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.
The correct answer is C.
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