w, x, y and z are positive integers such that y

Question

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

Sri21 · Answer

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

BrentGMATPrepNow · Answer

My bad. I meant for the target question to ask "What is the value of x?" (not z)

Cheers,
Brent

chetan2u · Answer

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..

BrentGMATPrepNow · Answer

You're absolutely right! 
Sorry, I edited the question while you were composing your response.

Cheers,
Brent

Tuanguyen248 · Answer

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..

BrentGMATPrepNow · Answer

A thousand apologies. 
I meant for the target question to ask "What is the value of x" (not z)

shawnxnee · Answer

(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

PKN · Answer

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.

BrentGMATPrepNow · Answer

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

Paugustin90 · Answer

Aren't the factors x, 02,and 6? If not, how did you come up with 0,2,4?  GMATPrepNow

BrentGMATPrepNow · Answer

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

I've edited my solution accordingly. Cheers, Brent

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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.

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