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# If a, c, d, x, and y are positive integers such that ay < x and c/d is

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If a, c, d, x, and y are positive integers such that ay < x and c/d is  [#permalink]

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04 May 2015, 06:13
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If a, c, d, x, and y are positive integers such that ay < x and c/d is the lowest-terms representation of the fraction (x+ay)/(x-ay) , then c is how much greater than d? (If c/d is an integer, let d = 1.)

(1) x/y is an odd integer.

(2) a = 4

Kudos for a correct solution.

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Re: If a, c, d, x, and y are positive integers such that ay < x and c/d is  [#permalink]

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07 May 2015, 22:39
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Bunuel wrote:
If a, c, d, x, and y are positive integers such that ay < x and c/d is the lowest-terms representation of the fraction (x+ay)/(x-ay) , then c is how much greater than d? (If c/d is an integer, let d = 1.)

(1) x/y is an odd integer.

(2) a = 4

Kudos for a correct solution.

This question is an application of the concept of factors. Too many variables has made the question cumbersome but its logic is good to know.

To understand it, think: Will 247 and 251 have any common factors (other than 1)? Don't try to factorize them. Both numbers are odd so they have only odd factors. 251 is 4 away from 247. Any odd factor of 247 will not be a factor of 251 because 4 does not have any odd factors. For example, 3 doesn't go into 4 so if 3 were a factor of 247, it could not be a factor of 251. Similarly, 247 and 255 can have no common factors because these two odd numbers have a difference of 8 in them. On the same lines, 247 and 253 could share a common factor (3) because they are 6 apart. If 3 were a factor of 247, it would be a factor of 253 too - (it is not but that doesn't concern us).

Now go on to this question: If a, c, d, x, and y are positive integers - good
such that ay < x - what is the relevance of this? Note we have a factor x - ay in the denominator. This means that x - ay is positive. x + ay will obviously be positive so basically, it's a positive ratio with larger number as the numerator and smaller as the denominator.

$$\frac{x+ay}{x - ay} = \frac{c}{d}$$

We need c - d.

(1) x/y is an odd integer.
Multiply and divide left fraction by y.

$$\frac{x/y+a}{x/y - a}= \frac{c}{d}$$

$$\frac{Odd+a}{Odd - a}= \frac{c}{d}$$
Value of c - d will depend on value of Odd and a. Insufficient

(2) a = 4

$$\frac{x+4y}{x - 4y} = \frac{c}{d}$$
Value of c - d will depend on value of x and y. Insufficient

Both together,

$$\frac{Odd+4}{Odd - 4}= \frac{c}{d}$$
The numerator and denominator are different odd numbers (Odd + 4 = Odd_N, Odd- 4 = Odd_D) which are 8 away from each other. They will have only Odd factors and can have no common factors since 8 has no odd factor. So (Odd + 4)/(Odd - 4) is in its lowest form.
Hence c is 8 more than d. Sufficient

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Re: If a, c, d, x, and y are positive integers such that ay < x and c/d is  [#permalink]

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07 May 2015, 16:23
2
Hi All,

This DS question is thick with variables and restrictions, but it can be solved by TESTing VALUES. You just have to make sure to be organized with your work.

We're told a number of facts:
1) A, C, D, X and Y are POSITIVE INTEGERS
2) (A)(Y) < X
3) (X+AY)/(X-ZY) = C/D where C and D are 'reduced' to the lowest possibilities.

We're asked for the difference between C and D.

Fact 1: X/Y is an ODD INTEGER

IF...
X = 3
Y = 1
A = 1
(X+AY)/(X-AY) = 4/2 = 2/1
C = 2
D = 1
The answer to the question is 2-1 = 1

IF...
X = 3
Y = 1
A = 2
(X+AY)/(X-AY) = 5/1
C = 5
D = 1
The answer to the question is 5-1 = 4
Fact 1 is INSUFFICIENT

Fact 2: A = 4

IF...
X = 5
Y = 1
A = 4
(X+AY)/(X-AY) = 9/1
C = 9
D = 1
The answer to the question is 9-1 = 8

IF...
X = 6
Y = 1
A = 4
(X+AY)/(X-AY) = 10/2 = 5/1
C = 5
D = 1
The answer to the question is 5-1 = 4
Fact 2 is INSUFFICIENT

