McGraw-Hill Education GRE 2019

(singke) #1

  1. D Since lines l and k are perpendicular, their slopes are negative reciprocals.
    Since the slope of line l is 3, the slope of line k is −^13. The equation for line
    k now reads: y = −^13 x + b. Since the lines intersect at the point (3,4), those
    coordinates must satisfy the equation for line k. To solve for b, plug those
    coordinates into the equation:
    y = −^13 x + b
    4 = −^13 3 + b
    4 = −1 + b
    5 = b
    The equation for line k is thus y = −^13 x + 5.

  2. B Isolate y so that the equation is in y = mx + b form:
    ax + by = c
    by = −ax + c
    y = –ax + cb
    y = –bax + cb
    The slope of the line is thus −ab. Since the slope of the line is negative,

    • a
      b < 0. Thus




a
b > 0.

Quantitative Comparison Questions



  1. A Since the lines go downward, both slopes are negative. Line m is steeper,
    meaning its slope is more negative than the slope of l. Thus the slope of l is
    greater. The correct answer is A.

  2. B Since line m contains the points (0,0) and (3,5), its slope is (5 – 0)/(3 − 0) =^53.
    Line m is steeper than line l, so the slope of line l must be less than^53 , meaning
    its slope is less than 2. The correct answer is B.

  3. C Since line q passes through the origin, it must contain the point (0,0). The
    slope of line q is thus b a – 0– 0 = ba. The two quantities are equal. The answer is C.

  4. B The slope of the line is b 0 – – 0d = –bd. Since the line points downward, its slope
    must be negative. Thus –bd < 0. Multiply both sides by −1 (remember to flip the
    inequality!): bd > 0. Since m < 0 and db > 0, Quantity B is greater.

  5. B Manipulate the equation to be in y = mx + b form:
    2 y – 3x + 7 = 12
    2 y – 3x = 5
    2 y = 3x + 5
    y =^32 x +^52
    The slope is^32 and the y-intercept is^52. Quantity B is greater.


446 PART 4 ■ MATH REVIEW

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