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If x and y are positive integers, is x/y < (x+2)/(y+3)?

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

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

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Given: x,y >0; x,y are integers.
Question: is x/y<(x+2)/(y+3)? or if we cross-multiply (we can do this because x and y are positive) and simplify we get is 3x<2y?

(1) Insufficient. If x is small then the answer to the question will be yes, but if x is very big and y=21, then answer will be no.
(2) Insufficient. No info about y, so y can be either very small or very big which will yield two different answers.

(1)+(2) Sufficient. If y=21 and x=4, then we get 12< 42, so the answer to the question is no.

Answer C

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

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Bunuel wrote:
If x and y are positive integers, is x/y < (x+2)/(y+3)?

(1) y > 20
(2) x < 5

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If we simplify the original prompt the question reworded becomes 3x < 2y ?

Statement 1: nothing is given about x; insufficient

Statement 2: nothing is given about y; insufficient

Combined, we know that 3x is always smaller than 2y because y is 21 at its smallest value. x is only 4 at its greatest value.
Sufficient

Answer: C

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

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Bunuel wrote:
If x and y are positive integers, is x/y < (x+2)/(y+3)?

(1) y > 20
(2) x < 5

Kudos for a correct solution.


Answer should be C.
As i thought, this problem cannot be dependent on value of any one of the variable. I mean even if y > 20, for x = y given equation does not hold true, and conclusion varies for condition x < y and x > y. So, option A, B and D are out.
So, we need values or range for x and y both, and keep an eye on relation between x and y, since the validity of the equation varies depending on conditions: x = y, x > y and x < y.
For the given ranges combined, we get only one relation between x and y, i.e. y > x.
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Re: If x and y are positive integers, is x/y < (x+2)/(y+3)? [#permalink]

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Bunuel wrote:
If x and y are positive integers, is x/y < (x+2)/(y+3)?

(1) y > 20
(2) x < 5

Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

Statement #1: We are adding 2 to the numerator and 3 to the denominator, so we know the resultant fraction will move closer to 2/3. If all we know is that the denominator of the starting fraction is greater than 20, then we have no idea what the size of the starting fraction is: it could be much greater than 2/3, or much smaller than 2/3, depending on the numerator, of which we have no idea. We can draw no conclusion right now. This statement, alone, by itself, is insufficient.

Statement #2: Now, all we know is that the numerator of the starting fraction is less than 5 — it could be 4, 3, 2, or 1. We have no idea of the denominator. If y = 50, then we get a very small fraction. But if x = 4 and y = 1, the fraction equals 4, much larger than 2/3. In this statement, we have no information about the denominator, and since we know nothing about the denominator, we know nothing about the size of the starting fraction: it could be either greater or less than 2/3. Therefore, we can draw no conclusion. This statement, alone, by itself, is also insufficient.

Now, combine the statements. We know y > 20 and x < 5. Well no matter what values we choose, we are going to have a denominator much smaller than the numerator. The larger possible fraction we could have under these constraints would be 4/21 (largest possible numerator with smallest possible denominator). The fraction 4/21 is much smaller than 1/2, so it’s definitely smaller than 2/3. Any fraction with y > 20 and x < 5 will be less than 2/3. Therefore, adding 2 to the numerator and 3 to the denominator will move the resultant fraction closer to 2/3, which has the net effect of increasing its value. Therefore, the answer to the prompt question is “yes.” Because we can give a definite answer to the prompt, we have sufficient information.

Neither statement is sufficient individually, but together, they are sufficient. Answer = C.
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If x and y are positive integers, is x/y < (x+2)/(y+3)? [#permalink]

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New post 15 Aug 2016, 01:59
If x and y are positive integers, is x/y < (x+2)/(y+3)?

(1) y > 20
(2) x < 5
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Re: If x and y are positive integers, is x/y < (x+2)/(y+3)? [#permalink]

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stonecold wrote:
1) If x and y are positive integers, is {x/y} <{{x+2}/{y+3}}?

Statement #1: y > 20
Statement #2: x < 5



HAPPY INDEPENDENCE DAY...STONY..

Given {x/y} <{{x+2}/{y+3}}

cross multiply or mulitiply individual variables we get the same equation...

x(y+3) < y(x+2)
xy + 3x < yx + 2y

3x < 2y
=> x/y < 2/3 ? we need to find this.

Stat 1: y > 20..no information about x...not sufficient...

Stat 2: x < 5..no information about y...not sufficient..

Stat 1 + Stat 2: Now take least y value and greater number from x i.e y = 21 and x = 4 ...since the value is not divisible consider 4/24 = 1/6 < 2/3.

Similarly consider some other value let x = 3 and y = 18..these values are exactly divisible, hence consider 1/6 < 2/3

or x = 4 and y = 21 we get result < 2/3....Sufficient..

Hence C is sufficient..

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

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

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New post 29 Aug 2017, 00:34
Option C, since the fraction gets closer to 2/3, we need to know values of x and Y to know if 2/3 is greater than or less than x/y.
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Re: If x and y are positive integers, is x/y < (x+2)/(y+3)?   [#permalink] 29 Aug 2017, 00:34
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