Combined, we know....
X/Y is an ODD INTEGER
A = 4

IF...
X = 5
Y = 1
A = 4
(X+AY)/(X-AY) = 9/1
C = 9
D = 1
The answer to the question is 9-1 = 8

IF...
X = 13
Y = 1
A = 4
(X+AY)/(X-AY) = 17/9
C = 17
D = 9
The answer to the question is 17-9 = 8

IF...
X = 15
Y = 3
A = 4
(X+AY)/(X-AY) = 27/3 = 9/1
C = 9
D = 1
The answer to the question is 9-1 = 8

With 3 different examples, it certainly appears that the answer is ALWAYS 8.
Combined, SUFFICIENT

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Re: If a, c, d, x, and y are positive integers such that ay < x and c/d is  [#permalink]

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07 May 2015, 18:08
1
Are tutors meant to post solutions to these? There is a very clean and fast solution here, but I don't want to spoil anyone's fun!
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Re: If a, c, d, x, and y are positive integers such that ay < x and c/d is  [#permalink]

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08 May 2015, 03:45
IanStewart wrote:
Are tutors meant to post solutions to these? There is a very clean and fast solution here, but I don't want to spoil anyone's fun!

Hi Ian! Welcome Back!!!

Yes, tutors are allowed to post to these questions. I'd wait for couple of days after the question is published though.
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Re: If a, c, d, x, and y are positive integers such that ay < x and c/d is  [#permalink]

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08 May 2015, 06:31
My solution is essentially the same as Karishma's, with a slightly different presentation. Let's use both statements together, and as we do, we'll see why we need them both:

Using S1, x/y = ø, where ø is odd, so x = øy

Plug this into the fraction, and cancel the y:

(x + ay) / (x - ay) = (øy + ay) / (øy - ay) = (ø + a) / (ø - a)

From S2 (a = 4) the fraction becomes (ø + 4) / (ø - 4). To answer the question, we need to reduce this fraction to lowest terms, then subtract the denominator from the numerator. But ø+4 and ø-4 differ by 8, and if two integers differ by 8, their GCD must be a factor of 8; the GCD of ø+4 and ø-4 can only be 1, 2, 4 or 8. But both ø+4 and ø-4 are odd (because ø is odd), so their GCD cannot be 2, 4 or 8; it must be 1. If their GCD is 1, the fraction is already reduced, so the answer is 8.

We genuinely needed both statements along the way (the value of a is crucial, and without S1, we cannot be sure the fraction is reduced), so the answer is C.
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Re: If a, c, d, x, and y are positive integers such that ay < x and c/d is  [#permalink]

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11 May 2015, 05:30
Bunuel wrote:
If a, c, d, x, and y are positive integers such that ay < x and c/d is the lowest-terms representation of the fraction (x+ay)/(x-ay) , then c is how much greater than d? (If c/d is an integer, let d = 1.)

(1) x/y is an odd integer.

(2) a = 4

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

The given fact can be rewritten:

(x+ay)/(x-ay)=c/d

This equation, though, can’t be solved for the target expression (c – d).

Because all values are positive integers and ay < x, then x + ay will be bigger than both x and ay. In addition, x – ay will still be positive. Given these facts, try testing cases.

(1) NOT SUFFICIENT:

Try x/y=3. Recall that ay < x, so a< x/y or a < 3. Therefore, a is either 1 or 2.
If a = 1, and x/y=3, then x = 3y. The equation becomes:

(3y+1y)/(3y-1y)={4y}/{2y}=2/1, so c = 2 and d = 1. Thus c – d = 1.

If a = 2, and x/y=3, then x = 3y. The equation becomes:

(3y+2y)/(3y-2y)={5y}/y=5/1, so c = 5 and d = 1. Thus c – d = 4.

There are at least two different values of c – d, so the statement is not sufficient.

(2) NOT SUFFICIENT:

If a = 4, then the inequality becomes 4y < x. Say y = 1; then 4(1) < x, or x > 4. Try some values for x:

If a = 4, x = 5 and y = 1, then (5+4)/(5-4)=9/1, so c = 9 and d = 1. Thus c – d = 8.
If a = 4, x = 6 and y = 1, then the fraction is (6+4)/(6-4)=10/2=5/1, so c = 5 and d = 1. Thus c – d = 4.

There are at least two different values of c – d, so the statement is not sufficient.

