Problem Solving

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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