practice aSVaB core teSt 1
part 4: mathematics Knowledge 1. This is a geometry problem
b. P–— Qand R–—S are intersecting lines. The fact that POS is a 90-degree angle means that P–— and R–SQ — are perpendicular, indicating that all the angles formed by their intersec-tion, including ROQ, measure 90°. Choice a is incorrect because this is half of the cor-rect measure. Choice c is incorcor-rect because this is double the correct measure. Choice d is incorrect because this is 360° minus the measure of ROQ.
2. This is a fractions problem.
a. Incorrect answers include adding both the numerator and the denominator and not converting fifths to tenths properly. Choice b is incorrect because you also multiplied the numerator of ––103 by 2 when converting all fractions to tenths, but this fraction is already in terms of tenths. Choice c is incor-rect because you did not convert fifths to tenths properly. Choice d is incorrect because you added the fractions incorrectly;
you need to get a common denominator.
3. This is a percents problem.
d. To convert a fraction to a percent, change the denominator to 100 with multiplication.
(Multiply the denominator and the numera-tor by the same number, so that you do not change the value of the original fraction.) For example, 4–5 × ––2020 = 10080––, which is equivalent to 80%. Another way to consider this prob-lem is to change it to a decimal first by dividing the numerator, 4, by the denomina-tor, 5: 4.00 ÷ 5 = 0.80 = 80%. Choice a is incorrect because 0.45 = 10045, which does not simplify to 4–5. Choice b is incorrect because hundredths place (two decimal places) to get the answer 0.33. Choice a is incorrect because this equals 10013, which is not equal to
1– to add or subtract fractions, they must have common denominators. Since both of the denominators (3 and 5), are factors of 15, use 15 as your common denominator.
numerator of the first fraction, 11, is smaller than the numerator of the second fraction, 14, borrow one whole number from the 7 in 711––15, changing it to 6, and add 15––15to 11––15to get 26––15. Therefore, 7––1115 = 626––15. Lastly, 626––15 – 214––15 = 412––15 = 44–5. Choice b is incorrect because you did not subtract the fractional parts correctly;
it should be 11––15 –14––15, not 14––15 –11––15. Choice c is incorrect because you added all three frac-tions. Choice d is incorrect because this is too large since you are subtracting a quantity from a sum that is approximately 8.
6. This is a decimals problem.
d. Choice a is incorrect because it reads 276.
Choice b is incorrect because it reads 2,706.
Choice c is incorrect because it reads 20,076.
7. This is a decimals problem.
a. The first box is one greater than –5; the second is one greater than 0. Choice b is incorrect because the first box is to the right of –5 by one unit, so it should be –4. Choice c is incorrect because the second box is to the right of 0 by one unit, so it should be 1.
Choice d is incorrect because you are not using the position of 0 on the number line as a reference point.
8. This is an algebra problem.
d. 168 is the only number that can be divided by 3, 7, and 8. 168 ÷ 3 = 56, 168 ÷ 7 = 24, 168 ÷ 8 = 21. Choice a is incorrect because 8 does not divide evenly into this number.
Choice b is incorrect because 7 does not divide evenly into this number. Choice c is incorrect because 3 does not divide evenly into this number.
9. This is an algebra problem.
c. The meaning of 43 is 4 times itself 3 times.
Choice a is incorrect because this does not mean the product of three 4s. Choice b is incorrect because this equals 2 × 4 × 4, not 4 × 4 × 4. Choice d is incorrect because this equals 3 × 3 × 3 × 3.
10. This is a fractions problem.
b. Use 35 for C; F = (9–5 × 35) + 32. Therefore, F = 63 + 32, or 95°. Choice a is incorrect because you added incorrectly; you mistak-enly carried a one to the tens position.
Choice c is incorrect because you forgot to add 32. Choice d is incorrect because you subtracted 32 instead of adding it.
