# Full-Length PSAT 10 Math Practice Test-Answers and Explanations

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## PSAT 10 Math Practice Test Answers and Explanations

PSAT 10 Math Practice Tests Explanations
PSAT 10 Math Practice Test 1:
Section 1 – No Calculator

1- Choice D is correct
$$12x-14>10x+5-3x+1$$→Combine like terms: $$12x-14>7x+6$$→ Subtract $$7x$$ from both sides: $$5x-14>6$$, Add 14 both sides of the inequality. $$5x>20$$, Divide both sides by $$5$$.
$$\frac{20}{5}<x→<x>4$$

2- Choice A is correct
$$x^2=169→x=13$$ (positive value) Or $$x=-13$$ (negative value)
Since x is positive, then: $$f(169)=f(13^2 )=10-5(13)=10-65=-55$$

3- Choice D is correct
$$8x^2-5x+8=7x^2+x+15⇒x^2-6x-7=0$$, Find the factors of the quadratic equation.
$$→(x-7)(x+1)=0→x=7 or x=-1, a>b$$, then: $$a=7$$ and $$b=-1$$, $$\frac{a}{b}=\frac{7}{-1}=-7$$

4- Choice D is correct
First, find the equation of the line. All lines through the origin are of the form $$y=mx$$, so the equation is $$y=-\frac{2}{3}x$$. Of the given choices, only choice $$C (6,-4)$$, satisfies this equation:
$$y=-\frac{2}{3}x→-4=-\frac{2}{3}(6)=-4$$

5- Choice D is correct
$$𝑥$$ is the number of all John’s sales per month and $$4\%$$ of it is: $$4\%×x=0.04x$$
John’s monthly revenue: $$0.04x+3,600$$

6- Choice D is correct
The input value is $$-3$$. Then: $$x=-3$$, $$f(x)=3x^2+3x+2=3(-3)^2+3(-3)+2=20$$

7- Choice A is correct
To rewrite $$\frac{1}{\frac{1}{x-6}+\frac{1}{x+2}}$$, first simplify $$\frac{1}{x-6}+\frac{1}{x+2}$$
$$\frac{1}{x-6}+\frac{1}{x+2}=\frac{1(x+2)+1(x-6)}{(x-6)(x+2)}=\frac{2x-4}{(x-6)(x+2)}$$
Then:$$\frac{1}{\frac{1}{x-6}+\frac{1}{x+2}}=\frac{1}{\frac{2x-4}{(x-6)(x+2)}}=\frac{(x-6)(x+2)}{2x-4}$$ (Remember, $$\frac{1}{\frac{1}{x}}=x$$)
This result is equivalent to the expression in choice A.

8- Choice C is correct
Of the $$39$$ employees, there are $$10$$ females under age $$45$$ and $$8$$ males age $$45$$ or older. Therefore, the probability that the person selected will be either a female under age $$45$$ or a male age $$45$$ or older is: $$\frac{10}{39}+\frac{8}{39}=\frac{18}{39}=\frac{6}{13}$$

9- Choice A is correct
Plug in the values of x and y of the point (-2, 8) in the equation of the parabola. Then:
$$8=a(-2)^2+6(-2)+16→8=4a-12+16→8=4a+4$$
$$→4a=8-4=4→a=4/4=1→a^2=(1)^2=1$$

10- Choice A is correct
$$4a=6b→b=\frac{2a}{3} and 4a=10c→c=\frac{2a}{5}$$
$$2a+3b-4c=2a+3(\frac{2a}{3})-4(\frac{2a}{5})=2a+2a-\frac{8a}{5}=\frac{(20-8)a}{5}=\frac{12a}{5}=2.4a$$

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11- Choice C is correct
Two triangles ∆BAE and ∆BCD are similar. Then: $$\frac{AE}{CD}=\frac{AB}{BC}→\frac{10}{16}=\frac{x}{20}→200=16x→x=12.5$$

12- Choice C is correct
$$\begin{cases}\frac{x}{3}+\frac{4y}{6}=1 \\y-x=6 \end{cases}$$→ Multiply the top equation by 3. Then, $$\begin{cases}x+2y=3 \\y-x=6 \end{cases}$$
$$(3y=9→y=3)$$ , plug in the value of (y) into the first equation →$$(x=-3)$$

13- Choice C is correct
To solve this problem, first recall the equation of a line:
Where, and Remember that slope is the rate of change that occurs in a function and that the intercept is the  value corresponding to .
Since the height of John’s plant is 4 inches tall when he gets it. Time (or ) is zero. The plant grows 8 inches per year. Therefore, the rate of change of the plant’s height is 4. The intercept represents the starting height of the plant which is 4 inches.

