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Arithmetic

by BTGmoderatorRO » Sun Jan 14, 2018 7:54 am
If x and y are positive, which of the following must be greater than $$\frac{1}{\sqrt{x+y}}$$ ?

1. $$\frac{\sqrt{x+y}}{2x}$$

2. $$\frac{\sqrt{x}+\ \sqrt{y}}{x+y}$$

3. $$\frac{\sqrt{x}-\ \sqrt{y}}{x+y}$$

(A) None
(B) 1 only
(C) 2 only
(D) 1 and 3 only
(E) 2 and 3 only

OA is E
2 and 3 my answers are near 1 can I get a deep understanding of this from an Expert. Thanks in advance
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Querying the questions's answer

by DrMaths » Mon Jan 15, 2018 5:06 am
Surely the answer cannot include answer (3)?
Because, if x = y = 1
then answer (3) could give

$$\sqrt{1}-\sqrt{1}\ =\ 0\ <\ \frac{1}{\sqrt{1\ +\ 1}}\ for\ positive\ values\ of\ \ roots$$
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by DrMaths » Mon Jan 15, 2018 5:17 am
The answer must be NONE [A], because all answers include square roots which could generate a negative answer.
As $$\sqrt{x}\ =\ +\ or\ -\ \sqrt{x}$$
Therefore the "MUST" statement is contravened.
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by GMATGuruNY » Mon Jan 15, 2018 5:37 am
DrMaths wrote:The answer must be NONE [A], because all answers include square roots which could generate a negative answer.
As $$\sqrt{x}\ =\ +\ or\ -\ \sqrt{x}$$
Therefore the "MUST" statement is contravened.
This line of reasoning is incorrect.
On the GMAT, √ = the positive root ONLY.
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by GMATGuruNY » Mon Jan 15, 2018 5:43 am
Roland2rule wrote:If x and y are positive, which of the following must be greater than $$\frac{1}{\sqrt{x+y}}$$ ?

1. $$\frac{\sqrt{x+y}}{2x}$$

2. $$\frac{\sqrt{x}+\ \sqrt{y}}{x+y}$$

3. $$\frac{\sqrt{x}-\ \sqrt{y}}{x+y}$$

(A) None
(B) 1 only
(C) 2 only
(D) 1 and 3 only
(E) 2 and 3 only
Let x=1 and y=1.
1/√(x+y) = 1/√(1+1) = 1/√2.
Eliminate any expression not greater than 1/√2.

Scanning options I, II and III, we can quickly see that the numerator of III is equal to 0:
(√x-√y)/(x+y) = (√1-√1)/(1+1) = 0.
Eliminate any answer choice that includes III (D and E).

I: √(x+y)/2x
√(1+1)/(2*1) > 1/√2
√2/2 > 1/√2.
Cross-multiplying, we get:
√2 *√2 > 2*1
2>2.
Doesn't work.
Eliminate any remaining answer choice that includes I (B).

To compare fractions:
Multiply the denominator in each fraction by the NUMERATOR in the OTHER fraction.
The numerator that yields the greater product belongs to the greater fraction.

Multiplying the denominator in II by the numerator in the prompt, we get:
(x+y)(1) = x+y.
Multiplying the denominator in the prompt by the numerator in II, we get:
√(x+y) * (√x + √y)

= √(x² + xy) + √(xy + y²)

= √(more than x²) + √(more than y²)

= (more than x) + (more than y).

Since the product in blue is greater than the product in red, the numerator in II yields the greater product.
Thus, the fraction in II must be greater than the fraction in the prompt.
Eliminate any remaining answer choice that does not include II (A).

The correct answer is C.
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