Inequality DS Question 1

[Param800]

Is xy < x^2*y^2?
1) xy>0
2) x+y=1

The OA is C

Please help me to understand a quick and easy method to solve such questions.
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[gander123]

Hey Param,

Is xy < x^2*y^2?
1) xy>0
2) x+y=1

Consider (1): Given xy > 0 we know that x and y must be either both positive or negative and x and y must not equal 0. However, x and y could still equal 1. If x = y = 1, then xy = x^2*y^2. On the other hand, if x = 3 and y = 4, then xy = 3*4 = 12 < 3^2*4^2 = 144. Statement 1 alone is NOT SUFFICIENT.

Consider (2): Given that x + y = 1, it follows that x = - y + 1. Taking x = 0 and y = 1 yields x*y = 0*1 = 0 = x^2*y^2 = 0^2*1^2 = 0. On the other hand, if x = -2 and y = 3, then x*y = -2*3 = -6 < -2^2*3^2 = 4 * 9 = 36. Statement 2 alone is NOT SUFFICIENT.

Taking (1) and (2) together, x + y = 1 can never be fulfilled if x and y were negative values. Thus, from (1) it follows that x and y both have to be positive, non-zero numbers. The only way in which to positive, non-zero nummbers add to 1 is x, y < 1. Thus both, x and y have to be fractions. Then for any pair of fractions x and y, x*y > x^2*y^2. Try this for 1/2 and 1/2 --> 1/4 > 1/16 and so on. Both Statements together SUFFICIENT.

HTH,

Tobi

[Param800]

Thank You Tobi