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must z and 18 have a common

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by sanju09 » Fri Aug 16, 2013 9:51 pm
If x, y, and z are positive integers, and x is a multiple of 3, for which of the following must z and 18 have a common factor greater than 1?

I. (x/9) + (y/6) = (z/18)
II. (x/6) + (y/9) = (z/18)
III. (x/6) + (y/5) = (z/30)

A. I only
B. II only
C. I and II only
D. I an III only
C. I, II, and III


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Source: — Problem Solving |

by lunarpower » Sat Aug 17, 2013 5:42 am
Some hints:

* Kill the fractions! (Multiply I and II by 18, and III by 30.)

* Once you've done that, substitute "3n" or "3(integer)" for x, since x is a multiple of 3.

* Then see what you can factor out.
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by Uva@90 » Sat Aug 17, 2013 6:24 am
Hi Ron/Sanju,

Is Answer: D ?

Regards,
Uva.
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by Brent@GMATPrepNow » Sat Aug 17, 2013 6:43 am
sanju09 wrote:If x, y, and z are positive integers, and x is a multiple of 3, for which of the following must z and 18 have a common factor greater than 1?

I. (x/9) + (y/6) = (z/18)
II. (x/6) + (y/9) = (z/18)
III. (x/6) + (y/5) = (z/30)

A. I only
B. II only
C. I and II only
D. I an III only
C. I, II, and III
IMPORTANT: Since we're told that x is a multiple of 3, we can say that x = 3k for some integer k.

Here's one approach.
I. (x/9) + (y/6) = (z/18)
Multiply both sides by 18 to get: 2x + 3y = z
Now replace x with 3k to get: 2(3k) + 3y = z
Factor out the 3 to get: 3(2k + y) = z
This tells us that z is a multiple of 3, which means z and 18 must have a common factor greater than 1.
So, statement I works

II. (x/6) + (y/9) = (z/18))
Multiply both sides by 18 to get: 3x + 2y = z
Let's test possible values of x and y
If x = 3 and y = 2, then z = 13
In this case, z and 18 do not have a common factor greater than 1.
So, statement II doesn't work.

III. (x/6) + (y/5) = (z/30)
Multiply both sides by 30 to get: 5x + 6y = z
Now replace x with 3k to get: 5(3k) + 6y = z
Factor out the 3 to get: 3(5k + 2y) = z
This tells us that z is a multiple of 3, which means z and 18 must have a common factor greater than 1.
So, statement III works

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by GMATGuruNY » Sun Aug 18, 2013 3:16 pm
sanju09 wrote:If x, y, and z are positive integers, and x is a multiple of 3, for which of the following must z and 18 have a common factor greater than 1?

I. (x/9) + (y/6) = (z/18)
II. (x/6) + (y/9) = (z/18)
III. (x/6) + (y/5) = (z/30)

A. I only
B. II only
C. I and II only
D. I an III only
C. I, II, and III
To see the situation more clearly, we can plug in values.
Be sure to satisfy the constraint that x is multiple of 3.
Since statement I is included in four of the five answer choices, it is almost certain that statement I must be included in the correct answer choice.
To save time, start with statement II.

II: (x/6) + (y/9) = (z/18)
To clear the fractions, multiply both sides of the equation by the GREATEST of the 3 denominators (18).
18 (x/6 + y/9) = 18 (z/18)
3x + 2y = z.

Case 1: If x=3 and y=1, then z = 3(3) + 2(1) = 11.
z=11 and 18 do not have a common factor greater than 1.
Here, it does not have to be true that z and 18 have a common factor greater than 1.
Eliminate any answer choice that includes statement II.
Eliminate B, C and E.

III: (x/6) + (y/5) = (z/30)
To clear the fractions, multiply both sides of the equation by the GREATEST of the 3 denominators (30).
30 (x/6 + y/5) = 30 (z/30)
5x + 6y = z.

Case 1: If x=3 and y=1, then z = 5(3) + 6(1) = 21.
z=21 and 18 are both divisible by 3.

Case 2: If x=6 and y=2, then z = 5(6) + 6(2) = 42.
z=42 and 18 are both divisible by 3 and 6.

In each case, z and 18 have a common factor greater than 1.
One more to be safe:
Case 3: If x=9 and y=5, then z = 5(9) + 6(5) = 75.
z=75 and 18 both divisible by 3.

In every case, z and 18 have a common factor greater than 1.
Thus, the correct answer choice must include statement III.
Eliminate A.

The correct answer is D.
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