Is 1/x 1/y? (1) x 0

Question

Is  \frac{1}{x}>\frac{1}{y} ?

(1)  x < y 
(2)  x > 0

chetan2u · Answer

\frac{1}{x}-\frac{1}{y}>0.........\frac{y-x}{xy}>0 So two case A) if xy>0 that is both X and y are of same sign, y-x>0.....y>X B) if xy<0 that is both are of different sign, y-x<0.....y0 Nothing about y Insufficient Combined X0.... So 0>X>y So case A Sufficient C

BrentGMATPrepNow · Answer

Target question : Is 1/x > 1/y? This is a good candidate for rephrasing the target question . Take: 1/x > 1/y Subtract 1/y from both sides to get: 1/x - 1/y > 0 Rewrite with common denominators: y/xy - x/xy > 0 Combine: (y - x)/xy > 0 REPHRASED target question : Is (y - x)/xy positive? Statement 1 : x < y Let's TEST some values. There are several values of x and y that satisfy statement 1. Here are two: Case a : x = 1 and y = 2. In this case, (y - x)/xy = 1/2. So, the answer to the REPHRASED target question is YES, (y - x)/xy IS positive Case b : x = -1 and y = 2. In this case, (y - x)/xy = 3/-2 = -3/2. So, the answer to the REPHRASED target question is NO, (y - x)/xy is NOT positive Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT Statement 2 : x > 0 Since we have not information about y there's no way to answer the REPHRASED target question with certainty. Statement 2 is NOT SUFFICIENT Statements 1 and 2 combined Statement 1 tells us that x < y Statement 2 tells us that 0 < x When we combine the statements, we get: 0 < x < y If x < y, then y - x > 0. In other words, (y - x) is POSITIVE Also, if x and y are both greater than 0, we know that the product xy is POSITIVE So, (y - x)/xy = POSITIVE/POSITIVE = POSITIVE So, the answer to the REPHRASED target question is YES, (y - x)/xy IS positive Since we can answer the REPHRASED target question with certainty, the combined statements are SUFFICIENT Answer: C Cheers, Brent RELATED VIDEO FROM OUR COURSE https://www.youtube.com/watch?v=MyG1K3ee69w

MathRevolution · Answer

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The question may be modified as follows: 
 \frac{1}{x} >\frac{1}{y} 
=>  xy^2 > x^2y  by multiplication by  x^2y^2 
=>  xy^2 - x^2y > 0 
=>  xy(y-x) > 0

Since we have 2 variables ( x  and  y ) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Applying both conditions together yields  y > x > 0 . So,  x>0, y>0  and  y-x>0. 
It follows that the product  xy(y – x)  is positive. Therefore, both conditions are sufficient when considered together.

Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
If  x = 2  and  y = 3 , then  \frac{1}{x} = \frac{1}{2} ,  \frac{1}{y} = \frac{1}{3} , and the answer is ‘yes’.
If  x = -2  and y = 3 , then  \frac{1}{x} = -\frac{1}{2} ,  \frac{1}{y} = \frac{1}{3} , and the answer is ‘no’.
Condition 1) is not sufficient on its own.

Condition 2)
This condition provides us with no information about the variable y, so it is not sufficient.

Therefore, the answer is C.
Answer: C

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provides an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

PLUTO · Answer

Hang on a minute. As per inequalities formula, if y>x, then 1/x > 1/y. So if I see a statement "Is 1/x>1/y?" why can i not assume that the target question is "if y>x"?

randomavoidplease · Answer

Can anyone answer this, because even I am confused.

chetan2u · Answer

Hi, 
You cannot do this because you are not sure whether x and y are positive or negative..
Say x =2 and y=-3....1/2>1/-3, but does it mean -3>2..NO

SO NEVER, cross-multiply in an inequality unless you are sure that the variable is positive. If it is negative, the inequality sign will change..

randomavoidplease · Answer

But when we are multiplying with a negative on the other side, we would change the inequality sign as well. Which would lead to -3<2 . Wouldn't it?

ARIEN3228 · Answer

S(1) alone insufficient.

Two cases arise x and y can be positive , or negative.

S(2) alone insufficient.
Nothing about y.

S1 and S2 sufficient.

We know x>0.

This also means that y is always be positive.
Ans : C

chetan2u · Answer

Yes, you are correct, and that is why you do not cross multiply in 1/x>1/y to get y>x, as you do not know whether x or y is negative or positive.

ueh55406 · Answer

Ok,  chetan2u   BrentGMATPrepNow , I have a doubt.

1) x < y 
 2) x > 0

let x= 1/2 and y= 2

is  1/1/2 > 2 ? 
 2>2  No.

let x= 3 and y=4

is  1/3 >1/4  yes.

should be E, right ?

BrentGMATPrepNow · Answer

Not quite. I have highlighted the mistake above.

It should be:  is  \frac{1}{\frac{1}{2}} > \frac{1}{2} ?  
  2>\frac{1}{2}  YES

ueh55406 · Answer

Arghh ! darn it ! Silly mistakes all over again :/

IanStewart · Answer

Using Statement 1 alone, certainly x can be negative, and y positive, and then 1/x < 1/y for sure. But if x = 1 and y = 2, then 1/x > 1/y. So Statement 1 is not sufficient, and Statement 2 clearly is not alone. Together, we know y > x > 0, and since we now know the letters are all positive, we can safely divide or multiply by them (we don't need to worry about whether to reverse the inequality now). So we can just divide by xy on both sides of y > x, to get 1/x > 1/y, and the answer is C.

I'd add that if a test taker is "counting equations and counting unknowns" in inequality questions that look like this one, as one post above suggests doing, that test taker will do just as well on real GMAT questions by guessing randomly. We don't even have equations here, so the principle doesn't even make sense in this context.

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Is 1/x > 1/y? (1) x < y (2) x > 0

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