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Bunuel
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.

Answer: B.

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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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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[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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Thanks for the elaboration, really helpful !
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Great! Explanation and elaboration really helpful !
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Bunuel
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.

Answer: B.

Hope it's clear.

I took a different approach but I'm curious if it it's fail proof or I lucked out:

First, I cross multiplied "up" to read "Is (2 +x)(3+x) > (2+y)(3+y)?"

Then I used FOIL to read "Is x^2 +5x +6 > y^2 +5y+6?" or essentially "Is x>y? " (FOILing may have been unnecessary but I was quickly able to see the x>y this way)

Statement 1:
X+Y= 3 ; NS as x can be either 1 or 2; same for y

Statement 2:
x>y; This explicitly answers my question SUFFICIENT
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one can also just add one to both sides of the inquality

(2+x)/3+y > 2+y / 3+x

adding one to both sides

(2+x) (3+y) /(3+y) > (2+y) (3+x) / (3+x)

=> transpose and cancel the numerators

3+x>3+y
question stem becomes
is x>y ?

B/D are possible options
check for 1

final answer is B
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Here it is given : (2 + x)/(3 + y) > (2 + y)/(3 + x) where they are positive integers

Adding 1 on both sides we get :

(2+x)/(3+y) + 1 > (2+y)/(3+x) + 1
(2+x+y+3)/(3+y) > (2+y+x+3)/(3+x) => (5+x+y)/(3+y) > (5+x+y)/(3+x) we can divide (5+x+y) on both sides

1/((3+y) > 1/(3+x) ..cross multiply we get (3+x) > (3+y) => x>y
so the actual question is is x>y ?

(1) x + y = 3 .. Not sufficient as we cant say which is greater
(2) x > y this only we need -- sufficient

Ans B
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Let’s break down the question stem and try to simplify it.
Is \(\frac{(2+x) }{ (3+y)}\) > \(\frac{(2+y) }{ (3+x)}\)?

Since x and y are positive integers, all the expressions are positive; therefore, we can cross multiply without having to worry about changing the inequality sign. When we do so, the question stem can be rephrased as,

Is (2+x) (3+x) > (2+y) (3+y)?

Simplifying both sides of the expression, the question stem now becomes,

Is \(x^2\) + 5x + 6 > \(y^2\) + 5y + 6?

Simplifying further, we have,

Is \(x^2\) – \(y^2\) + 5x – 5y >0 ?
Or.

Is (x-y) (x+y) + 5(x-y) > 0?

Factoring (x-y) as it is common to both terms, we have,
Is (x-y) (x+y+5) > 0?

The product of the two expressions will be greater than zero only if both are positive or both are negative. However, since the question data is clear that both x and y are positive integers, (x+y+5) cannot be negative. This being the case, the entire expression will be greater than 0 if (x-y) >0.

So, the final stage is
Is (x-y) > 0 OR Is x > y?

This is the question we need to now answer using the statements.

From statement I alone, x+y = 3.

If x = 2, y = 1 and x > y; the question can be answered with a Yes
If x = 1, y = 2 and x < y; the question can be answered with a No.

Statement I alone is insufficient. Answer options A and D can be eliminated.

From statement II alone, x>y.

This is a direct YES to the question asked.
Statement II alone is sufficient to answer the question with a definite Yes. Answer options C and E can be eliminated.

The correct answer option is B.

Hope that helps!
Aravind BT
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