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X=Y

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by vinay1983 » Thu Sep 26, 2013 9:14 pm
Is x=y?

1. (x+y)* [1/x + 1/y] = 4

2. (x-50)^2 = (y-50)^2
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Source: — Data Sufficiency |

by [email protected] » Thu Sep 26, 2013 10:23 pm
Hi vinay1983,

This DS question combines a variety of algebra steps (and Classic Quadratics) and TESTing values.

The question: Is x = y? This is a YES/NO question.

Fact 1: (x + y){1/x + 1/y] = 4

Lots of algebra steps to simplify this:

(x + y)[y/xy + x/xy] = 4
(x + y)[(x + y)/xy] = 4
(x + y)^2 = 4xy
x^2 + 2xy + y^2 = 4x
x^2 - 2xy + y^2 = 0
(x - y)^2 = 0

So, x MUST = y
The answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT

Fact 2: (x - 50)^2 = (y - 50^2

Here we can TEST values:

If x = 0 and y = 0 then the answer to the question is YES
If x = 0 and y = 100 then the answer to the question is NO
Fact 2 is INSUFFICIENT

Final Answer: A

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Rich
Contact Rich at [email protected]
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by GMATGuruNY » Fri Sep 27, 2013 2:33 am
vinay1983 wrote:Is x=y?

1. (x+y)* [1/x + 1/y] = 4

2. (x-50)^2 = (y-50)^2
Plug in a value for x and solve for y.

Statement 1: (x+y)* [1/x + 1/y] = 4
Case 1: x=1
(1+y) * (1/1 + 1/y) = 4
1 + 1/y + y/1 + 1 = 4
1/y + y/1 = 2.
Only y=1 will satisfy this equation.
In this case, x=y.

Case 2: x=5
(5+y) * (1/5 + 1/y) = 4
1 + 5/y + y/5 + 1 = 4
5/y + y/5 = 2.
Only y=5 will satisfy this equation.
In this case, x=y.

The implication of these two random cases is that -- regardless of the value of x -- x=y.
SUFFICIENT.

Statement 2: (x-50)² = (y-50)²
If x=51, we get:
(51-50)² = (y-50)²
1 = (y-50)²
y-50 = 1 or y-50 = -1
y=51 or y=49.
Since it's possible that x=y or that x≠y, INSUFFICIENT.

The correct answer is A.

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by Java_85 » Fri Sep 27, 2013 1:19 pm
IMO also A is the answer. didn't get until I wrote the equations :)
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by vinay1983 » Fri Sep 27, 2013 5:14 pm
Thank you Rich and Mitch(Sort of rhyming isn't it?)

Actually I did what Rich as shown as the method. I got A, but was sceptical about this

(x-y)^2 = 0 Whether I could conclude that x=y.

Now satisfied!
You can, for example never foretell what any one man will do, but you can say with precision what an average number will be up to!
Join the discussion