Math practice
40. What is the perimeter of the polygon?
5"
5"
4"
2"
2"
6"
a. 24'' b. 25'' c. 27'' d. 32''
answers
arithmetic reasoning 1. This is a fractions problem.
c. The portion spent on transportation is
4.00
64.00 = ––161. Choice a is incorrect because that would correspond to $2 per day on trans-portation. Choice b is incorrect because this is the portion NOT spent on transportation.
Choice d is incorrect because that would correspond to $8 per day on transportation.
2. This is a fractions problem.
c. Raise the fractional parts to sixths and then subtract: 61–2 – 51–3 = 63–6 – 52–6 = 11–6. Choice a is incorrect because you would have only 6 cups after adding this amount of flour.
Choice b is incorrect because you subtracted 52–3 from 61–2. Choice d is incorrect because this is slightly too much flour; this would be 6––127 cups, not 61–2 cups.
3. This is an averages problem.
a. $956.58 ÷ 4 = $239.145. Choice b is incor-rect because the decimal portion is wrong.
Choice c is incorrect because the tenths digit is wrong. Choice d is incorrect because the tens digit is wrong.
4. This is a percents problem.
c. Using 32% = 0.32, multiply to get
($880,600)(0.32) = $281,792.00. Choice a is incorrect because this is 3.2%, not 32%.
Choice b is incorrect because you misinter-preted how to compute a percent. Choice d is incorrect because this is 68% of the amount and corresponds to the amount the company didn’t spend on labor to make the fabrics.
5. This is a percents problem.
a. Subtract the current amount minus the original amount to get $2.80 – $2.50 =
$0.30. Then, divide this increase by the origi-nal amount to get 0.302.50 = 0.12. So, the percent increase is 12%. Choice b is incorrect
because you forgot to divide the difference in price by the original amount. Choice c is incorrect because you divided the difference in price by the current amount instead of the original amount. Choice d is incorrect because you computed the difference in price as 0.20, not 0.30.
6. This is a fractions problem.
c. (41–4 in.) × 20 miles1 in. = (41–4 × 20) miles = (––17
4 × 20) miles = 85 miles. Choice a is incor-rect because this corresponds to 21–4 inches, not 41–4 inches. Choice b is incorrect because you converted 41–4 to 16––4, not 17––4. Choice d is incorrect because this corresponds to 41–2 inches, not 41–4 inches.
7. This is a fractions problem.
a. Set up a proportion: 152 = 750p . Solve this for p by cross-multiplying and then dividing by the coefficient of p: 15p = 1,500, so p = 100.
Choice b is incorrect because you divided incorrectly. Choice c is incorrect because you used the ratio in-state to out-of-state as 15:3, not 15:2. Choice d is incorrect because you set up the proportion incorrectly—you flipped one of the fractions.
8. This is an averages problem.
c. Add the amounts and divide by 4:
$80+ $1684+$52 +$52 = $88. Choice a is incor-rect because it is the mode, not the mean.
Choice b is incorrect because it is the median, not the mean. Choice d is incorrect because this is only the average of the two highest priced pieces of art.
9. This is a fractions problem.
a. There are 26 + 19 = 45 ways to get either a yellow jelly bean or a green jelly bean. The total number of jelly beans in the bag is 105.
So, the probability of getting yellow or green is 10545 = 37. Choice b is incorrect because you didn’t compute the correct quotient. Choice c is incorrect because this is the probability of NOT getting yellow or green. Choice d is incorrect because there are unequal amounts of the different colors of jelly bean, but the computation leading to this choice assumes that all colors are equally likely to be chosen.
10. This is a fractions problem.
c. The probability of choosing a peanut equals 1 minus the probability of choosing either an almond or a cashew. This equals 1 – (1–4 + ––101)
= ––2640 = 13––20. So, the total number of peanuts is
––13
20 × 200 = 130. Choice a is incorrect because this is the number of cashews and almonds.
