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Miscellaneous Math Practice Problems Set 8 (page 2)

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Updated on Jul 15, 2011

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  1. 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.
  2. 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.
  3. 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.
  4. d.   This series actually has two alternating sets of numbers. The first number is doubled, giving the third number. The second number has 4 subtracted from it, giving the fourth number. Therefore, the blank space will be 12 doubled, or 24.
  5. d.   Figure the amounts by setting up the following equations: First, S = $3 + $23 = $26. Now, B = ($1 × 5) + ($2 × 2) or $5 + $4 = $9. MR = $1 × 2 = $2; and D = $4 × 1 = $4. Now, add: $9 + $2 + $4 = $15. Now subtract: $26 – $15 = $11.
  6. 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.
  7. c.   Follow the order of operations and evaluate the exponent first. Since any nonzero number raised to the zero power is equal to 1, then 40 = 1. Now multiply 5(1) = 5 to get the simplified value.
  8. a.   It is easiest to use trial and error to arrive at the solution to this problem. Begin with choice a: After the first hour, the number would be 20; after the second hour, 40; after the third hour, 80; after the fourth hour, 160; and after the fifth hour, 320. Fortunately, in this case, you need go no further. The other answer choices do not have the same outcome.
  9. a.   The unreduced ratio is 8,000:5,000,000; reduced, the ratio is 8:5,000. Now divide: 5,000 ÷ 8 = 625, for a ratio of 1:625.
  10. 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.
  11. a.   Since 4 and 6 are factors of 24, and 4 is a perfect square,
  12. d.   Irrational numbers are nonrepeating, nonterminating decimals. They also cannot be written in fraction form without approximating. Choice d has a pattern, but the same exact pattern does not repeat each time, making it irrational. Choice a is a repeating decimal, choice b is a fraction, and choice c is a terminating decimal, so they are rational, not irrational.
  13. d.   First, write the problem in columns:
  14. Now subtract, beginning with the right-most 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.

  15. c.   This is a problem of addition. You may simplify the terms: M = F + 10; then substitute: M = 16 + 10, or 26.
  16. c.   76 ÷ 19 = 4; the other division operations will not end in whole numbers.
  17. a.   The mode of a set of numbers is the number that appears the most. Four appears in this set three times. Since it appears more often than any other number in the list, 4 is the mode.

More practice problems can be found at:

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