(1) AND (2) SUFFICIENT:

Statement 1 says that x/y is an odd integer. Call this odd integer p (that is p=x/y). Then x = py, and so the fraction becomes:

(py+4y)/(py-4y)={p+4}/{p-4}

The odd integer p equals x/y . In addition, a<{x/y} so a is also less than p. Because a = 4, p has to be at least 5.
If p = 5, then (5y+4y)/(5y-4y)={9y}/y=9/1, so c – d = 9 – 1 = 8.
If p = 7, then (7y+4y)/(7y-4y)={11y}/{3y}=11/3, so c – d = 11 – 3 = 8.
If p = 9, then (9y+4y)/(9y-4y)={13y}/{5y}=13/5, so c – d = 13 – 5 = 8.

There’s a pattern here; the difference will always be 8. You can trust the pattern—test enough cases to convince yourself!—but a theory-based explanation follows.

The numerator and denominator of the fraction both “start” from the same number: py. The problem states that p is odd. It’s also the case that y must be odd. Why? x/y is also an odd integer. If y were even, then x would also have to be even in order to result in an integer, in which case x/y couldn’t be odd.

Okay, so py is odd. In addition, the numerator and the denominator of the fraction differ by 8 (because, in the numerator, you add 4y and in the denominator, you subtract 4y).

The value c – d represents this difference between the numerator and the denominator. The issue is whether any of the possible fractions can be reduced; if they can, then a smaller value of c – d will result (because c and d will be closer together). If they cannot, then the difference will always be 8.

Consider that the fraction can be written as {denominator +8}/denominator=denominator/denominator+8/denominator

The first fraction reduces to 1. The second fraction will reduce only if the 8 on top shares a factor with the denominator. Remember, though, that the denominator is odd. An odd number and the number 8 do not share any factors (other than 1); the fraction can’t be reduced.

Therefore the fraction cannot ever be reduced, and so c – d = 8 always. The two statements together are sufficient.

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Re: If a, c, d, x, and y are positive integers such that ay < x and c/d is  [#permalink]

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20 May 2015, 07:50
for statement 2

Since a=4
assuming
x= 9
y=2
Hence ay<x because 4*2<9 => 8<9

(x+4y)/(x-4y) => (9+8)/(9-8) => 17/1

17 -1 => 16.

Which makes statement 2 not reliable.

What am i missing here??

EMPOWERgmatRichC wrote:
Hi All,

This DS question is thick with variables and restrictions, but it can be solved by TESTing VALUES. You just have to make sure to be organized with your work.

We're told a number of facts:
1) A, C, D, X and Y are POSITIVE INTEGERS
2) (A)(Y) < X
3) (X+AY)/(X-ZY) = C/D where C and D are 'reduced' to the lowest possibilities.

We're asked for the difference between C and D.

Fact 1: X/Y is an ODD INTEGER

IF...
X = 3
Y = 1
A = 1
(X+AY)/(X-AY) = 4/2 = 2/1
C = 2
D = 1
The answer to the question is 2-1 = 1

IF...
X = 3
Y = 1
A = 2
(X+AY)/(X-AY) = 5/1
C = 5
D = 1
The answer to the question is 5-1 = 4
Fact 1 is INSUFFICIENT

Fact 2: A = 4

IF...
X = 5
Y = 1
A = 4
(X+AY)/(X-AY) = 9/1
C = 9
D = 1
The answer to the question is 9-1 = 8

IF...
X = 6
Y = 1
A = 4
(X+AY)/(X-AY) = 10/2 = 5/1
C = 5
D = 1
The answer to the question is 5-1 = 4
Fact 2 is INSUFFICIENT

Combined, we know....
X/Y is an ODD INTEGER
A = 4

IF...
X = 5
Y = 1
A = 4
(X+AY)/(X-AY) = 9/1
C = 9
D = 1
The answer to the question is 9-1 = 8

IF...
X = 13
Y = 1
A = 4
(X+AY)/(X-AY) = 17/9
C = 17
D = 9
The answer to the question is 17-9 = 8

IF...
X = 15
Y = 3
A = 4
(X+AY)/(X-AY) = 27/3 = 9/1
C = 9
D = 1
The answer to the question is 9-1 = 8

With 3 different examples, it certainly appears that the answer is ALWAYS 8.
Combined, SUFFICIENT

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Rich
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Joined: 24 Jun 2008
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Re: If a, c, d, x, and y are positive integers such that ay < x and c/d is  [#permalink]

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20 May 2015, 10:00
2
kelvind13 wrote:
for statement 2

Since a=4
assuming
x= 9
y=2
Hence ay<x because 4*2<9 => 8<9

(x+4y)/(x-4y) => (9+8)/(9-8) => 17/1

17 -1 => 16.

