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Number Properties- help on this DS question

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Number Properties- help on this DS question

by riteshpatnaik » Mon Oct 17, 2016 6:26 am
Is xy≤1/2?

(1) x2+y2=1

(2) x2-y2=0

here also I chose E but the answer is A

Explanation given is (1) x2+y2=1. Recall that (x-y)2≥0(x-y)2≥0 (square of any number is more than or equal to zero). Expand: x2-2xy+y2≥0 and since x2+y2=1 then: 1-2xy≥0. So, xy≤1/2. Sufficient.

(2) x2-y2=0. Re-arrange and take the square root from both sides: |x|=|y|. Clearly insufficient.


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Source: — Data Sufficiency |

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by [email protected] » Mon Oct 17, 2016 10:42 am
Hi riteshpatnaik,

You've stated that you chose Answer E, but you did not explain WHY you chose that answer. If you explain your approach/logic or show your work, then I can help you to figure out what you did incorrectly.

We're asked if (X)(Y) is ≤ 1/2? This is a YES/NO question.

1) X^2 + Y^2 = 1

Since the sum of these two squares is 1, the individual values of X and Y have to be relatively small.

IF... X=0, Y=1, then (X)(Y) = 0 and the answer to the question is YES.

As the values of X and Y get closer and closer together, we still end up with a fraction that is less than or equal to 1/2...

IF... X=(root 1/2), Y=(root 1/2), then (X)(Y) = 1/2 and the answer to the question is YES.
The answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT

2) X^2 - Y^2 = 0

IF... X=(root 1/2), Y=(root 1/2), then (X)(Y) = 1/2 and the answer to the question is YES.
IF... X=2, Y=2, then (X)(Y) = 4 and the answer to the question is NO.
Fact 2 is INSUFFICIENT

Final Answer: A

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by Jay@ManhattanReview » Wed Dec 28, 2016 4:51 am
riteshpatnaik wrote:
Explanation given is (1) x2+y2=1. Recall that (x-y)2≥0(x-y)2≥0 (square of any number is more than or equal to zero). Expand: x2-2xy+y2≥0 and since x2+y2=1 then: 1-2xy≥0. So, xy≤1/2. Sufficient.
Please help
I think that you have a challenge in following the explanation of statement 1. Let us redo the same.

S1: x^2 + y^2 = 1.

Subtracting and adding 2xy, we get,

x^2 + y^2 - 2xy + 2xy = 1

=> (x-y)^2 + 2xy = 1

=> (x-y)^2 = 1- 2xy

Since (x-y)^2 is a non-negative number, (x-y)^2 ≥ 0.

=> 1 - 2xy ≥ 0

=> 1 ≥ 2xy

=> 1/2 ≥ xy

Sufficient!

Hope this helps!

-Jay

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