Mathematics Knowledge for ASVAB Power Practice Problems (page 2)

Updated on Aug 12, 2011


  1. b.   Convert the mixed number to the improper fraction and then invert to .
  2. a.   Since the area of one tile is 12 inches × 12 inches = 144 square inches, which is one square foot, one tile is needed for each square foot of the floor. The square footage of the room is 10 × 15 = 150 square feet, so 150 tiles are needed.
  3. d.   Although there are other factors that 54 and 108 share, dividing both by the greatest common factor, 54, will result in the lowest terms in one step. The correct answer is .
  4. b.   Substitute the values of each letter and simplify. The expression becomes , which simplifies to after performing multiplication. Add 3 + 18 in the numerator to get , which simplifies to 7.
  5. c.   First, combine like terms on the left side of the equation to get 2p – 10 = 16. Add 10 to both sides of the equation: 2p = 26. Divide both sides of the equation by 2: p = 13.
  6. b.   The third side must measure between the difference and the sum of the two known sides. Since 7 – 5 = 2 and 7 + 5 = 12, the third side must measure between 2 and 12 units.
  7. d.   The correct answer is 175%.
  8. d.   Find the answer by changing the fractions to decimals: = 0.333; = 0.25; = 0.286. The decimal 0.286, or , is between the other two.
  9. d.   The order of operations dictates addressing the exponents first. Then perform the operations within the parentheses. Finally, perform the operation outside the parentheses: Raising exponents by exponents is done by multiplying: 4 × 2 = 8; 3(36x8); the correct answer is 108x8.
  10. c.   One of the most vital steps in solving for an unknown in any algebra problem is to isolate the unknown on one side of the equation.
  11. b.   Area is equal to base times height: 2 × 4 = 8 square units.
  12. b.   One hundred cases is five times twenty cases, so the cost is 20 times $11.39, or $227.80.
  13. a.   The 8 is two places to the right of the decimal point, so the correct answer is eight hundredths.
  14. a.   The square of a number is the number times itself. When you are looking at the answer choices, the only ones that would be possible are either choice a or c. Trial and error shows that 9 × 9 = 81. Then, multiply 81 by 9. The product is 729, so the correct answer is 9.
  15. d.   The key word product tells you to multiply. Therefore, multiply the coefficients of 2 and 3, and multiply the variables by adding the exponents of like bases. Keep in mind that x = x1; (3x2y) × (2xy2) becomes 3 × 2 × x2 × x × y × y2. This simplifies to 6x3y3.
  16. c.   Since odd integers, such as 3, 5 and 7, are two numbers apart, add 2 to the expression: x – 1 + 2 simplifies to x + 1.
  17. a.   Take the square root of 676 to find the length of each side; therefore, each side measures 26 inches.
  18. c.   The correct answer is 2.52.
  19. c.   Since the ships are going west and north, their paths make a 90° angle. This makes a right triangle where the legs are the distances the ships travel, and the distance between them is the hypotenuse. Using the Pythagorean theorem, 4002 + 3002 = distance2. The distance is 500 miles.
  20. b.   First, change the percentage to a fraction: . Then, change it to a mixed number and reduce to lowest terms: .
  21. a.   A prime number is a number that has exactly two factors: one and itself. The first four prime numbers are 2, 3, 5, and 7. The sum of these numbers is 17.
  22. a.   Since the solution to the problem x + 25 = 13 is –12, choices b, c, and d are all too large to be correct.
  23. c.   The decimal –0.16 is less than –0.06, the smallest number in the range.
  24. c.   Move the decimal point four places to the right in 4.32 to get the correct answer of 43,200.
  25. c.   In order to solve for x, get x alone on one side of the equation. First, add r to both sides of the equation: s + r = 2xr + r. The equation becomes s + r = 2x. Then divide each side of the equation by 2: . Cancel the 2s on the right side of the equation to get a result of , which is equivalent to answer choice c.
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