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If x^2-y^2=1, x=?

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by Max@Math Revolution » Thu Jan 28, 2016 5:53 pm
If x^2-y^2=1, x=?

1) x+y=1/2
2) y^2=9/16


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by manik11 » Fri Jan 29, 2016 6:20 am
The question can be rephrased as (x+y)(x-y) = 1 , x=?

Statement 1 : x+y=1/2

If x+y=1/2 then (x-y) has to be 2.
Now we have two equations and two unknowns. Solvable for x. --> SUFFICIENT
If we chose to solve for X it would equal 5/4

Eliminate B,C,E

Statement 2 : y^2=9/16

Substitute the value of y^2 in original prompt (x^2-y^2=1)
We would get:
x^2-9/16=1
--> x^2=25/16
--> x = +5/4 or -5/4
--> INSUFFICIENT
Eliminate D

Answer : A
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by Max@Math Revolution » Sun Jan 31, 2016 6:26 pm
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x^2-y^2=1, x=?
1) x+y=1/2
2) y^2=9/16


In the original condition, there are 2 variables(x,y) and 1 equations(x2-y2=1), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), (x-y)(x+y)=1 -> x-y=2 and x+y=1/2 -> x=5/4, which is unique and sufficient.
For 2), y=3/4, -3/4, which is not unique and nor is x. Since it is not sufficient, the answer is A.
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