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Re: If x=2y=4z for positive integers x, y, and z, which of the following c [#permalink]

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30 Jun 2017, 01:39

Given problem : If x=2y=4z for positive integers x, y, and z, find the average of x,y,3z?

Since x=2y=4z, assume x=4x,y=2x and z=x So the average of x,y,3z will be \(\frac{(4x+2x+3x)}{3} = 3x\) Of the answer options, only 15(Option A) is a multiple of 3 and is our correct answer.
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Re: If x=2y=4z for positive integers x, y, and z, which of the following c [#permalink]

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30 Jun 2017, 02:16

1

This post was BOOKMARKED

MathRevolution wrote:

If x=2y=4z for positive integers x, y, and z, which of the following can be the average (arithmetic mean) of x, y and 3z?

A. 15 B. 20 C. 25 D. 40 E. 47

x = 2y = 4z

x, y, 3z

Average= Sum/Count

Average = (4z+2z+3z)/3

= 9z/3

= 3z

As we can see the answer should be multiple of 3.

Answer is A
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I'm Sorry for very silly mistake.But 3x/4, does ans should be multiple of 3. Can you please explain bit.

Since the given options are all integers, it needs to be assumed that \(\frac{3x}{4}\) will be an integer. Therefore, the resultant will 3 * (some number). In the given options, the only option that is a multiple of 3 is Option A. That is why it is the right option. Kapish?

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