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22 Jul 2020, 23:17
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
71% (01:38) correct
29%
(01:55)
wrong
based on 183
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Not Attempted Yet
If |x - y| = 8, what is the value of x + y ?
(1) x^2 - y^2 = 32
(2) x + y > 0
(1) x^2 - y^2 = 32
(2) x + y > 0
23 Jul 2020, 00:05
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jefferyk
If |x - y| = 8, what is the value of x + y ?
(1) x^2 - y^2 = 32
(2) x + y > 0
(1) x^2 - y^2 = 32
(2) x + y > 0
Statement 1: Insufficient
x^2 - y^2 = 32
=> (x-y)(x+y) = 32
If x-y = 8, then x+y = 4.
If x-y = -8, then x+y = -4
Statement 2: Insufficient
Nothing about x+y value.
Combining, Sufficient
We get x+y = 4.
Hence, OA is (C).
23 Jul 2020, 00:38
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jefferyk
If |x - y| = 8, what is the value of x + y ?
(1) x^2 - y^2 = 32
(2) x + y > 0
(1) x^2 - y^2 = 32
(2) x + y > 0
|x-y|=8....
a) If x>y, then x-y=8.
b) If x<y, then x-y=-8
We are looking for the value of x+y.
(1) \(x^2 - y^2 = 32\)
\((x-y)(x+y)=32.....x+y=\frac{32}{x-y}\)
a) If x>y, then \(x+y=\frac{32}{x-y}=\frac{32}{8}=4......x+y=4\), and
b) If x<y, then \(x+y=\frac{32}{x-y}=\frac{32}{-8}=-4......x+y=-4\)
Insuff
(2) \(x + y > 0\)
Nothing at all
Combined
x+y>0, so it cannot be -4.
x+y=4
Suff
C
