Bunuel wrote:
In an integer division operation, the divisor is x, the quotient is y, the dividend is z, and remainder is r. Is x > y?
(1) z = 151
(2) r = 11
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:Quick hypothetical situations should prove statement 1 to be insufficient. If x is 151, then you'd be dividing the same thing by itself and y would be 1 with no remainder, and x would be much larger than y. But you could flip that and make x = 1, in which you'd just be dividing 151 by 1 and the quotient (y) would be 151, much larger than x.
Recognizing the algebra of division can be quite helpful here, too. Since
Quotient = Dividend * Divisor + Remainder, just knowing statement 1 tells you that 151 = xy + r, and clearly that's not enough information to solve.
Statement 2 is also insufficient. Keep in mind here that, without knowing statement 1, you essentially just have an undefined problem with a remainder of 11. Algebraically that's:
z = xy + 11
And clearly not enough information to solve.
But taken together, the information is sufficient. Why, if it still leaves two variables?
151 = xy + 11, so:
140 = xy
The other crucial element of this being a division problem is that the divisor can't be less than the remainder! If you're dividing 151 by, say, 7, you'd never have 11 left over. Even though 7 * 20 = 140, you wouldn't stop there and say that the remainder is 11, because 11 can still divide one more time by 7 (making it 21, remainder 4). So since 140 = xy AND x > 11, there's no way for an integer y to be greater than x. Therefore the two statements together are sufficient.