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

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Re: Is 1/x > 1/y? (1) x < y (2) x > 0 [#permalink]
=>

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.

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.
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Re: Is 1/x > 1/y? (1) x < y (2) x > 0 [#permalink]
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

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

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"?
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Re: Is 1/x > 1/y? (1) x < y (2) x > 0 [#permalink]
PLUTO wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

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

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"?

Can anyone answer this, because even I am confused.
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Re: Is 1/x > 1/y? (1) x < y (2) x > 0 [#permalink]
beingink wrote:
PLUTO wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

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

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"?

Can anyone answer this, because even I am confused.

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..
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Re: Is 1/x > 1/y? (1) x < y (2) x > 0 [#permalink]
chetan2u wrote:
beingink wrote:
PLUTO wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

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

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"?

Can anyone answer this, because even I am confused.

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..

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?
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Re: Is 1/x > 1/y? (1) x < y (2) x > 0 [#permalink]
S(1) alone insufficient.

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

S(2) alone insufficient.

S1 and S2 sufficient.

We know x>0.

This also means that y is always be positive.
Ans : C
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Re: Is 1/x > 1/y? (1) x < y (2) x > 0 [#permalink]
beingink wrote:
chetan2u wrote:

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..

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?

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.
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Re: Is 1/x > 1/y? (1) x < y (2) x > 0 [#permalink]
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

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

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 ?
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Re: Is 1/x > 1/y? (1) x < y (2) x > 0 [#permalink]
Top Contributor
ueh55406 wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

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

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 ?

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
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Re: Is 1/x > 1/y? (1) x < y (2) x > 0 [#permalink]
BrentGMATPrepNow wrote:
ueh55406 wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

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

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

Arghh ! darn it ! Silly mistakes all over again :/
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Re: Is 1/x > 1/y? (1) x < y (2) x > 0 [#permalink]
1
Kudos
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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Re: Is 1/x > 1/y? (1) x < y (2) x > 0 [#permalink]
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Re: Is 1/x > 1/y? (1) x < y (2) x > 0 [#permalink]
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