11. This is a geometry problem.
c. 5(3)(8) = 120; 120 ÷ 3 = 40. Choice a is incorrect because you added the dimensions instead of multiplying them, and disre-garded the 1–3 in the formula. Choice b is incorrect because you used 1–2 instead of 1–3 in the formula. Choice d is incorrect because you forgot to multiply by 1–3.
12. This is a fractions problem.
b. To solve this problem, you must first convert yards to inches. There are 36 inches in a yard; 36(31–3) = 120. Choice a is incorrect because you converted 3 yards to inches, not 31–3 yards. Choice c is incorrect because you assumed that 1 yard equals 48 inches, not 36 inches. Choice d is incorrect because you assumed that 1 yard equals 48 inches, and converted 3 yards to inches, not 31–3 yards.
13. This is a decimals problem.
d. To change a fraction into decimal, divide the numerator by the denominator: 13 ÷ 4 = 3.25. Another way to consider this problem is to change 13––4 into a mixed fraction by Choice c is incorrect because this equals 33–4 = 15––4, not 13––4.
14. This is a percents problem.
b. Percent means “out of 100.” In order to turn a percent into a decimal, divide it by 100:
125% = 125–––100 = 1.25. (When dividing a num-ber by a power of 10 such as 10, 100, or 1,000, simply move the decimal point of the numerator one place to the left for every zero in the denominator.) Choice a is incor-rect because you moved the decimal point 3 places to the left, not 2. Choice c is incorrect because you moved the decimal point only 1 place to the left, not 2. Choice d is incorrect because you didn’t move the decimal point 2 places to the left.
15. This is a geometry problem.
d. An isosceles triangle has two equal legs and one base. The perimeter of the triangle is 3 feet, which is equivalent to 36 inches (12 inches in every foot). The base is 14 inches, so the sum of the two legs is 36 inches – 14 inches = 22 inches. Since both legs are of equal length, 22 ÷ 2 = 11 inches for each leg. Choice a is incorrect because this is the perimeter of the triangle, not the length of a leg. Choice b is incorrect because if both legs were 18 inches long, there could be no base because the perimeter of the entire triangle is 36 inches. Choice c is incor-rect because this is the sum of the lengths of both legs.
16. This is an algebra problem.
c. Subtract 25 from both sides to get x = –12.
Choice a is incorrect because –13 + 25 = 12, not 13. Choice b is incorrect because –11 + 25 = 14, not 13. Choice d is incorrect because you added 25 to both sides instead of subtracting it.
17. This is a geometry problem.
b. A cube has four sides and two bases (a top and a bottom), which means that it has six faces. Choice a is incorrect because you forgot the top and bottom. Choice c is incorrect because the number of faces does not equal the number of bases. Choice d is incorrect because this is the number of edges, not faces.
18. This is a geometry problem.
d. To solve this problem, you should use the formula A = lw, or 117 = 9l. Next, you must divide 117 by 9 to find the answer. Choice a is incorrect because you mistakenly divided by 90. Choice b is incorrect because you divided the correct length by 2. Choice c is incorrect because you doubled the length.
19. This is a geometry problem.
d. A square is a special case of all of these fig-ures except the trapezoid. A square is not a trapezoid because a trapezoid has only two sides parallel. Choice a is incorrect because a square is a parallelogram because its oppo-site sides are parallel. Choice b is incorrect because a square is a rectangle because it is a quadrilateral with 90-degree angles. Choice c is incorrect because a square is a rhombus because it is a parallelogram with all sides equal in length.
20. This is a geometry problem.
b. The Pythagorean theorem states that the square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides, so we know that 12 + x2 = ( 10)2, so that 1 + x2 = 10, and so, x2 = 10 – 1 = 9, so x = 3. Choice a is incorrect because if this were true, 12 + 22 = 5, so that the hypotenuse would have length 5, not 10. Choice c is incorrect because you used the Pythagorean theorem assuming that 10 was the length of one of the legs, not the hypotenuse. Choice d is incorrect because you forgot to take the square root.
