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

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.

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!

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)

Question asked:-
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

C is the answer

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.

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.

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.

The correct answer is C.

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.

Hence answer should be E.

What am i missing here??

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.

My Answer: "C"

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

Hi
a < x/y, a is not equal to x/y

(x+ay)/(x-ay)=c/d
x/ay=c+d/c-d
x/ay=(c/d)+1/(c/d)-1
x/ay=n+1/n-1, where n=c/d

Value of n is clearly dependent on (x/y) and a. Hence Statement 1 and 2 are clearly insufficient.

Combining both equations.
x/y=2k+1, k is positive integer greater than 2
and a=4

\(\frac{2k+1}{4}\)=\(\frac{n+1}{n-1}\)
n=1+\(\frac{8}{2k-3}\)
c/d=1+\(\frac{8}{2k-3}\)

We can clearly see that for any value of k greater than 2, difference between c and d will remain 8. Also HCF(2k-3,8)=1, hence numerator and denominator are in the lowest terms representation.
Sufficient!!!

Here's my approach, I am stuck at one step though. Please help -

Statement A : x/y is odd
(x+ay)/(x-ay) = c/d
Dividing numerator and denominator by y : (x/y + a) / (x/y - a) = c/d
let's say q is a common multiple of the num and den. Therefore :
c -d = q(x/y + a) - q(x/y - a)
c - d = qx/y +aq -qx/y +aq
c - d = 2aq

Since we do not have any information about a or q, Statement A alone is not sufficient

Statement B : Plugging a = 4 in the same expression:
c - d = 8q

Now q could be any integer from 1 to infinity, how do we come to the conclusion that q = 1 ?

Many thanks!

If x = 14,y=2,a=4 ; it comes to 22/6 or 11/8 and 11-8=3
Pls correct me if I’m wrong

If x = 14,y=2,a=4 ; it comes to 22/6 or 11/8 and 11-8=3
Pls correct me if I’m wrong

If x = 14,y=2,a=4 ; it comes to 22/6 or 11/8 and 11-8=3
Pls correct me if I’m wrong

You have a small error when you canceled down the fraction: 22/6 is equal to 11/3, not to 11/8.