GMATPrepNow wrote:
w, x, y and z are positive integers such that y < z < x < w. When w is divided by x, the quotient is y, and the remainder is z. What is the value of x?
(1) The least common multiple of w and x is 30
(2) x³ – 8x² + 12x = 0
Target question: What is the value of x? Given: w, x, y and z are positive integers such that y < z < x < w. When w is divided by x, the quotient is y, and the remainder is z. Statement 1: The least common multiple of w and x is 30 There are several values of w and x that satisfy statement 1. Here are two:
Case a: w = 15 and x = 6. In this case, we get 15 divided by 6 equals 2 with remainder 3. In other words, w = 15, x = 6, y = 2 and z = 3. Here, the answer to the target question is
x = 6Case b: w = 15 and x = 10. In this case, we get 15 divided by 10 equals 1 with remainder 5. In other words, w = 15, x = 10, y = 1 and z = 5. Here, the answer to the target question is
x = 10Since we cannot answer the
target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: x³ – 8x² + 12x = 0 Factor to get: x(x² - 8x + 12) = 0
Factor again to get: x(x - 2)(x - 6) = 0
So, there are 3 possible solutions: x = 0, x = 2 and x = 6
We're told that x is a POSITIVE INTEGER, so
x cannot equal 0There's also a problem with the solution
x = 2.
We're told that 0 < y < z < x < w
So, we get: 0 < y < z <
2 < w
Since there are no integer values of y and z that can satisfy this inequality,
x cannot equal 2So, it
must be the case that
x = 6Since we can answer the
target question with certainty, statement 2 is SUFFICIENT
Answer: B
Cheers,
Brent