AAPL wrote: ↑Thu Nov 12, 2020 9:32 am
GMAT Prep
If x and y are positive, which of the following must be greater than \(\dfrac{1}{\sqrt{x+y}}?\)
I. \(\dfrac{\sqrt{x+y}}{2x}\)
II. \(\dfrac{\sqrt{x}+\sqrt{y}}{x+y}\)
III. \(\dfrac{\sqrt{x}-\sqrt{y}}{x+y}\)
A. None
B. I only
C. II only
D. I and III only
E. II and III only
OA
C
Solution:
Let’s test each Roman numeral choice.
I. √(x + y)/(2x) > 1/√(x + y) ?
[√(x + y)]^2 > 2x ?
x + y > 2x ?
y > x ?
Since we are not given any information about the values of x and y except that they are positive, we can’t say y is definitely greater than x. Therefore, I is not true.
II. (√x + √y)/(x + y) > 1/√(x + y) ?
(√x + √y)/(x + y) > [√(x + y)]/(x + y) ?
√x + √y > √(x + y) ?
(√x + √y)^2 > [√(x + y)]^2 ?
x + 2√(xy) + y > x + y ?
2√(xy) > 0 ?
Since we are given that x and y are both positive, 2√(xy) is definitely greater than 0. Therefore, II is true.
II. (√x - √y)/(x + y) > 1/√(x + y) ?
(√x - √y)/(x + y) > [√(x + y)]/(x + y) ?
√x - √y > √(x + y) ?
If x ≤ y, we see that √x - √y is non-positive, but √(x + y) is positive. So √x - √y will not be greater than √(x + y). If x > y, let’s square both sides, and we have:
(√x - √y)^2 > [√(x + y)]^2 ?
x - 2√(xy) + y > x + y ?
-2√(xy) > 0 ?
Since x and y are both positive, -2√(xy) will be negative and will not be greater than 0. Therefore, III is not true.
Answer: C
