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The value of x is one quarter of z. The sum ....

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by factor26 » Sun Sep 18, 2011 3:11 pm
The value of x is one quarter of z. The sum of x, y, and z is equal to 26. If the value of y is twice the value of z, what is the largest factor of the sum of y and z?
A- 2

B- 3

C- 8

D- 12

E- 24

ANSWER IS E CAN SOMEONE PLEASE EXPLAIN THE LOGIC BEING TESTED HERE?
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by pemdas » Sun Sep 18, 2011 3:17 pm
find largest factor (y+z)
x=z/4
x=26-y-z
y=2z

z/4=26-2z-z, z=104-8z-4z, 13z=104, z=8 and y=16
z+y=24, the largest factor is 24
factor26 wrote:The value of x is one quarter of z. The sum of x, y, and z is equal to 26. If the value of y is twice the value of z, what is the largest factor of the sum of y and z?
A- 2

B- 3

C- 8

D- 12

E- 24

ANSWER IS E CAN SOMEONE PLEASE EXPLAIN THE LOGIC BEING TESTED HERE?
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by GMATGuruNY » Mon Sep 19, 2011 11:51 am
factor26 wrote:The value of x is one quarter of z. The sum of x, y, and z is equal to 26. If the value of y is twice the value of z, what is the largest factor of the sum of y and z?
A- 2

B- 3

C- 8

D- 12

E- 24

ANSWER IS E CAN SOMEONE PLEASE EXPLAIN THE LOGIC BEING TESTED HERE?
Plug in a value for z in order to determine the ratio of x:y:z.
Let z= 4.

The value of x is one quarter of z:
Thus, x = (1/4)*4 = 1.

The value of y is twice the value of z:
Thus, y = 2*4 = 8.

Thus, x:y:z = 1:8:4.

The sum of x, y, and z is equal to 26:
x+y+z = 1+8+4 = 13.
All the values need to double.
Thus, x =2, y=16, and z=8, so that x+y+z = 2+16+8 = 26.

Thus, y+z = 16+8 = 24.
The largest factor of 24 is 24.

The correct answer is E.
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by factor26 » Tue Sep 20, 2011 6:46 pm
Thanks Guru! it makes much more sense now! i set up the equation without plugging in an actual value for z rather i just turned it into an expression given the info provided in the question.

the value of x is 1/4 --- x =1/4z
the value of z = z
the value of y is 2 times the value of z --- y = 2z

thus 3.25z=26
z=8
x=2
y=16

thus use the values for y and z to get 24 (8+16)

the largest factor for a number is itself ... the answer must be E
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