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w, x, y and z are positive integers such that y < z < x < w.
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Updated on: 13 Jun 2018, 10:27
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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
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Brent Hanneson – GMATPrepNow.com
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Originally posted by GMATPrepNow on 13 Jun 2018, 07:15.
Last edited by GMATPrepNow on 13 Jun 2018, 10:27, edited 1 time in total.



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Re: w, x, y and z are positive integers such that y < z < x < w.
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13 Jun 2018, 10:18
Hello, Can you please explain how B is correct Given: When w is divided by x, the quotient is y, and the remainder is z , so w=xy+z Solving B gives x=0,2 or 6 Since y<z<x and all are positive integers x has to be atleast 3. so x=6. w=xy+z, if y=1, z=2 then w=8 if y=2 z=3 then w=15 if y=3, z=4 then w=22 if y=4,z=5 then w= 29 So, z can be 2,3,4 or 5



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w, x, y and z are positive integers such that y < z < x < w.
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13 Jun 2018, 10:28
My bad. I meant for the target question to ask "What is the value of x?" (not z) Cheers, Brent
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Re: w, x, y and z are positive integers such that y < z < x < w.
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13 Jun 2018, 10:30
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 z?
(1) The least common multiple of w and x is 30 (2) x³ – 8x² + 12x = 0 y<z<X<w.... 1) LCM of w and X is 30.. W and X can take many values and so can z.. Insufficient 2) \(x^38x^2+12x=0.........x(x6)(x2)=0\) So X can be 0,6 or 2 Since y<z<x<w and all are positive integers, X has to be 3 at least.. So X is 6.. Now values fitting can be 1<2<6<8 or 1<3<6<9.. So z can be anything from 2 to 5 Insufficient Combined.. w can be 10,15 or 30.. 10/6 gives y as 1 and z as 4.. 15/6 gives y as 2 and z as 3.. 30/6 not possible.. So z can be 3 or 4.. Insufficient.. GMATPrepNow pl check.. I am sure you wanted to ask value of X..
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Re: w, x, y and z are positive integers such that y < z < x < w.
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13 Jun 2018, 10:34
chetan2u wrote: 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 z?
(1) The least common multiple of w and x is 30 (2) x³ – 8x² + 12x = 0 y<z<X<w.... 1) LCM of w and X is 30.. W and X can take many values and so can z.. Insufficient 2) \(x^38x^2+12x=0.........x(x6)(x2)=0\) So X can be 0,6 or 2 Since y<z<x<w and all are positive integers, X has to be 3 at least.. So X is 6.. Now values fitting can be 1<2<6<8 or 1<3<6<9.. So z can be anything from 2 to 5 Insufficient Combined.. w can be 10,15 or 30.. 10/6 gives y as 1 and z as 4.. 15/6 gives y as 2 and z as 3.. 30/6 not possible.. So z can be 3 or 4.. Insufficient.. GMATPrepNow pl check.. I am sure you wanted to ask value of X.. You're absolutely right! Sorry, I edited the question while you were composing your response. Cheers, Brent
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Re: w, x, y and z are positive integers such that y < z < x < w.
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13 Jun 2018, 10:48
we got the formula: w=xy+z (w>x>z>y)
From (1) (w,x) = {1,2,3,5,6,10,15}
if w=15 => x= {6,10} => y= {1,2} & z = {3,5} if w=10 => x = 6 => y= 1 & z= 4 => insufficient From (2) => x = {0,2,6} => x= 6. => cannot find z => insufficient.
(1)+(2) we still got problem from (1) => Answer is E i think..



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Re: w, x, y and z are positive integers such that y < z < x < w.
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13 Jun 2018, 10:54
Tuanguyen248 wrote: we got the formula: w=xy+z (w>x>z>y)
From (1) (w,x) = {1,2,3,5,6,10,15}
if w=15 => x= {6,10} => y= {1,2} & z = {3,5} if w=10 => x = 6 => y= 1 & z= 4 => insufficient From (2) => x = {0,2,6} => x= 6. => cannot find z => insufficient.
(1)+(2) we still got problem from (1) => Answer is E i think.. A thousand apologies. I meant for the target question to ask "What is the value of x" (not z)
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w, x, y and z are positive integers such that y < z < x < w.
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13 Jun 2018, 19:51
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 (1) Factors of 30: 1,2,3,5,6,10,15,30 x could be 10 and w could be 15, or x could be 6 and w could be 10. Multiple values of x allowed so INSUFFICIENT(2) \(x^{3}\) – \(8x^{2}\) + 12x = 0 x(\(x^{2}\) – 8x + 12) = 0 x(x6)(x2) = 0 In the question stem, it says that all numbers are positive, so it rules out that x = 0 Additionally, question stem also specifies that x must have at least 2 positive numbers that are less in value, so that rules out x = 2 Thus we have that x = 6 SUFFICIENTB



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w, x, y and z are positive integers such that y < z < x < w.
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14 Jun 2018, 09:09
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 Given, (i) w, x, y and z are positive integers , (ii) y < z < x < w and (iii)w=xy+z Question stem, x=? st1, LCM(w,x)=30, we have more than one combination of (w,x) which yields more than one value of x. Hence st1 is not sufficient. st2, x³ – 8x² + 12x = 0 Here we have 3 roots of x, viz, 0,2 & 6. As per the given constraints the value of x can't be 0(x is positive) and 2(x is the 3rd positive number in the sequence) .(from given data (i) & (ii)) Hence , x=6. So, st2 is sufficient. Ans .(B) P.S: Let's check the validity of given data(iii) in st2, we have x=6, and we have to validate w=xy+z. The possible combinations are: y=1, z=2,x=6, w=8 y=2, z=3,x=6, w=15 y=3, z=4,x=6, w=22 y=4, z=5,x=6, w=29 So, all the given conditions met.
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w, x, y and z are positive integers such that y < z < x < w.
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Updated on: 26 Jun 2018, 11:41
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
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Originally posted by GMATPrepNow on 15 Jun 2018, 09:55.
Last edited by GMATPrepNow on 26 Jun 2018, 11:41, edited 1 time in total.



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Re: w, x, y and z are positive integers such that y < z < x < w.
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26 Jun 2018, 11:36
Aren't the factors x, 02,and 6? If not, how did you come up with 0,2,4? GMATPrepNow



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Re: w, x, y and z are positive integers such that y < z < x < w.
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26 Jun 2018, 11:42
Paugustin90 wrote: Aren't the factors x, 02,and 6? If not, how did you come up with 0,2,4? GMATPrepNowOops  good catch!!! Looks like I don't know my multiplication tables! I've edited my solution accordingly. Cheers, Brent
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Re: w, x, y and z are positive integers such that y < z < x < w. &nbs
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