Let L be the length of the rectangular and W be the with of the rectangular. Then, $$L=2W+8$$
The perimeter of the rectangle is 88 meters. Therefore: $$2L+2W=88, L+W=44$$
Replace the value of L from the first equation into the second equation and solve for W:
$$(2W+8)+W=44→3W+8=44→3W=36→W=12$$
The width of the rectangle is 12 meters and its length is: $$L=2W+8=2(12)+8=32$$
The area of the rectangle is: length × width $$= 12 × 32 = 384$$

$$\frac{2y}{9}=x-\frac{1}{9} x+4$$, Multiply both sides of the equation by $$9$$. Then: $$9×\frac{2y}{9}=9×(x-\frac{1}{9} x+4)→2y=9x-x+36→2y=8x+36→y=4x+18 →y-4x=18$$
Now, subtract 4x from both sides of the equation. Then: $$y-4x=18$$

First, factorize the numerator and simplify.$$\frac{x^2-4}{x+2} +4(x+3)=20 \frac{(x-2)(x+2)}{(x+2)}+4(x+3)=20$$
Divide both sides of the fraction by $$(x+2)$$. Then: $$x-2+4x+12=20→5x+10=20$$
Subtract 10 from both sides of the equation. Then: →$$5x=20-10=10→x=\frac{10}{5}=2$$

17- The answer is $$\frac{4}{3}$$
First, simplify the numerator and the denominator. $$\frac{(6(x)(y^3 ))^2}{x^2 (3y^2)^3}$$
Remove $$x^2 y^6$$ from both numerator and denominator.$$\frac{36x^2 y^6}{27x^2 y^6}=\frac{36}{27}=\frac{4}{3}$$

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#### PSAT 10Math Practice Tests ExplanationsPSAT 10Math Practice Test 2:Section 2 – No Calculator

18- Choice B is correct
Choices A, C and D are incorrect because $$72\%$$ of each of the numbers is a non-whole number.
A. 40 $$72\% of 40 = 0.72×40=28.8$$
B. 50 $$72\% of 50=0.72×50=36$$
C. 55 $$72\% of 55=0.72×55=39.6$$
D. 60  $$72\% of 60=0.72×60=43.2$$
Only choice B gives a whole number.

19- Choice B is correct
$$\frac{5}{6}×30=\frac{150}{6}=25$$

20- Choice C is correct
The slop of line A is: $$m=\frac{y_2-y_1}{x_2-x_1}=\frac{20-4}{6-(-2)}=2$$
Parallel lines have the same slope and only choice $$C (y=x)$$ has slope of 2.

21- Choice C is correct
Substituting 3 for x and 22 for y in $$y = nx+4$$ gives $$22=(n)(3)+4$$,
which gives $$n=6$$. Hence, $$y=6x+4$$. Therefore, when $$x = 4$$, the value of y is: $$y=(6)(4)+4 = 28$$

22- Choice A is correct
The description $$4+4x$$ is $$8$$ more than $$12$$ can be written as the equation $$4+4x=8+12$$, which is equivalent to $$4+4x=20$$. Subtracting $$4$$ from each side of $$4+4x=20$$ gives
$$4x=16$$. Since $$8x$$ is $$2$$ times $$4x$$, multiplying both sides of $$4x=16$$ by $$2$$ gives $$8x=32$$

23- Choice D is correct
Let $$x$$ be equal to $$0.5,-0.5$$, then: $$x=-0.5,0.5$$
$$\sqrt{x^2+1}=\sqrt{(0.5)^2+1}=\sqrt{1.25}≈1.12 . \sqrt{x^2} +1=\sqrt{(0.5)^2}+1=0.5+1=1.5$$
$$\sqrt{x^2+1}=\sqrt{(-0.5)^2+1}=\sqrt{1.25}≈1.12. \sqrt{x^2} +1=\sqrt{-0.5)^2}+1=0.5+1=1.5$$
Then, option D is correct. D. $$\sqrt{x^2 }+1>\sqrt{x^2+1}>x$$

24- Choice B is correct
The smallest number is $$-16$$. 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: $$-100=(-16)+(-15)+(-14)+(-13)+(-12)+(-11)+(-10)+x→ -100=-91+x, →x=-100+91=-9$$

25- Choice C is correct
The capacity of a red box is $$40\%$$ bigger than the capacity of a blue box and it can hold $$35$$ books. Therefore, we want to find a number that $$40\%$$ bigger than that number is $$35$$. Let $$x$$ be that number. Then: $$1.4×x=35$$, Divide both sides of the equation by $$1.4$$. Then: $$x=\frac{35}{1.40}=25$$

26- 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: $$8, 6, 7, 9, 5$$
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 (𝑚𝑒𝑎𝑛) $$=\frac{sum \ of \ terms}{number \space of \space terms}=\frac{8+6+7+9+5}{5}=\frac{35}{5}=7$$
Median is the number in the middle. To find median, first list numbers in order from smallest to largest. $$5, 6, 7, 8, 9,$$ Median of the data is 7. 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, c=a

27- Choice D is correct
Percent of cities in the type of pollution A: $$\frac{8}{10}×100=80\%$$
Percent of cities in the type of pollution C: $$\frac{7}{10}×100=70\%$$
Percent of cities in the type of pollution E: $$\frac{5}{10}×100=50\%$$

28- Choice D is correct
Let x be the number of cities need to be added to type of pollutions E. Then:
$$\frac{x+5}{9}=0.8→x+5=9×0.8→x+5=7.2→x=2.2$$

29- Choice B is correct
Since $$f(x)$$ is linear function with a negative slop, then when $$x=-4,f(x)$$ is maximum and when $$x=5,f(x)$$ is minimum. Then the ratio of the minimum value to the maximum value of the function is: $$\frac{f(5)}{f(-4)}=\frac{-2(5)+2}{-2(-4)+2}=\frac{-8}{10}=\frac{-4}{5}$$

30- Choice B is correct
AB=15 And BC=36, AC=$$\sqrt{15^2+36^2} =\sqrt{225+1296}=\sqrt{1521}=39$$
Perimeter =$$15+36+39=90$$ , Area =$$\frac{36×15}{2}=270$$
In this case, the ratio of the perimeter of the triangle to its area is: $$\frac{90}{270}=\frac{1}{3}$$
If the sides AB and AC become one third shorter, then: AB=5 And AC=13
BC=$$\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$$, Perimeter =$$5+12+13=30$$
Area =\9\frac{12×5}{2}=30\), In this case the ratio of the perimeter of the triangle to its area is: $$\frac{30}{30}=1$$

31- Choice B is correct
The ratio of boy to girls is 5:9. Therefore, there are 5 boys out of 14 students. To find the answer, first divide the total number of students by 14, then multiply the result by 5.
$$84 ÷ 14 = 6 ⇒ 5 × 6 = 30$$, There are 30 boys and $$54 (84 – 30)$$ girls. So, 24 more boys should be enrolled to make the ratio 1:1

32- Choice A is correct
Ratio of women to men in city A: $$\frac{400}{455}=0.88$$
Ratio of women to men in city B: $$\frac{620}{868}=0.71$$
Ratio of women to men in city C: $$\frac{600}{700}=0.86$$
Ratio of women to men in city D: $$\frac{650}{800}=0.81$$
Choice A provides the maximum ratio of women to men in the four cities.

33- Choice A is correct
Percentage of men in city A = $$\frac{455}{855}×100=56.87\%$$
Percentage of women in city C = $$\frac{600}{1300}×100=46.15\%$$
Percentage of men in city A to percentage of women in city C = \$$frac{56.87}{46.15}=1.23$$

34- Choice A is correct
Let the number of women should be added to city D be $$x$$, then:
$$\frac{650+x}{800}=1.4→650+x=800×1.4=1120→x=470$$

35- Choice C is correct
If $$f(x)=-2x+2(2x+3)+1$$ , then find $$f(-3x)$$ by substituting $$4x$$ for every $$x$$ in the function. This gives: $$f(-3x)=-2(-3x)+2(2(-3x)+3)+1=6x-12x+6+1$$ ,
It simplifies to: $$f(-3x)=6x-12x+6+1=-6x+7$$

36- Choice C is correct
The amount of petrol consumed after $$x$$ hours is: $$7.5×x=7.5x$$
Petrol remaining after x hours driving: $$60-7.5x$$

37- Choice A is correct
In the figure angle A is labeled $$(9x-2)$$ and it measures 43. Thus, $$9x-2=43$$ and $$9x=45$$ or $$x=5$$. That means that angle B, which is labeled $$(12x)$$, must measure $$12×5=60$$.
Since the three angles of a triangle must add up to 180, $$43+60+y-12=180$$, then: $$y+91=180→y=180-91=89$$

38- Choice A is correct
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 (𝑚𝑒𝑎𝑛) $$= \frac{sum \space of \space terms}{number \space of \space terms}=\frac{8+12+15+15+8+14}{6}=12$$

39- Choice D is correct
The perimeter of the rectangle is: $$2x+2y=40→x+y=20→x=20-y$$
The area of the rectangle is: $$x×y=96→(20-y)(y)=96→y^2-20y+96=0$$
Solve the quadratic equation by factoring method.
$$(y-8)(y-12)=0→y=8$$ (Unacceptable, because y must be greater than 10) or $$y=12$$
If $$y=12 →x×y=96→x×12=96→x=8$$

