|=|-17|-|-6|=17-6=11\)

41- Choice C is correct
A probability is the likelihood of a successful event occurring divided by the total number of events possible. In this case, a successful event is selecting either a red or a yellow marble and the total number of events possible is the total number of marbles. Combine the number of red and yellow marbles: 8 + 5 = 13, and divide this by the total number of marbles: 6 + 8 + 5 = 19. The probability is 13 out of 19.

42- Choice C is correct
A factor must divide evenly into its multiple. 14 cannot be a factor of 80 because 80 divided by 16 = 5.71

43- The answer is 16.
The total cost of the phone call can be represented by the equation:
\(TC = $4.00 + $0.4x\), where \(x\) is the duration of the call after the first five minutes. In this case, \(x = 30\). Substitute the known values into the equation and solve: \(TC = $4.00 + $0.4 × 30\)
\(TC = $4.00 + $12.00, TC = $16.00\)

44- Choice C is correct
Find the difference of the two expression: \((5x+8)-(5x-3)=5x+8-5x+3=11\)

45- The answer is 45.
First, find the number. Let \(x\) be the number. Write the equation and solve for \(x\). \(150\%\) of a number is 75, then:
\(1.5×x=75→x=75÷1.5=50, 90\%\) of 50 is: \(0.9×50=45\)

46- Choice A is correct
The length of MN is equal to: \(3x+5x=8x, Then: 8x=40→x=\frac{40}{8}=5\)
The length of ON is equal to: \(5x=5×5=25\) cm

47- Choice A is correct
The general slope-intercept form of the equation of a line is \(y = mx + b\), where \(m\) is the slope and \(b\) is the \(y\)-intercept. By substitution of the given point and given slope, we have:
\(-2 = (2)(7) + b\), So, \(b = –2 – 14 = -6\), and the required equation is \(y = 2x – 16\).

48- Choice D is correct
Multiplying each side of \(-3x-y=6\) by \(2\) gives \(-6x -2y = 12\). Adding each side of \(-6x – 2y = 12\) to the corresponding side of \(6x+4y=10\) gives \(2y=22\) or \(y=11\). Finally, substituting \(11\) for \(y\) in \(6x+4y=10\) gives \(6x+4(11)=10\) or \(x=-\frac{17}{3}\).

49- Choice B is correct
\(x_{1,2} =\frac{-b ± \sqrt{b^2-4ac}}{2a}\),
\( ax^2 + bx + c = 0, 4x^2 + 14x + 6 = 0 \) ⇒ then: \(a = 4\), \(b = 14\) and \(c = 6\),
\(x = \frac{-14 +\sqrt{14^2 – 4.4.6}}{2.4} = – \frac{1}{2}, x =\frac{-14 – \sqrt{14^2 – 4.4.6}}{2.4} = – 3\)

50- Choice A is correct
Use Pythagorean theorem: \(a^2+b^2=c^2→s^2+h^2=(5s)^2→s^2+h^2=25s^2\)
Subtracting s^2 from both sides gives: \(h^2=24s^2\)
Square roots of both sides: \(h=\sqrt{24s^2}=\sqrt{4×6×s^2} =\sqrt{4}×\sqrt{6}×\sqrt{s^2 }=2×s×\sqrt{6}=2s\sqrt{6}\)

51- Choice D is correct
Let \(x\) be the number of purple marbles. Let’s review the choices provided:
A. \(\frac{1}{10}\), if the probability of choosing a purple marble is one out of ten, then:
Probability \(=\frac{number \space of \space desired \space outcomes}{number \space of \space total \space outcomes}=\frac{x}{20+30+40+x}=\frac{1}{10}\)
Use cross multiplication and solve for \(x\).
\(10x=90+x→9x=90→x=9\)
Since, number of purple marbles can be 9, then, choice be the probability of randomly selecting a purple marble from the bag.
Use same method for other choices.
B. \(\frac{1}{4}\)
\(\frac{x}{20+30+40+x}=\frac{1}{4}→4x=90+x→3x=90→x=30\)
C. \(\frac{2}{5}\)
\(\frac{x}{20+30+40+x}=\frac{2}{5}→5x=180+2x→3x=180→x=60\)
D. \(\frac{7}{15}\)
\(\frac{x}{20+30+40+x}=\frac{7}{15}→15x=630+7x→8x=630→x=78.75\)
Number of purple marbles cannot be a decimal.

52- Choice C is correct
Area = w × h
Area = 138 × 83 = 11,454

53- Choice D is correct
\(4x^2y^3 + 5x^3y^5 – (5x^2y^3 – 2x^3y^5) = 4x^2y^3 + 5x^3y^5 – 5x^2y^3 + 2x^3y^5 = – x^2y^3 + 7x^3y^5\)

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