abhicoolmax wrote:
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test
Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200
Is x^2 + y^2 > 100?(1) 2xy < 100 --> clearly insufficient: if \(x=y=0\) then the answer will be NO but if \(x=10\) and \(y=-10\) then the answer will be YES.
(2) (x + y)^2 > 200 --> \(x^2+2xy+y^2>200\). Now, as \((x-y)^2\geq{0}\) (square of any number is more than or equal to zero) then \(x^2+y^2\geq{2xy}\) so we can safely substitute \(2xy\) with \(x^2+y^2\) (as \(x^2+y^2\) is at least as big as \(2xy\) then the inequality will still hold true) --> \(x^2+(x^2+y^2)+y^2>200\) --> \(2(x^2+y^2)>200\) --> \(x^2+y^2>100\). Sufficient.
Answer: B.
Are you sure the OA is C?
I am feeling proud about myself
. I got just the ONE question wrong in my Veritas CAT test today, and this was the one. I marked
B as I proved it during the test - took 3.42 mins though. I should have got 100% correct otherwise. Good to know Veritas guys were wrong! I was a bit baffled by their explanation.
My proof (lengthy BUT conceptual proof):
(x+y)^2 > 200
=> |x + y| > 10*Sqrt(2)
=> For x^2 + y^2 to MINIMUM x=y : Why? Because squaring a number SPREADS the value exponentially. For given {x,y} such that x+y is known, x^2 + y^2 will ONLY spread MORE as we spread x and y away from their mean which is (x+y)/2 - regardless of the signs of x and y; in-fact opposite signs will spread the sum of squares even further. Hence x,y both MUST be > 5*sqrt(2). Hence MIN(x^2 + y^2) MUST be > 25*2 + 25*2 = 100. Hence,
x^2 + y^2 > 100.
If there is any Veritas Instructor here, please let me know if I got it wrong in some freakish way.
Your reasoning is fine. That's good thinking. Think of it in another way:
When 2 numbers are equal, their Arithmetic Mean = Geometric Mean
AM is least when it is equal to GM and GM is greatest when it is equal to AM.
So sum of the terms is least when the numbers are equal; product is maximum when they are equal.
For minimum value of \(x^2 + y^2\), we need \(x^2 = y^2\) or |x| = |y|
On the same line, if product is given to be constant, sum is minimum when numbers are equal.
If the sum is given to be constant, the product is maximum when the numbers are equal.