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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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New post 02 Mar 2015, 07:20
3
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A
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C
D
E

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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 08 Mar 2015, 14:51
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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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Re: If x and y are positive integers, is x/y < (x+2)/(y+3)?  [#permalink]

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New post 02 Mar 2015, 11:50
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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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New post 02 Mar 2015, 16:43
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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.


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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New post 03 Mar 2015, 02:03
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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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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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New post 28 Aug 2017, 23:34
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]

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New post 18 Dec 2017, 18:46
msk0657 wrote:
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..


sorry can you explain why we need to simplify the x/y < (x+2)/(y+3) to become 3x < 2y, does if we just put the number directly we could get the answear.
Thanks
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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 18 Dec 2017, 21:52
ryanpri wrote:
msk0657 wrote:
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..


sorry can you explain why we need to simplify the x/y < (x+2)/(y+3) to become 3x < 2y, does if we just put the number directly we could get the answear.
Thanks



Hi

The word simplification in almost every task does what its supposed to do: - it 'simplifies' the question or makes our task 'simple'.

Eg, if here we look at statement 1, and try to plug in the values straightaway, we will have to put a value of y>20 in the question. So we will write: LHS = x/21. RHS = (x+2)/24
We can see that this on its own doesnt tell us anything unless we know something about x. Or we will have to cross multiply to get 24x on LHS and 21x+42 on RHS.

From statement 2, similarly we will take a value of x less than 5, say 4. So we will substitute x=4 and then again this statement will become insufficient unless we know something about y.

But as you can see, this method of substituting numbers is slightly complicated. Instead if we can just simplify the question stem (as explained by others), all we have to then find is whether x/y < 2/3. And for this we need to know something about both x as well as y. So individual statements are not sufficient on their own.
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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 18 Dec 2017, 23:08
Hi

The word simplification in almost every task does what its supposed to do: - it 'simplifies' the question or makes our task 'simple'.

Eg, if here we look at statement 1, and try to plug in the values straightaway, we will have to put a value of y>20 in the question. So we will write: LHS = x/21. RHS = (x+2)/24
We can see that this on its own doesnt tell us anything unless we know something about x. Or we will have to cross multiply to get 24x on LHS and 21x+42 on RHS.

From statement 2, similarly we will take a value of x less than 5, say 4. So we will substitute x=4 and then again this statement will become insufficient unless we know something about y.

But as you can see, this method of substituting numbers is slightly complicated. Instead if we can just simplify the question stem (as explained by others), all we have to then find is whether x/y < 2/3. And for this we need to know something about both x as well as y. So individual statements are not sufficient on their own.[/quote]

Thanks for the explanation,
btw if we crop multiply we can get
Y=21
24x>21x+42

can i move the "21x" like this:
24x-21x>42
x=42/3

need your enlighment.thanks
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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 19 Dec 2017, 01:05
ryanpri wrote:
Hi

The word simplification in almost every task does what its supposed to do: - it 'simplifies' the question or makes our task 'simple'.

Eg, if here we look at statement 1, and try to plug in the values straightaway, we will have to put a value of y>20 in the question. So we will write: LHS = x/21. RHS = (x+2)/24
We can see that this on its own doesnt tell us anything unless we know something about x. Or we will have to cross multiply to get 24x on LHS and 21x+42 on RHS.

From statement 2, similarly we will take a value of x less than 5, say 4. So we will substitute x=4 and then again this statement will become insufficient unless we know something about y.

But as you can see, this method of substituting numbers is slightly complicated. Instead if we can just simplify the question stem (as explained by others), all we have to then find is whether x/y < 2/3. And for this we need to know something about both x as well as y. So individual statements are not sufficient on their own.


Thanks for the explanation,
btw if we crop multiply we can get
Y=21
24x>21x+42

can i move the "21x" like this:
24x-21x>42
x=42/3

need your enlighment.thanks[/quote]

Hi Ryan

In this particular question, you are being asked Whether x/y < (x+2)/(y+3) . We are not given this already.

But suppose we are already given that x/y < (x+2)/(y+3) and then we decide to play around with this by substituting the value of y as 21. Then we will write:

x/21 < (x+2)/24. Then cross multiplying 24x < 21x + 42. After that Yes we can do as you did. We will write:
24x-21x < 42 or 3x < 42 or x < 42/3 or x < 14 (you have written x=42/3, its not equal it will be x < 42/3)

I suggest you first go through the basics of algebra and inequalities.

You can go to the following thread by Bunuel and go through the topics of your choice.

https://gmatclub.com/forum/ultimate-gma ... 44512.html

In this particular case, I was talking about Topic no 7 (algebra) and Topic no 9 (Inequalities)
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Re: If x and y are positive integers, is x/y < (x+2)/(y+3)? &nbs [#permalink] 19 Dec 2017, 01:05
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