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Math Knowledge for ASVAB Power Practice Problems (page 2)

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Updated on Oct 30, 2012

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  1. a.   Square roots can be multiplied and divided, but they cannot be added or subtracted.
  2. d.   Start by adding 16 to 20. The sum is 36. Then, divide 36 by 6. The correct answer is 6.
  3. c.   An angle that is more than 90° is an obtuse angle.
  4. d.   This is the only choice that includes a 90° angle.
  5. b.   The simplest way to solve this problem is to cancel the a term that occurs in both the numerator and denominator. This leaves . This is , which simplifies to .
  6. d.   The symbol means that a is "less than or equal to" 5. The only choice that makes this statement false is 6.
  7. a.   Multiply the whole number by the fraction's denominator: 5 × 2 = 10. Add the fraction's numerator to the answer: 1 + 10 = 11. Now place that answer over the fraction's denominator: .
  8. d.   In this problem, 2 is the base and 5 is the exponent. Two raised to the power of 5 means to use 2 as a factor five times: 2 × 2 × 2× 2× 2 = 32.
  9. b.   Since y × y = y2, then 2y(y) is equal to 2y2.
  10. d.   To solve this problem, you must first find the common denominator, which is 6. The equation then becomes ; then, ; and then 4x = 24, so x = 6.
  11. b.   In a fractional exponent, the numerator (number on top) is the power, and the denominator (number on the bottom) is the root. Since 2 is the denominator, take the square root of 16. Since 4 × 4 = 16, 4 is the square root of 16. Then the numerator (power) is 1, so 41 = 4.
  12. c.   A common error is to think that 1 is a prime number. But, it is not, because one has only 1 factor: itself. The correct answer is 2.
  13. b.   Isolate the variable by subtracting 50 from both sides of the equation. Then divide both sides by 5. The correct answer is 20.
  14. d.   If the figure is a regular decagon, it can be divided into 10 equal sections by lines passing through the center. Two such lines form the indicated angle, which includes three of the 10 sections; of 360° is equal to 108°.
  15. c.   The number 112 is divisible by both 7 and 8 because each can divide into 112 without a remainder; 112 ÷ 7 = 16 and 112 ÷ 8 = 14. Choice a is divisible only by 7, choice b is not divisible by either, and choice d is divisible only by 8.
  16. d.   x2 + 4x + 4 factors into (x + 2)(x + 2). Therefore, one of the (x + 2) terms can be canceled with the denominator. This leaves x + 2.
  17. c.   Ask the question, What number divided by 72 equals 2? The correct answer is 144.
  18. b.   Find the correct answer by dividing the area by the given side length, because area = length  × width: 72 ÷ 12 = 6 inches.
  19. c.   An algebraic equation must be used to solve this problem. The shortest side can be denoted s. Therefore, s + (s + 2) + (s + 4) = 24; 3s + 6 = 24, and s = 6.
  20. b.   Two equations are used: T = 4O, and T + O = 10. This gives 5O = 10, and O = 2. Therefore, T = 8. The number is 82.
  21. d.   The first step in solving this problem is to add the fractions to get the sum of . This fraction reduces to x.
  22. d.   Supplementary angles add up to 180°. When given one and asked to find its supplement, subtract the given angle from 180°. The correct answer is 96°.
  23. d.   The angles labeled 120° are alternate interior angles of the lines c and d. When the alternate interior angles are congruent (the same measure), the lines are parallel. Therefore, lines c and d are parallel.
  24. d.   56,515 ÷ 4 = 14,128.75, or, rounded to the nearest whole number, 14,129.
  25. c.   Let x equal the number of oranges left in the basket. Three more than seven times as many oranges as five is 7(5) + 3 = 38. Removing five leaves x = 38 – 5 = 33 oranges.
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