Which makes statement 2 not reliable.

What am i missing here??

You're demonstrating why Statement 2 alone cannot be sufficient (since as you prove, c-d can be 16, but can also have other values). But that only means the answer is not B or D. When you combine the two statements, it is no longer possible that x = 9 and y = 2, because Statement 1 tells us that x/y must be an odd integer, and 9/2 is not an odd integer.
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Re: If a, c, d, x, and y are positive integers such that ay < x and c/d is  [#permalink]

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20 May 2016, 11:55
1
Bunuel wrote:
If a, c, d, x, and y are positive integers such that ay < x and c/d is the lowest-terms representation of the fraction (x+ay)/(x-ay) , then c is how much greater than d? (If c/d is an integer, let d = 1.)

(1) x/y is an odd integer.

(2) a = 4

Kudos for a correct solution.

Giving it a try. Experts please suggest if I am wrong at any point

Given information :-
a, c, d, x, and y are positive integers
ay < x (or x is greater than both a and y)
c/d is the lowest-terms representation of the fraction (x+ay)/(x-ay) (Since x>ay, this fraction will yield positive results)

c is how much greater than d? Need to know the values of c and d

(1) x/y is an odd integer

x and y can both be even or odd, but we don't know the values to put in given expression to get the value of c and d - (x+ay)/(x-ay)

(2) a = 4
x+4y/x-4y
we don't know the value of a and y

Combining statement 1 and 2 with the given information, I tried putting some values
let's assume y=1
since x is >ay, x will be > 4. let's say 5

5+4/5-4= 9/1 (9-1=8)

assume x=10
10+4/10-4= 14/6 (14-6=8)

assume y=2
minimum value of x can be 10 (4*2=8 , and since y is even x will also be even )
10+8/10-8= 18/2= 9/1 (9-1=8)

For any value we get c-d= 8

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Re: If a, c, d, x, and y are positive integers such that ay < x and c/d is  [#permalink]

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21 May 2016, 02:50
Hi Bunuel
It is hard question. Will we see such question in real Gmat test ?
I mean do I need to practice similar hard questions ?
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Re: If a, c, d, x, and y are positive integers such that ay < x and c/d is  [#permalink]

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21 May 2016, 03:07
hatemnag wrote:
Hi Bunuel
It is hard question. Will we see such question in real Gmat test ?
I mean do I need to practice similar hard questions ?

Practicing hard questions are almost always good, especially if you are aiming for a high quant score.
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Re: If a, c, d, x, and y are positive integers such that ay < x and c/d is  [#permalink]

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17 Jul 2017, 01:41
Bunuel wrote:
If a, c, d, x, and y are positive integers such that ay < x and c/d is the lowest-terms representation of the fraction (x+ay)/(x-ay) , then c is how much greater than d? (If c/d is an integer, let d = 1.)

(1) x/y is an odd integer.

(2) a = 4

Kudos for a correct solution.

Given: a,c,d,x,y -> + int, and
c/d = min (x+ay)/(x-ay)

To find: c-d
Stmt 1: x/y =odd
c/d = (x/y + a)/(x/y - a) = (odd+a)/(odd-a)
min value of c/d will depend on the value of a. "a" is unknown -> insufficient

Stmt 2: a=4
c/d=min (x+4y)/(x-4y)
x & y are unknown; -> insufficient

Stmt 1 & 2:
c/d = min (odd+4)/(odd-4)
For c/d to take minimum value odd = next odd integer to "4" (which is 5) and the common factor of the fraction should be "1".
c/d= 9/1 => c=9 and d=1

c-d=8-> Sufficient
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Re: If a, c, d, x, and y are positive integers such that ay < x and c/d is  [#permalink]

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13 Apr 2019, 08:38
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Re: If a, c, d, x, and y are positive integers such that ay < x and c/d is   [#permalink] 13 Apr 2019, 08:38
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