PSAT 10 Math Practice: Answers to 48 Legacy Questions

$PSAT 10 Math Practice: Answers to 48 Legacy Questions$

Answers to the corrected legacy practice questions

These explanations match the 48-question legacy skill-practice set. It is not an official digital PSAT 10 simulation. Do not convert the raw number correct into a PSAT score.

Return to the questions and current official-practice options.

Question 1: D

Combine like terms to get 12 x-14>7 x+6. Subtract 7 x and add 14 to obtain 5 x>20, so x>4.

Question 2: A

Since x is positive and x²=169, use x=13. Then f (169)=10-5(13)=-55.

Question 3: D

Rearrange to x²-6 x-7=0, which factors as (x-7)(x+1)=0. Thus a=7 and b=-1, giving a/b=-7.

Question 4: C

The line is y=-2 x/3. Substituting x=6 gives y=-4, so (6, -4) lies on the line.

Question 5: D

Four percent of x dollars is 0.04 x dollars. Add the fixed salary to get 0.04 x+3600.

Question 6: D

Substitute -3: f (-3)=3(9)+3(-3)+2=27-9+2=20.

Question 7: A

The inner sum is (2 x-4)/[(x-6)(x+2)]. Taking its reciprocal gives (x-6)(x+2)/(2 x-4). The stated exclusions keep both the original fractions and their reciprocal defined.

Question 8: C

The two disjoint groups contain 10+8=18 employees out of 39. The probability is 18/39=6/13.

Question 9: A

Substitute (-2, 8): 8=4 a-12+16=4 a+4. Hence a=1 and a²=1.

Question 10: A

From 4 a=6 b=10 c, b=2 a/3 and c=2 a/5. Then 2 a+3 b-4 c=(12/5) a=2.4 a.

Question 11: C

Triangles EAB and DCB are similar by their right angles and equal angles at B. Therefore AB/20=10/16, giving AB=12.5.

Question 12: C

Multiplying the first equation by 3 gives x+2 y=3. Add y-x=6 to obtain 3 y=9. Thus y=3 and x=-3.

Question 13: B

At the moment of purchase, time is 0 and the height is 4 inches. That starting height is the y-intercept. 8 inches per year is the slope.

Question 14: 384 square meters

Let the width be w. Then the length is 2 w+8 and the half-perimeter gives 3 w+8=44. Thus w=12 and the length is 32, so the area is 384 square meters.

Question 15: 18

Multiply by 9 to obtain 2 y=8 x+36. Dividing by 2 gives y=4 x+18, so y-4 x=18.

Question 16: 2

For x≠-2, the fraction simplifies to x-2. The equation becomes x-2+4 x+12=20, so 5 x=10 and x=2.

Question 17: $\frac43$

The numerator is 36 x² y⁶ and the denominator is 27 x² y⁶. Cancel the nonzero common factors to obtain 36/27=4/3.

Question 18: B

A whole number of wins requires 0.72 times the game count to be an integer. For 50 games this is 36 wins. 40, 55 and 60 give noninteger counts.

Question 19: B

Multiply the tank capacity by the filled fraction: 30(5/6)=25 gallons.

Question 20: C

The slope is (20-4)/(6-(-2))=16/8=2. A parallel line has slope 2, which gives y=2 x among the choices.

Question 21: C

Use 22=3 n+4 to find n=6. At x=4, y=6(4)+4=28.

Question 22: A

The statement gives 4+4 x=12+8=20. Thus 4 x=16, so 8 x=32.

Question 23: D

Let t=|x|, so 0<t<1. Then (t+1)²=t²+2 t+1>t²+1, giving t+1>sqrt (t²+1). Also sqrt (x²+1)>1>x. These establish the inequality in D for either sign of x.

Question 24: B

To maximize the last integer, make the other seven as small as allowed: -16, -15, -14, -13, -12, -11, -10. Their sum is -91, leaving -9. This is distinct from the other seven and achieves the required sum.

Question 25: C

If the blue box holds b books, the red box holds 1.4 b. Solve 1.4 b=35 to get b=25.

Question 26: C

The mean is (8+6+7+9+5)/5=7. The sorted list is 5, 6, 7, 8, 9, whose middle value is 7. Thus a=c.

Question 27: D

Divide each count by 10: 8/10=80%, 7/10=70%, and 5/10=50%.

Question 28: C

Let n be the additional type E cities. The requirement (5+n)/9≥0.8 gives n≥2.2. Since n must be a whole number, the minimum is 3. 2 is insufficient and 3 gives 8/9.

Question 29: B

The function is decreasing. Its minimum is f (5)=-8 and maximum is f (-4)=10, so the requested ratio is -8/10=-4/5.