Choice b is incorrect because you computed the probability of getting a peanut as being
––11
20, not 13––20. Choice d is incorrect because you didn’t account for the number of cashews in the can.
11. This is a percents problem.
c. The 8% tax is ($60)(0.08) = $4.80. The total bill is $60 + $4.80 = $64.80. So, the change she gets back is $100 − $64.80 = $35.20.
Choice a is incorrect because you computed 8% tax on $100, not the cost of the item.
Choice b is incorrect because you inter-changed the digits in the ones and tenths places. Choice d is incorrect because this is the total cost, not the amount of change she gets back.
12. This is an algebra problem.
b. Let x = the number of apples she has to begin with. Then, 1–2x = number of apples used for applesauce; 3–4(1–2x) number of apples used for pie; 3 = number of apples left over after making applesauce and pie. So, x = 1–2x the 3 apples left over from the correct total.
Choice c is incorrect because you added 3 apples to the total. Choice d is incorrect because this is twice too many apples. You likely added the fractions incorrectly.
13. This is a percents problem.
c. First, compute 35% of 80: (80)(0.35) = 28.
This is the number who did work overtime.
Now, subtract this from 80 to conclude that 52 did NOT work overtime. Choice a is incorrect because this is the number who did work overtime. Choice b is incorrect because 35% of 80 is not 35; you computed 35% of 100. Choice d is incorrect because you used 30% instead of 35%.
14. This is an algebra problem.
d. Multiply 2–a times b inches to get that 2b––a inches is Lydia’s height. Choice a is incorrect because you divided 2–a by b; you should have multiplied these two quantities. Choice b is incorrect because you didn’t treat the frac-tion 2–a correctly. Choice c is incorrect because you divided in the wrong order.
15. This is a geometry problem.
c. Use the area formula A = 1–2(base)(height): area formula incorrectly. Choice b is incor-rect because you used 3–4 instead of 1–2 in the area formula. Choice d is incorrect because you used 1–4 instead of 1–2 in the area formula.
16. This is a geometry problem.
a. Use w = length of the short side. Then, 2w = length of the longer side. Since the perimeter is 42, we have 2(w) + 2(2w) = 42, which simplifies to 6w = 42. Solving for w yields w = 7. So, the short side is 7 feet and the long side is 14 feet. Choice b is incorrect because the perimeter would be 48 feet, not 42 feet. Choice c is incorrect because you forgot to count the long side and short side each twice in the perimeter calculation.
Choice d is incorrect because the perimeter would be 72 feet, not 42 feet.
17. This is a geometry problem.
b. Use the formula V = (edge)3 to find the length of an edge: 1,000 = (edge)3, so that taking the cube root of both sides, we see that an edge is 10 inches. So, the area of one of the sides is (edge)2 = (10 in.)2 = 100 square inches. Choice a is incorrect because this is the length of an edge. Choice c is incorrect because you divided the vol-ume by 3 instead of taking the cube root to find an edge. Choice d is incorrect because you divided the volume by 3 and multiplied by 2 to find the area—both computations are wrong.
mathematics Knowledge 18. This is a fractions problem.
d. Divide each of the other five rectangles into equal parts of the same size as those into which the second rectangle is divided. Then, you can tell that 1 part of 10 is the portion shaded. This corresponds to ––101. Choice a is incorrect because this is too big; this corre-sponds to two full rectangles being shaded.
Choice b is incorrect because this would cor-respond to both small rectangles being shaded. Choice c is incorrect because you assumed all 6 regions are the same size.
19. This is a fractions problem.
c. ––164 = 1–4. Choice a is incorrect because you divided 4 by the difference 16 – 4 instead of 16. Choice b is incorrect because you divided 4 by 10, not 16. Choice d is incorrect because you likely divided 4 by 32, which is the number of ounces in 2 pounds, not 1 pound.
20. This is an algebra problem.
d. We want the value to the furthest left of 0.
−7–6 is equivalent to −11–6. The correct arrange-ment of these numbers from least to greatest is −7–6, −1, −1–2, 0. So, the least is −7–6.
