x, q, r are positive integers. x/17 = q and remainder is r.

X=17q+r

X=34A + R What is R

(1).

q is odd, no information about r and hence we don't know what X is?
INSUFFICIENT

(2).

r=5 => X=17q+5 = > X can be 5,22,39,56 and so on but the remainders of the listed numbers with 34 are different
INSUFFICIENT

Combining

X=17q+5

X can be 22,56, and so on REM(X/34) =22 Consistent with values listed

Hence (C) it is !

So we have x = 17q + r, what is remainder when x/34?

OK, let's kill this

Statement 1

q is odd

Well we dunno the remainder so insuff

For instance if x=17 then remainder 17
If x = 51 remainder 12

Statement 2

Remainder is 5
Insuff, we have two cases if x=5 then remainder 5
If x = 22 then remainder is 22

Both together

Well we know that q is odd and that the remainder is 5, therefore

17 (2k+1)+5 / 34 = 34k +22 / 34 will always have remainder of 22

C

Hi All,

This DS question is built around a pattern that you might have to do some work to discover. TESTing VALUES will help you to find it and select the correct answer:

We're told that X, Q and R are POSITIVE INTEGERS and that X/17 = Q remainder R. We're asked for the remainder when X is divided by 34....

Fact 1: Q is ODD

This doesn't tell us anything about X or R, so we can quickly TEST VALUES to prove an insufficiency

IF....
Q = 1
R = 1
X = 18
18/34 = 0 remainder 18, so the answer is 18

IF....
Q = 1
R = 2
X = 19
19/34 = 0 remainder 19, so the answer is 19
Fact 1 is INSUFFICIENT

Fact 2: R = 5

IF....
Q = 1
R = 5
X = 22
22/34 = 0 remainder 22, so the answer is 22

IF....
Q = 2
R = 5
X = 39
39/34 = 1 remainder 5, so the answer is 5
Fact 2 is INSUFFICIENT

Combined, we know:
Q is ODD
R = 5

IF....
Q = 1
R = 5
X = 22
22/34 = 0 remainder 22, so the answer is 22

IF....
Q = 3
R = 5
X = 56
56/34 = 1 remainder 22, so the answer is 22

IF....
Q = 5
R = 5
X = 90
90/34 = 2 remainder 22, so the answer is 22

From this work, we have found a pattern - the remainder is ALWAYS 22.
Combined, SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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