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# More Arithmetic Reasoning Practice Questions for ASVAB Power Practice Problems (page 2)

By LearningExpress Editors
LearningExpress, LLC
Updated on Jan 26, 2012

1. b.   The total number of pages assigned is 80;
2. c.   This is a three-step problem involving multiplication, subtraction, and addition. First, find out how many fewer minutes George jogged this week than usual: 5 hours × 60 minutes = 300 minutes – 40 minutes missed = 260 minutes jogged. Now add back the number of minutes George was able to make up: 260 minutes + 20 + 13 minutes = 293 minutes. Now subtract again: 300 minutes – 293 = 7 minutes jogging time lost.
3. a.   The train's speed can be found using the formula Distance = rate × time. From this we get the formula , since we are looking for the speed. By substituting, 6 , which simplifies to 50. The speed is 50 miles per hour.
4. b.   To find the percentage of people who said they rode at least three times a week, divide 105 by 150: 105 ÷ 150 = 0.7, which is 70%. 0.7× 100,000 = 70,000.
5. d.   According to the table, a pound in weight makes a difference of \$0.64, or \$0.04 per ounce over 4 pounds. Fruit that weighs 4 pounds 8 ounces will cost 8 × 0.04 or \$0.32 more than fruit that costs 4 pounds. Therefore, the cost is \$1.10 + 0.32 = \$1.42.
6. b.   To find the answer, begin by adding the cost of the two sale puppies: \$15 + \$15 = \$30. Now subtract this amount from the total cost: \$70 – \$30 = \$40 paid for the third puppy.
7. b.   First, divide to determine the number of 20-minute segments there are in 1 hour: 60 ÷ 20 = 3. Now multiply that number by the number of times Rita can circle the block: 3 × 5 = 15.
8. c.   Between 8:14 and 9:00, 46 minutes elapse, and between 9:00 and 9:12, 12 minutes elapse, so this becomes a simple addition problem: 46 + 12 = 58.
9. a.   Let x equal the number of pounds of chocolate to be mixed. You know the mixture's total cost is the cost of the chocolates plus the cost of the caramels, or M = A + B. In terms of x, M = 3.95(x + 3), A = 5.95x, while B = 2.95(3). Combine terms: 3.95(x + 3) = 5.95x + 2.95(3). Simplify: 3.95x + 11.85 = 5.95x + 8.85, or 11.85 – 8.85 = (5.95 – 3.95)x, which becomes 2x = 3. Thus, x = 1.5 pounds.
10. c.   This is a problem of addition. You may simplify the terms: M = F + 10; then substitute: M = 16 + 10, or 26.
11. d.   First, add the weight of Ralph's triplets: , or (after finding the least common denominator) , or , or . Now find the weight of Harvey's twins: . Now subtract: . So Harvey's twins outweigh Ralph's triplets by pounds. (No further reduction of the fraction is possible.)
12. d.   There are four quarters for every dollar; \$12 × 4 quarters per dollar = 48 quarters.
13. c.   From 5:16 P.M. to 7:16 A.M. the next day is 14 hours. An additional 42 minutes occurs between 7:16 A.M. and 7:58 A.M: (58 – 16 = 42).
14. b.   Carmella worked 15 hours per week for 8 weeks: 15 × 8 = 120. In addition, she worked 15 hours for Mariah for one week, so 120 + 15 = 135.
15. c.   Take apart the statement and translate each part. The word twice tells you to multiply the quantity by two. In the second part of the statement, the word sum is a key word for addition. So the sum of six and four is translated as 6 + 4. The whole statement becomes 2(6 + 4).
16. c.   Divide the amount of cod by the number of crates: 400 ÷ 20 = 20.
17. c.   The speed of the train, obtained from the table, is 60 miles per hour. Therefore, the distance from Chicago would be equal to 60t. However, as the train moves on, the distance decreases from Los Angeles, so there must be a function of –60t in the equation. At time t = 0, the distance is 2,000 miles, so the function is 2,000 – 60t.
18. b.   Divide \$350 by \$25; 350 ÷ 25 = 14 weeks.
19. d.   This is a division problem. First, change the mixed number to a fraction: . Next, invert and multiply: .
20. b.   The total value of the supplies and instruments is found by adding the cost of each item: 1,200 + (2 × 350) + (3 × 55) + 235 + 125 + 75 = 1,200 + 700 + 165 + 235 + 125 + 75. The total is \$2,500.
21. d.   If Salwa (S) is 10 years older than Roland (R), then S is equal to R + 10. Therefore, the equation is S = R + 10.
22. c.   January is approximately \$38,000; February is approximately \$41,000; and April is approximately \$26,000. These added together give a total of \$105,000.
23. b.   Visualize a number line. The drop from 31° to 0° is 31°. The are still nine more degrees to drop. They will be below zero, so –9°F is the temperature at midnight.
24. a.   This answer is in the correct order and is "translated" correctly: Rachel had (=) 3 apples and ate (–) 1.
25. d.   Each number is divided by 2 to find the next number: 40 ÷ 2 = 20, so 20 is the next number.
26. c.   First you must subtract the percentage of the installation cost during construction (1.5%) from the percentage of the installation cost after construction (4%). To do this, begin by converting the percentages into decimals: 4% = 0.04; 1.5% = 0.015. Now subtract: 0.04 – 0.015 = 0.025. This is the percentage of the total cost that the homeowner will save. Multiply this by the total cost of the home to find the dollar amount: 0.025 × \$150,000 = \$3,750.
27. b.   First, simplify the problem: K = 5 × 17 = 85, so Kara made 85 cents; R = 7 × 14 = 98, so Rani made 98 cents, the higher amount of money; R – K = 98 – 85 = 13. Therefore, Rani made 13 cents more than Kara.
28. d.   Because Linda pays with a check only if an item costs more than \$30, the item Linda purchased with a check in this problem must have cost more than \$30. If an item costs more than \$30, then it must cost more than \$25 (choice d), as well.
29. d.   First, write the problem in columns:
30. Now subtract, beginning with the rightmost column. Since you cannot subtract 11 inches from 5 inches, you must borrow 1 foot from the 6 in the top left column, then convert it to inches and add: 1 foot = 12 inches; 12 inches + 5 inches = 17 inches.

The problem then becomes:

So the answer is choice d, 6 inches.

31. a.   To find the fraction, compare the down payment with the total cost of the car; 630/6,300 reduces to .

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