21. This is a decimals problem.
b. Divide the top by the bottom to see that
5–
8 = 5 ÷ 8 = 0.625. Choice a is incorrect because you are not interpreting the mean-ing of a fraction correctly; the digits of the equivalent decimal are not simply the top number followed by the bottom number.
Choice c is incorrect because this is 7–8, not 5–8. Choice d is incorrect because this is larger than 1, and so cannot be equivalent to a proper fraction.
22. This is a fractions problem.
b. Solve the proportion: 5–9 = ––36?. Multiply 5 by 4 to see that ? = 20. So, 5–9 = 20––36. Choice a is
incorrect because this is equivalent to 1–2, not 5–9. Choice c is incorrect because this is equivalent to 2–3, not 5–9. Choice d is incorrect because you multiplied 5 by 6 instead of 4.
23. This is a fractions problem.
a. 24–5 ÷ 7 = ––145 ÷ 7 = ––145 × 1–7 = 2–5. Choice b is incorrect because you multiplied instead of dividing the fractions. Choice c is incorrect because you took the reciprocal of both frac-tions. Choice d is incorrect because you divided 2 by 7, but then multiplied that result by 4–5.
24. This is a fractions problem.
a. 4 – 14–5 = 35–5 – 14–5 = 21–5. Choice b is incorrect because once you borrowed, you didn’t sub-tract the fractional parts correctly. Choices c and d are incorrect because you didn’t bor-row correctly.
25. This is a fractions problem.
a. 5–8 × ––154 = 1–6. Choice b is incorrect because once you canceled common factors, you added the bottom numbers instead of multiplying them. Choice c is incorrect because you added top numbers and bottom numbers separately to get this answer. Choice d is incorrect because you divided instead of multiplying the fractions.
26. This is a fractions problem.
c. 12 × 161 × 38 = 31 = 3. Choice a is incorrect because this is the reciprocal of the correct answer. Choice b is incorrect because you treated 16 as if it were ––161 and proceeded to add the fractions instead of multiplying them. Choice d is incorrect because you added the top numbers and bottom num-bers instead of multiplying the fractions.
27. This is a fractions problem.
d. 133–4 ÷ 5 = 55––4 ÷ 5 = ––554 × 1–5 = 11––4 =23–4. Choice a is incorrect because you simplified 133–4 as 40––4, not 55––4. Choice b is incorrect because you mistakenly simplified 133–4 as 45––4, not ––554. Choice c is incorrect because you mistakenly simplified 133–4 as 50––4, not ––554.
28. This is a decimals problem.
b. Since the digit in the thousandths place is 4, which is less than 5, 0.7849 rounded to the nearest hundredth is 0.78. Choice a is incor-rect because you rounded to the nearest tenth. Choice c is incorrect because you rounded to the nearest thousandth. Choice d is incorrect because you rounded incorrectly.
Since the digit in the thousandths place is less than 5, don’t increase the digit in the hundredths place when rounding.
29. This is a decimals problem.
c. Line up the decimal points and add zero place holders so that all have the same num-ber of digits after the decimal point:
2.360
14.000
+0.083
16.443
Choice a is incorrect because you did not line up the decimal points correctly. Choice b is incorrect because you didn’t carry 1 when adding the hundredths digits. Choice d is incorrect because you mistakenly carried a 1 when adding the tenths.
30. This is a decimals problem.
d. Line up the decimal points and add zero place holders. When subtracting, you must borrow from the tenths place:
1.500 —→ 1.5\4 0\910
–0.188 –0.1 8 8
1.3 1 2
Choice a is incorrect because you didn’t line up the decimal points. Choice b is incorrect because you borrowed 2 tenths instead of 1 tenth. Choice c is incorrect because you bor-rowed 2 tenths instead of 1 tenth, and forgot about the thousandths digit.