21. This is a decimals problem.
d. 2–3 = 0.6666 repeating, so 52–3 is equivalent to 5.6666 repeating, or about 5.67. Choice a is incorrect because you are not treating the fraction 2–3 correctly; it is not equal to 0.23.
Choice b is incorrect because 1–3 is approxi-mately 0.33 while 2–3is approximately 0.67.
Choice c is incorrect because this is the fur-thest away from 52–3 because its whole part is 0.
22. This is a geometry problem.
d. If the figure is a regular decagon, it can be divided into ten equal sections by lines pass-ing through the center. Two such lines form the indicated angle, which includes three of the ten sections: ––103 of 360° = 108°. Choice a is incorrect because this would correspond to the angle of one sector, not three. Choice b is incorrect because this would be the angle for one sector if the figure had 8 sides, not 10. Choice c is incorrect because this would be the angle for two sectors if the fig-ure had 8 sides, not 10.
23. This is an algebra problem.
b. The 7 is in the hundredths place, therefore, 0.07 is equal to –––1007 and 2.07 = 2–––1007 . “Nega-tive 2.07” is equal to –2–––1007 . Choice a is incor-rect because ––107 = 0.7, not 0.07. Choice c is incorrect because 1,000––––7 = 0.007, not 0.07.
Choice d is incorrect because –2.07 does not equal –2.7 because 0.7 is not equal to 0.07.
24. This is a percents problem.
b. 62.5% is 62.5100. You should multiply both the numerator and denominator by 10 to move the decimal point, resulting in 1,000625 , and then factor both the numerator and denom-inator to find out how far you can reduce the fraction; = × × ×
× × × 625
1,000
5 5 5 5
5 5 5 8. If you cancel the three 5s that are in both the numerator and denominator, you will get 5–8. Choice a is incorrect because this equals 6.25%, not 62.5%. Choice c is incorrect because this equals 625%, not 62.5%. Choice d is incor-rect because this equals 640%, not 62.5%.
25. This is a geometry problem.
d. A line that intersects two parallel lines forms supplementary angles on either side of it.
Supplementary angles are angles whose measures add up to 180°; 180 – 40 = 140.
Choice a is incorrect because P and Q are not congruent. Choice b is incorrect because P and Q are supplementary, not complimentary, angles. Choice c is incorrect because angle Q is obtuse and supplemen-tary to P; since 40° + 80° ≠ 180°, this choice is not
correct.
Here are the steps you should take, depending on your AFQT score on the first practice test:
■ If your AFQT is below 29, you need more help in reading and/or math. You should spend plenty of time reviewing the lessons and practice questions found in this book.
■ If your AFQT is 29–31, be sure to focus on your weakest subjects in the review lessons and prac-tice questions that are found in this book.
■ If your AFQT is above 31, review the areas that give you trouble, and then take the second prac-tice test in Chapter 12 to make sure you are able to get a passing score again.
Write your raw score (the number you got right) for each test in the following blanks. Then turn to Chapter 3 to find out how to convert these raw scores into the scores the armed services use.
1. Arithmetic Reasoning: ______ right out of 30 2. Word Knowledge: ______ right out of 35 3. Paragraph Comprehension: ______ right out of 15 4. Mathematics Knowledge: ______ right out of 25
chapter
T
wo subtests of the ASVAB—Arithmetic Reasoning and Mathematics Knowledge—cover math skills.Arithmetic Reasoning is basically math word problems. Mathematics Knowledge tests your knowledge of math concepts, principles, and procedures. You don’t have to do a lot of calculation in the Mathe-matics Knowledge subtest; you need to know basic terminology (like sum and perimeter), formulas (such as the area of a square), and computation rules. Both subtests cover the subjects you probably studied in school. This chapter reviews concepts you will need for both Arithmetic Reasoning and Mathematics Knowledge. Chapter 7 gives you more of these types of problems for extra practice.