8th Grade SBAC Math Practice Test Questions

A. \(a<b<c\)

B. \(b<a<c\)

C. \(a=c\)

D. \(b<c=a\)

8-

What percent of cities are in the type of pollution \(A, C,\) and \(D\) respectively?

A. \(60\%, 40\%, 90\%\)

B. \(30\%, 40\%, 90\%\)

C. \(30\%, 40\%, 60\%\)

D. \(40\%, 60\%, 90\%\)

9-

How many cities should be added to the type of pollution \(B\) until the ratio of cities in the type of pollution \(B\) to cities in the type of pollution \(E\) will be 0.625?

A. 2

B. 3

C. 4

D. 5

10- In the following right triangle, if the sides \(AB\) and \(AC\) become twice longer, what will be the ratio of the perimeter of the triangle to its area?

A. \(\frac{1}{2}\)

B. 2

C. \(\frac{1}{3}\)

D. 3

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Answers:

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

3- D
\(\frac{2}{5}×25=\frac{50}{5}=10\)

4- D
The slope 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 a slope of 1.

5- C
\(x\) 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:
\(x=3y^2\)
\(x=75\)
\(75=3y^2→y^2=\frac{75}{3}=25→y=5\)

6- D
The amount of money that Jack earns for one hour: \(\frac{$616}{44}=$14\)
The number of additional hours that he works to make enough money is: \(\frac{$826-$616}{1.5×$14}=10\)
The number of total hours is: \(44+10=54\)

7- C
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: \(6, 3, 4, 9, 8\)
average\(= \frac{sum \space of \space terms}{number \space of \space terms}=\frac{6+3+4+9+8}{5}=6\)
Median is the number in the middle. To find the median, first list the numbers in order from smallest to largest.
\(3, 4, 6, 8, 9\)
The median of the data is 6.
Mode is the number that appears most often in a set of numbers. Therefore, there is no mode in the set of numbers.
\(Median = Mean, then, a=b\)

8- A
Percent of cities in the type of pollution A: \(\frac{6}{10} × 100=60\%\)
Percent of cities in the type of pollution C: \( \frac{4}{10} × 100 = 40\%\)
Percent of cities in the type of pollution D: \( \frac{9}{10}× 100 = 90\%\)

9- A
Let the number of cities be added to the type of pollution \(B\) be \(x\). Then:
\(\frac{x + 3}{8}=0.625→x+3=8×0.625→x+3=5→x=2\)

10- A
\(AB=12\) 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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