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# If x and y are positive integers, is (2 + x)/(3 + y) greater

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If x and y are positive integers, is (2 + x)/(3 + y) greater [#permalink]

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12 Mar 2013, 03:48
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If x and y are positive integers, is (2 + x)/(3 + y) greater than (2 + y)/(3 + x)?

(1) x + y = 3
(2) x > y

[Reveal] Spoiler:
Can we cross multiply in this question? Suppose it wasn't given positive then we move the variables to one side and solve?
[Reveal] Spoiler: OA

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Last edited by hazelnut on 04 May 2017, 04:07, edited 2 times in total.
EDITED THE QUESTION.

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Re: If x and y are positive integers, is (2 + x)/(3 + y) greater [#permalink]

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12 Mar 2013, 04:08
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If x and y are positive integers, is (2 + x)/(3 + y) greater than (2 + y)/(3 + x)?

Is $$\frac{2 + x}{3 + y}>\frac{2 + y}{3 + x}$$? Since both denominators are positive we can safely cross-multiply. Though we can solve the question without doing that.

(1) x + y = 3. If x=1 and y=2, then the answer is NO (3/5<4/4) but if x=2 and y=1, then the answer is YES (4/4>3/5). Not sufficient.

(2) x > y. This implies that the numerator of LHS is more than the numerator of RHS, and the denominator of LHS is less than the denominator of RHS, which means that LHS > RHS. Sufficient.

Hope it's clear.
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Re: If x and y are positive integers, is (2 + x)/(3 + y) greater [#permalink]

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17 Mar 2013, 08:36
Bunuel wrote:
If x and y are positive integers, is (2 + x)/(3 + y) greater than (2 + y)/(3 + x)?

Is $$\frac{2 + x}{3 + y}>\frac{2 + y}{3 + x}$$? Since both denominators are positive we can safely cross-multiply. Though we can solve the question without doing that.

(1) x + y = 3. If x=1 and y=2, then the answer is NO (3/5<4/4) but if x=2 and y=1, then the answer is YES (4/4>3/5). Not sufficient.

(2) x > y. This implies that the numerator of LHS is more than the numerator of RHS, and the denominator of LHS is less than the denominator of RHS, which means that LHS > RHS. Sufficient.

Hope it's clear.

Dear Bunuel,

Can you please elaborate on an Algebra approach?

So far I'd go like this:

Since we know that the denominators are positive, we can cross multiply:

$$x^2+5x<y^2+5y$$

$$x^2-y^2<5y-5x$$

$$(x+y)(x-y)<5(y-x)$$

$$(x+y)(x-y)<-5(x-y)$$

Here, can we divide by (x-y)? If not, how to continue?

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Re: If x and y are positive integers, is (2 + x)/(3 + y) greater [#permalink]

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18 Mar 2013, 22:57
Not sure if you would really like to take the algebra approach.

The question is pretty much clear about the usage of positive integer. So lets take a small set of positive integers {1,2,3,4}

Now, for us to get 2+x > 3+y we need to have only x>y; thats the only condition which can help us solve the equation. Since, that's given in B! Hence, B is the answer
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Re: If x and y are positive integers, is (2 + x)/(3 + y) greater [#permalink]

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20 Mar 2013, 03:02
[quote="fozzzy"]If x and y are positive integers, is (2 + x)/(3 + y) greater than (2 + y)/(3 + x)?

(1) x + y = 3
(2) x > y

x,y +ve intigers

from 1

x,y are in fact 1,2 but we dont know which is which...insuff

from 2

if x>y then : numerator 2+x >2+y ( numerator of each side) and 3+y<3+x (denominator of each side), thus larger numerator/smaller denominator is surely > smaller numerator/ larger denominator ..hope this makes sense

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Re: If x and y are positive integers, is (2 + x)/(3 + y) greater [#permalink]

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20 Mar 2013, 05:06
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LGOdream wrote:
Bunuel wrote:
If x and y are positive integers, is (2 + x)/(3 + y) greater than (2 + y)/(3 + x)?

Is $$\frac{2 + x}{3 + y}>\frac{2 + y}{3 + x}$$? Since both denominators are positive we can safely cross-multiply. Though we can solve the question without doing that.

(1) x + y = 3. If x=1 and y=2, then the answer is NO (3/5<4/4) but if x=2 and y=1, then the answer is YES (4/4>3/5). Not sufficient.

(2) x > y. This implies that the numerator of LHS is more than the numerator of RHS, and the denominator of LHS is less than the denominator of RHS, which means that LHS > RHS. Sufficient.

Hope it's clear.

Dear Bunuel,

Can you please elaborate on an Algebra approach?

So far I'd go like this:

Since we know that the denominators are positive, we can cross multiply:

x^2+5x<y^2+5y

$$x^2-y^2<5y-5x$$

$$(x+y)(x-y)<5(y-x)$$

$$(x+y)(x-y)<-5(x-y)$$

Here, can we divide by (x-y)? If not, how to continue?

First of all, is (2 + x)/(3 + y) greater than (2 + y)/(3 + x)? Means is $$\frac{2 + x}{3 + y}>\frac{2 + y}{3 + x}$$?

Cross-multiply: is $$(2+x)(3+x)>(2+y)(3+y)$$ --> is $$5x+x^2>5y+y^2$$? --> is $$(x-y)(x+y)>-5(x-y)$$? Here we cannot divide by x-y, since we don't know whether it's positive or negative.

What we can do is: $$(x-y)(x+y)>-5(x-y)$$? --> $$(x-y)(x+y)+5(x-y)>0$$? --> $$(x-y)(x+y+5)>0$$?

(1) x + y = 3. The question becomes: is $$(x-y)(3+5)>0$$? --> is $$x-y>0$$? We don't know that, thus this statement is not sufficient.

(2) x > y --> $$x-y>0$$. So, we can reduce by x-y and the question becomes: is $$x+y+5>0$$? Since x and y are positive then the answer to this question is YES. Sufficient.

Hope it helps.
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Re: If x and y are positive integers, is (2 + x)/(3 + y) greater [#permalink]

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08 Jun 2014, 06:40
Thanks for the elaboration, really helpful !

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Re: If x and y are positive integers, is (2 + x)/(3 + y) greater [#permalink]

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08 Jun 2014, 06:59
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Great! Explanation and elaboration really helpful !

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Re: If x and y are positive integers, is (2 + x)/(3 + y) greater [#permalink]

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31 Dec 2015, 04:34
Hello Bunuel,

Could you please assist me with this question:

(x−y)(x+y) > 5y- 5x ........(I)

Statement I- X+ Y =3

Put value in (I)

(X-Y)3 > 5Y-5X
3X - 3 Y > 5Y- 5X

8 X > 8 Y

or

X > Y

Which is same as statement 2. I understand statement 2 is sufficient. But with the solution I just mentioned I picked D. Please assist.

Thanks

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Re: If x and y are positive integers, is (2 + x)/(3 + y) greater [#permalink]

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Re: If x and y are positive integers, is (2 + x)/(3 + y) greater   [#permalink] 18 Jan 2017, 18:18
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