Question 30: B

The original hypotenuse is sqrt (15²+36²)=39. Scaling every length by 1/3 gives sides 5, 12, 13. Their perimeter is 30 and their area is (5)(12)/2=30, so the ratio is 1.

Question 31: C

There are 84(5/14)=30 boys and 84(9/14)=54 girls. Add 54-30=24 boys to make the counts equal.

Question 32: A

The women-to-men ratios are 400/455≈0.8791, 620/868≈0.7143, 600/700≈0.8571 and 650/800=0.8125. The greatest is the city A ratio, rounded to 0.88.

Question 33: C

The male share in A is 455/855 and the female share in C is 600/1300. Dividing those shares gives (455/855)/(600/1300)=1183/1026≈1.1530, which rounds to 1.15. Multiplying both shares by 100 would cancel in the ratio.

Question 34: A

With 800 men, a ratio of 1.4 requires 1.4(800)=1120 women. The additional count is 1120-650=470.

Question 35: C

First simplify f (x)=-2 x+4 x+6+1=2 x+7. Replacing x by -3 x gives f (-3 x)=-6 x+7.

Question 36: B

After x hours, 7.5 x liters have been used. Subtract this from 60 to get 60-7.5 x liters. The stated interval prevents a negative fuel amount.

Question 37: A

From 9 x-2=43, x=5. Angle B is 60 degrees, so angle C is 180-43-60=77 degrees. Since y-12=77, y=89.

Question 38: B

The total is 8+12+15+15+8+14=72. Divide by the six marks to obtain a mean of 12.

Question 39: D

The perimeter gives x+y=20 and the area gives xy=96. The two side lengths are 8 and 12. Since y>10, y=12 and x=8.

Question 40: D

Cross-multiply to obtain 9(a-b)=10 b. Then 9 a=19 b, so a/b=19/9. The original fraction requires b≠0.

Question 41: C

The hypotenuse is sqrt (a²+b²). Cos (beta) is a divided by that hypotenuse, so its reciprocal is sqrt (a²+b²)/a.

Question 42: B

Rewrite the expression as -10 x/3-6. The absolute-value inequality is -4<-10 x/3-6<4. Add 6, then divide by -10/3 and reverse the inequalities, giving -3<x<-3/5.

Question 43: D

Write x=ky². Using 18=k (3²) gives k=2. Then 288=2 y², so y²=144. The stated positive condition selects y=12.

Question 44: B

Double the second equation and subtract it from the first: -2 y=46, so y=-23. Substitute into x+3 y=-20 to obtain x=49.

Question 45: 6

Multiply by 2 x to get 2 x²+8 xy=-8 y²+12. Rearranging and dividing by 2 gives x²+4 xy+4 y²=6, which is (x+2 y)²=6.

Question 46: $10\sqrt2$ feet

A 45-45-90 triangle has equal legs and a hypotenuse sqrt (2) times either leg. The ladder is 10 sqrt (2) feet long, approximately 14.14 feet.

Question 47: $\frac32$ units

The numerical volume is s³ and the numerical surface area is 6 s². Thus s³=(1/4)(6 s²). Since an edge length is positive, divide by s² to get s=3/2 units.

Question 48: $\frac{19}{12}$

Use vertex form f (x)=a (x+3)²+11. The given intersection has f (3)=g (3)=5, so 36 a+11=5 and a=-1/6. Expanding gives b=6 a=-1 and c=9 a+11=19/2. Their product is 19/12.

Effortless Math Team

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PSAT 10 Math Practice: Answers to 48 Legacy Questions

Answers to the corrected legacy practice questions

Question 1: D

Question 2: A

Question 3: D

Question 4: C

Question 5: D

Question 6: D

Question 7: A

Question 8: C

Question 9: A

Question 10: A

Question 11: C

Question 12: C

Question 13: B

Question 14: 384 square meters

Question 15: 18

Question 16: 2

Question 17: \(\frac43\)

Question 18: B

Question 19: B

Question 20: C

Question 21: C

Question 22: A

Question 23: D

Question 24: B

Question 25: C

Question 26: C

Question 27: D

Question 28: C

Question 29: B

Question 30: B

Question 31: C

Question 32: A

Question 33: C

Question 34: A

Question 35: C

Question 36: B

Question 37: A

Question 38: B

Question 39: D

Question 40: D

Question 41: C

Question 42: B

Question 43: D

Question 44: B

Question 45: 6

Question 46: \(10\sqrt2\) feet

Question 47: \(\frac32\) units

Question 48: \(\frac{19}{12}\)

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