Data Sufficiency

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

Answer: B
Difficulty level: 700+
Source: www.gmatprepnow.com

A complicated problem so I'd play with the numbers and try to prove insufficiency by showing two different answers.

Evaluating Statement 1
Since the LCM of w and x is 30, and 30 has prime factors 2,3, and 5, use combinations of 2,3,and 5 to figure out what values are possible for w and x. Keeping w between x and 2x will ensure that y = 1 so we can fit the parameters of the problem that y < z < x < w

w=10 and x = 6. Fits! LCM = 30, y=1 and z=4
w=15 and x=10. Fits! LCM = 30, y=1 and z=5

With two possible answers for x, we've proven insufficiency.

Evaluate statement 2
Recognize that x is a factor of each term, so factor out an x and then factor the remaining quadratic to get (x)(x-6)(x-2) = 0. thus, x could equal 0, 2 or 6 according to this statement.
But x cannot equal zero, since 0 is not a positive integer.
And x cannot equal 2, because there wouldn't be possible values for both y and z, which must both be positive integers less than x.
So x would have to equal 6 and this statement is sufficient.

Brent@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

Answer: B
Difficulty level: 700+
Source: www.gmatprepnow.com

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 = 6
Case 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 = 10
Since 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 0

There'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 2

So, it must be the case that x = 6
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent