Did you take the PSAT 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.

## PSAT Math Practice Test Answers and Explanations

**PSAT Math Practice Tests Explanations****PSAT 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, \(\begin{cases}x+2y=3 \\y-x=6 \end{cases}\)

→Add two equations.

\((3y=9→y=3)\) , 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.

14- **The answer is 384**

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 \(= 12 × 32 = 384\)

15- **The answer is 18**

\(\frac{2y}{9}=x-\frac{1}{9} x+4\), 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: \(y-4x=18\)

16- **The answer is 2**

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 Math Practice Tests Explanations**

**PSAT Math 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**

\(Cosβ=\frac{Adjacent \space side}{hypotenuse}\), 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: \(x+12=10→x=-2\)

45- **The answer is 6**

\(x+4y=\frac{-8y^2+12}{2x}\), 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\)

46- **The answer is 14.14**

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\(=(-\frac{1}{6})×(-1)×\frac{57}{6}=\frac{57}{36}=\frac{19}{12}\)