31. This is a decimals problem.
c. First, add the positive decimals:
12.00
+4.60
16.60
Then, add –0.92 to this sum, which is equiv-alent to the following subtraction problem:
16.60 —→ –16\5.6\5 10
–0.92 – 0 .9 2
–15 .6 8
Choice a is incorrect because you omitted the negative sign in front of –0.92 and pro-ceeded to add all three numbers as if they were all positive. Choice b is incorrect because you neglected to borrow one from the ones place. Choice d is incorrect because you didn’t line up the decimal points when subtracting 0.92 from the sum of the other two decimals.
32. This is a decimals problem.
c. Move the decimal point 4 places to the right to get (2.39)(10,000) = 23,900. Choice a is incorrect because this is equivalent to multi-plying 2.39 by 100, not 10,000. Choice b is incorrect because this is equivalent to multi-plying 2.39 by 1,000, not 10,000. Choice d is incorrect because this is equivalent to multi-plying 2.39 by 100,000, not 10,000.
33. This is a decimals problem.
a. (0.0063)(5) = 0.0315. Choice b is incorrect because you moved the decimal point only three places instead of four after multiplying 63(5). Choice c is incorrect because you moved the decimal point only two places instead of four after multiplying 63(5).
Choice d is incorrect because you moved the decimal point only one place instead of four after multiplying 63(5).
34. This is a percents problem.
d. 45% = 10045–– = ––209. Choice a is incorrect because
4–
5 is equal to 0.80, which is 80%. Choice b is incorrect because 5–8 is equal to 0.625, which is 62.5%. Choice c is incorrect because 1–2 is equal to 0.50, which is 50%.
35. This is a percents problem.
b. Move the decimal point two places to the right to get 0.925 = 92.5%. Choice a is incor-rect because you moved the decimal point one too many places to the right. Choice c is incorrect because you moved the decimal point one too few places to the right. Choice d is incorrect because you moved the deci-mal point two places left instead of right.
36. This is a percents problem.
b. 12% of 60 = (60)(0.12) = 7.2. Choice a is incorrect because 12(5) = 60, but this is not a correct interpretation of computing a per-cent. Choice c is incorrect because this is 88% of 60, not 12%. Choice d is incorrect because this is 120% of 60, not 12%; you mistakenly said 12% = 1.2 instead of 0.12.
37. This is a percents problem.
c. There are 600 – 330 = 270 male freshmen.
So, the percentage is –––270600 = ––209 = 45%. Choice a is incorrect because you didn’t divide 270 by the number of students (600); rather, you divided by 1,000. Choice b is incorrect because you converted the fraction incor-rectly. Choice d is incorrect because this is the percentage of female freshmen.
38. This is a percents problem.
b. Let x = number of pages of the entire thesis.
Then, 28% of x is 42, which can be written symbolically as 0.28x = 42. So, x = 150.
Choice a is incorrect because you used 72%
(the percent left to go) in place of 28% in the computation. Choice c is incorrect because this is slightly too high and is likely the result of an arithmetic error. Choice d is incorrect because you incorrectly multiplied 28 times 42.
39. This is a geometry problem.
b. This is the only one whose measure is between 90° and 180°. Choice a is incorrect because this is an acute angle. Choice c is incorrect because this is a right angle. Choice d is incorrect because this is an acute angle.
40. This is a geometry problem.
a. Add the lengths of all sides:
2˝ + 5˝ + 5˝ + 4˝ + 2˝ + 6˝ = 24˝. Choices b, c, and d are incorrect because of arithme-tic errors. Add all sides together.
chapter
T
he Word Knowledge subtest of the ASVAB is basically a vocabulary test. Combined with the Paragraph Comprehension score, Word Knowledge helps make up your Verbal Expression score—it is one of the four subtests that determine whether you will be allowed to enlist. Your ability to understand your training materials depends in part on your reading comprehension and vocabulary skills.There are two different kinds of questions on the Word Knowledge subtest:
■ Synonyms—identifying words that mean the same as the given words
■ Context—determining the meaning of a word or phrase by noting how it is used in a sentence or paragraph