Is x x/y (2) y 100

Question

Is x < y ?

(1) y/x > x/y
(2) y > 100

mikemcgarry · Answer

Dear goodyear2013 I'm happy to help. :-)

This is a very tricky one. You may find this blog on arithmetic with inequalities helpful: https://magoosh.com/gmat/2013/gmat-quant ... qualities/ Statement #1 : y/x > x/y If x & y are both positive, then we can simply cross multiply, (y^2) > (x^2), and take a square root, y > x. But, of course, the catch is --- we don't know whether x or y is positive. Suppose y = -5 and x = -2. Then Statement #1 is true, but x > y This is that tricky idea that a negative number with a smaller absolute value is larger than a negative number with a larger absolute value. It's better to have $500 in your checking account than $100, but it's better to have $100 in debt on your credit card than $500. We can't answer the prompt with this alone. This statement, alone and by itself, is insufficient . Statement #2 : y > 100 Well, this one simply gives us no constraints on x at all, so x could be anything. We don't know. This statement, alone and by itself, is insufficient . Combined statements Think of this way Case One: x is negative --- then, since we know y is positive, any positive number is greater than any negative number. Therefore, it must be true that y > x. Case Two: x is positive. Then, as in statement #1 above, we can cross-multiply in the inequality, take a square-root, and get y > x. Thus, regardless of where x falls, y must be greater than x. Combined the statements are sufficient . OA = (C) Does all this make sense? Mike :-)

Bunuel · Answer

Is x < y ? (1) y/x > x/y. If y=2 and x=1 , then xy . Not sufficient. Notice that we cannot cross-multiply given inequality because we don't know the signs of x and y. (2) y > 100. We know nothing about x. Not sufficient. (1)+(2) If x is a negative number then obviously: (y=positive)>(x=negative) ; If x is a positive number (so if both x and y are positive) then we can cross-multiply (1) and get: y^2>x^2 --> take the square root: |y|>|x| --> y>x . Sufficient. Answer: C.

nide · Answer

1. y/x - x/y >0 y^2-x^2/xy >0 i. y^2-x^2>0 and xy>0 a. if xy>0 then x>0 and y>0 and the values of x,y should satisfy (x+y)(y-x)>0 => since x,y>0 x+y>0 => y-x >0 => y>x b. if xy>0 then x<0 and y<0 and the values of x,y should satisfy (x+y)(x-y)>0 => since x,y<0 x+y<0 => x-y <0 => x0 and y<0 => y0 => x0 => insufficient 1 and 2. y>100 => y>0 from ia,ib,iia,iib i. y^2-x^2>0 and xy>0 a. if xy>0 then x>0 and y>0 and the values of x,y should satisfy (x+y)(y-x)>0 => since x,y>0 x+y>0 => y-x >0 => y>x b. if xy>0 then x<0 and y<0 and the values of x,y should satisfy (x+y)(x-y)>0 => since x,y<0 x+y<0 => x-y <0 => x0 and y<0 => y0 => x

CrackverbalGMAT · Answer

This is a moderately difficult question on Inequalities. Beware of the trap element though – you cannot cross multiply the terms in an inequality until unless you are absolutely sure of the signs of the variables.

Let’s look at statement I alone. To simplify the inequality, let us take all the variable terms on to the LHS and keep the RHS as 0 (this is a sure shot strategy that you can adopt in inequalities’ questions to make your life simpler). When we do that, we have,

$\frac{y}{x}$ – $\frac{x}{y}$ > 0 which gives

$\frac{y^2 – x^2}{xy}$ > 0

$\frac{(y-x)(y+x)}{xy}$ > 0

If the LHS is positive, it only means:

Either, (y-x), (y+x) and xy should all be positive OR negative.

If we take y = 5 and x =4 , then $\frac{(5-4) (5+4)}{20}$ > 0. Here, x < y.

If we take y = -5 and x = -4, then $\frac{(-5+4) (-9)}{20}$ > 0. Here x > y.

This data is insufficient. Answer options A and D can be eliminated. Possible answer options are B, C and E.

From statement II alone, we have data about y only. We do not have any information about x. Therefore, this statement is insufficient. Answer option B can be eliminated.

Let’s combine both the statements now, by taking some simple values that satisfy both statements together.

If y = 200 and x = -400, $\frac{y}{x}$ = $\frac{-1}{2}$ and $\frac{x}{y}$ = -2. $\frac{y}{x}$ > $\frac{x}{y}$. Here, x < y ( remember, x is negative and y is positive).

If y = 200 and x = 10, $\frac{y}{x}$ = 20 and $\frac{x}{y}$ = $\frac{1}{20}$.$\frac{y}{x}$ > $\frac{x}{y}$. Here also, x<y.

Therefore, we get a definite YES as an answer when we combine the statements. So, the correct answer option is C.

The average time for this kind of a question can be about 2 minutes.

Hope this helps!

Is x < y ? (1) y/x > x/y (2) y > 100

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