40- Choice D is correct
The equation $$\frac{a-b}{2b}=\frac{5}{9}$$ can be rewritten as $$\frac{a}{2b}-\frac{b}{2b}=\frac{5}{9}$$, from which it follows that $$\frac{a}{2b}-\frac{1}{2}=\frac{5}{9}$$, or $$\frac{a}{2b}=\frac{5}{9}+\frac{1}{2}=\frac{19}{18}.$$⇒$$\frac{a}{b}=\frac{19}{9}$$, Subtract 12 from both sides of the equation. Then: $$x+12=10→x=-2$$

41- Choice D is correct
$$Cos⁡β=\frac{Adjacent \space side}{hypotenuse}$$, To find the hypotenuse, we need to use Pythagorean theorem.
$$a^2+b^2=c^2→c=\sqrt{a^2+b^2}, cos (β)=\frac{a}{c}=\frac{a}{\sqrt{a^2+b^2}}, sec(β)=\frac{1}{cos ⁡(β)} =\frac{\sqrt{a^2+b^2 }}{a}$$

42- Choice B is correct
$$|\frac{2x}{3}-4x-6|<4.⇒-4<\frac{2x}{3}-4x-6<4$$.Multiply both side by 3. ⇒ $$-12<2x-12x-18<12$$, add 18 from all sides of the inequality. $$18-12<-10x-18+18<12+18⇒6<-10x<30$$, Divide all sides by $$-10$$. $$\frac{6}{-10}>x>-3$$
(Remember that when you divide all sides of an inequality by a negative number, the inequality sing will be swapped. < becomes >)

43- Choice D is correct
$$x$$ is directly proportional to the square of $$y$$. Then: $$x=cy^2 18=c(3)^2→18=9c→c=18/9=2$$, The relationship between $$x$$ and $$y$$ is: $$x=2y^2 x=288, 288=2y^2→y^2=288/2=144→y=12$$

44- Choice B is correct
$$\begin{cases}2x+4y=6 \\x+3y=-20 \end{cases}$$
Multiply the second equation by $$-2$$ then,
$$\begin{cases}2x+4y=6 \\-2x-6y=40 \end{cases}$$
Add two equations , $$-2y=46→y=-23$$ , plug in the value of y into the second equation, $$x+3y=-20→x+4(-23)=-20→x=49$$
Subtract 12 from both sides of the equation. Then: $$x+12=10→x=-2$$

$$x+4y=\frac{-8y^2+12}{2x}$$, Multiply both sides by $$2x$$.
$$2x×(x+4y)=2x×((\frac{-8y^2+12}{2x})→2x^2+8xy=-8y^2+12 →2x^2+8xy+8y^2=12→2×(x^2+4xy+4y^2 )=12→x^2+4xy+4y^2=6 x^2+4xy+4y^2=(x+2y)^2, Then: →(x+2y)^2=6$$

The relationship among all sides of special right triangle
$$45^\circ-45^\circ- 90^\circ$$ is provided in this triangle: According to picture length of ladder is $$\sqrt{2} x⇒L=10\sqrt{2}=14.14$$

47- The answer is $$\frac{3}{2}$$
Let x be the length of an edge of cube, then the volume of a cube is: $$V=x^3$$
The surface area of cube is: $$SA=6x^2$$, The volume of cube A is $$\frac{1}{4}$$ of its surface area. Then:
$$x^3=\frac{6x^2)}{4}→x^3=\frac{3}{2}x^2$$, divide both side of the equation by $$x^2$$. Then: $$\frac{x^3}{x^2} =\frac{3x^2}{2x^2} →x=\frac{3}{2}$$

48- The answer is $$\frac{19}{12}$$
The intersection of two functions is the point with 3 for $$x$$. Then:
$$f(3)=g(3) and g(3)=(3+2)=5$$,
Then, $$f(2)=5→a(3)^2+b(3)+c=5.⇒9a+3b+c=5$$ (i)
The value of x in the vertex of the parabola is: $$x=-\frac{b}{2a}→-3=-\frac{b}{2a}→b=6a$$ (ii)
In the point $$(-3, 11)$$, the value of the $$f(x)$$ is 11.
$$f(-3)=11→a(-3)^2+b(-3)+c=11→9a-3b+c=11$$ (iii)
Using the first two equation: $$\begin{cases}9a+3b+c=5 \\ 9a-3b+c=11 \end{cases}$$
Equation 1 minus equation 2 is: (i)-(iii) →$$6b=-6→b=-1$$ (iv)
Plug in the value of b in the second equation: \9b=6a →a=\frac{b}{6}=-\frac{1}{6}\)
Plug in the values of a and be in the first equation. Then:
$$9(-\frac{1}{6})+3(-1)+c=5⇒c=5+3+\frac{9}{6}=\frac{57}{6}=9 \frac{1}{2}$$
The product of a, b and c$$=(-\frac{1}{6})×(-1)×\frac{57}{6}=\frac{57}{36}=\frac{19}{12}$$

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