buoyant wrote:If x and y are both positive integers and x>y , what is the remainder when x is divided by y ?

(1) y is a two-digit prime number.

(2) x= qy+9 , for some positive integer q

[spoiler]OA: C[/spoiler]

Statement 1 tells us that y is a two digit prime number. We know that x > y. If y = 11 and x = 22, the remainder would be 0. Alternatively, y could be 11 and x could be 21 and the remainder 10.

Insufficient

One might be tempted to say that Statement 2 is sufficient because it tells us that x = (a multiple of y) + 9. So one's initial inclination might be to jump and say the remainder is 9. The problem is that y might be equal to or smaller than 9, in which cases the remainder would be some number less than 9.

Insufficient

Together we know that x = (a multiple of y) + 9 and, since y is a two digit number, that y is greater than 9. So the remainder is 9.

Sufficient

Choose C.

buoyant wrote:If x and y are both positive integers and x>y , what is the remainder when x is divided by y ?

(1) y is a two-digit prime number.

(2) x= qy+9 , for some positive integer q

Statement 1:
No information about x.
INSUFFICIENT.

Statement 2:
Let q=y=1.
Then x = 1*1 + 9 = 10.
In this case, x/y = 10/1 = 10 R0.

Let q=1 and y=2.
Then x = 1*2 + 9 = 11.
In this case, x/y = 11/2 = 5 R1.

Since the remainder can be different values, INSUFFICIENT.

Statement combined:
Case 1: q=1 and y=11
Here, x = 1*11 + 9 = 20.
In this case, x/y = 20/11 = 1 R9.

Case 2: q=2 and y=13
Here, x = 2*13 + 9 = 35.
In this case, x/y = 35/13 = 2 R9.

R=9 in both cases.
One more random case to be safe.

Case 3: q=6 and y=31
Here, x = 6*31 + 9 = 195.
In this case, x/y = 195/31 = 6 R9.

In every case, R=9.
SUFFICIENT.

The correct answer is C.