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# w, x, y and z are positive integers such that y < z < x < w.

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w, x, y and z are positive integers such that y < z < x < w.  [#permalink]

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Updated on: 13 Jun 2018, 10:27
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Question Stats:

27% (02:33) correct 73% (02:38) wrong based on 103 sessions

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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

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.
Intern
Joined: 27 Oct 2016
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Re: w, x, y and z are positive integers such that y < z < x < w.  [#permalink]

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13 Jun 2018, 10:18
1
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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Joined: 12 Sep 2015
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w, x, y and z are positive integers such that y < z < x < w.  [#permalink]

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13 Jun 2018, 10:28
Top Contributor
My bad. I meant for the target question to ask "What is the value of x?" (not z)

Cheers,
Brent
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Brent Hanneson – GMATPrepNow.com

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Joined: 02 Aug 2009
Posts: 6795
Re: w, x, y and z are positive integers such that y < z < x < w.  [#permalink]

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13 Jun 2018, 10:30
1
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) $$x^3-8x^2+12x=0.........x(x-6)(x-2)=0$$
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..
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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CEO
Joined: 12 Sep 2015
Posts: 2865
Re: w, x, y and z are positive integers such that y < z < x < w.  [#permalink]

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13 Jun 2018, 10:34
Top Contributor
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) $$x^3-8x^2+12x=0.........x(x-6)(x-2)=0$$
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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Brent Hanneson – GMATPrepNow.com

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Joined: 10 Jun 2018
Posts: 12
Re: w, x, y and z are positive integers such that y < z < x < w.  [#permalink]

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13 Jun 2018, 10:48
1
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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Joined: 12 Sep 2015
Posts: 2865
Re: w, x, y and z are positive integers such that y < z < x < w.  [#permalink]

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13 Jun 2018, 10:54
Top Contributor
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.  [#permalink]

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13 Jun 2018, 19:51
2
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(x-6)(x-2) = 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 SUFFICIENT

B
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w, x, y and z are positive integers such that y < z < x < w.  [#permalink]

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14 Jun 2018, 09:09
1
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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PKN

Rise above the storm, you will find the sunshine

CEO
Joined: 12 Sep 2015
Posts: 2865
w, x, y and z are positive integers such that y < z < x < w.  [#permalink]

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Updated on: 26 Jun 2018, 11:41
Top Contributor
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 = 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

Cheers,
Brent
_________________

Brent Hanneson – GMATPrepNow.com

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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Joined: 16 Mar 2018
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Re: w, x, y and z are positive integers such that y < z < x < w.  [#permalink]

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26 Jun 2018, 11:36
1
Aren't the factors x, 02,and 6? If not, how did you come up with 0,2,4? GMATPrepNow
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Joined: 12 Sep 2015
Posts: 2865
Re: w, x, y and z are positive integers such that y < z < x < w.  [#permalink]

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26 Jun 2018, 11:42
Top Contributor
Paugustin90 wrote:
Aren't the factors x, 02,and 6? If not, how did you come up with 0,2,4? GMATPrepNow

Oops - good catch!!!
Looks like I don't know my multiplication tables!

I've edited my solution accordingly.

Cheers,
Brent
_________________

Brent Hanneson – GMATPrepNow.com

Re: w, x, y and z are positive integers such that y < z < x < w. &nbs [#permalink] 26 Jun 2018, 11:42
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