McGraw-Hill Education GRE 2019

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Exercise Answers


Discrete Quantitative Questions



  1. B To solve for (x − 5 ), subtract 12 from both sides of the equation: x − 5 = 2.
    Substitute 2 for (x − 5): 2^2 = 4.

  2. 30 To get rid of the denominators, multiply across the equation by 30 (which
    is a multiple of the values in all the denominators):
    30(^12 a +^13 a + 101 a = 28)
    15 a + 10a + 3a = 840


Combine like terms:
28 a = 840
a = 30


  1. A To express y in terms of x, you need to isolate y. Subtract 2y and 2x from
    both sides of the equation and arrive at y = 3 − 2x.

  2. B Notice that 4a + 6b = 2(2a + 3b). Therefore, to combine the equations,
    multiply the first equation by 2 and arrive at 4a + 6b = 8. Substitute 4a + 6b
    into the second equation and arrive at x = 8.

  3. D To express a 2 in terms of x, you need to isolate a 2. To do so, first subtract x
    from both sides:^92 a = 3 − x. To go from^92 a to a 2 , divide both sides by 9 and arrive
    at a 2 = 3 – 9 x.

  4. E Since you have two linear equations, you have to decide between
    combination and substitution. Since the coefficient in front of p in the second
    equation is 1, substitution is the preferred method. Write p in terms of q:
    p = 15 − 2q. Substitute (15 − 2q) for p in the first equation and arrive at
    3(15 − 2q) + 5q = 27. Distribute: 45 − 6q + 5q = 27. Solve for q. q = 18.

  5. 203 Since none of the variables has a coefficient of 1, use combination. You want
    to solve for b, so multiply across the first equation by −4:
    − 4(1. 5a + 3b = 20)
    −6a − 12b = −80


Now the equations are:
−6a − 12b = −80
6 a + 9b = 60

Add the equations and arrive at −3b = −20. Solve for b: b = 20/3.


  1. A What is the relationship between x^2 + 5x + 6 and x^2 + 5x − 12? x^2 + 5x + 6 is
    18 more than x^2 + 5 x − 12. Since x^2 + 5x + 6 = 32, x^2 + 5x − 12 = 32 − 18 = 14.

  2. 361 Use PEMDAS: First evaluate the expression within the parentheses:
    3 – 5 = –2:
    –2(–2) ÷ 12^2


CHAPTER 11 ■ ALGEBRA 267

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