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If x=2y=4z for positive integers x, y, and z, which of the following c

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Math Revolution GMAT Instructor
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If x=2y=4z for positive integers x, y, and z, which of the following c [#permalink]

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30 Jun 2017, 01:04
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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
[Reveal] Spoiler: OA

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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, 01:50
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

If x=2y=4z,
x + y + 3z = 4z + 2z + 3z = 9z
Average of 3 numbers =$$\frac{9z}{3}$$ = 3z

Since the average is multiple of 3, answer has to be a multiple of 3 as well.

Only option A is a multiple of 3.

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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
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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.

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

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30 Jun 2017, 09:35
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

Given $$x=2y=4z$$

$$x = 4z$$ --------- (i)

$$2y = 4z$$

$$y = 2z$$ ----------- (ii)

Average of $$x, y$$ and $$3z = \frac{x + y + 3z}{3}$$

Substituting the values of $$x$$ and $$y$$ from (i) and (ii) respectively, we get;

$$\frac{4z + 2z + 3z}{3} = \frac{9z}{3} = 3z$$

Average of $$x, y$$ and $$3z$$ must be a multiple of $$3$$.

Among the options Only (Option A) $$15$$ is multiple of $$3$$.

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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, 10:42
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

Let, x=2y=4z = 4

So, x = 4 , y = 2 & z = 1

Arithmetic Mean of x, y & 3z is $$\frac{4 + 2 + 3}{3} = 3$$

So, The number must be a multiple of 3, only (A) satisfies, thus answer will be (A) 15
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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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02 Jul 2017, 18:13
==> You get x=4z and y=2z, and the average=$$\frac{(x+y+3z)}{3}=\frac{(4z+2z+3z)}{3}=3z$$, so it always needs to be the multiple of 3.

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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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23 Jul 2017, 07:27
Hi Bunuel, Vyshak,

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?

x=2y ==> y=x/2

x=4z ==> z=x/4

arithmetic mean of x, y and 3z = [x+(x/2)+(3x/4)]/3

=10x/12 ==>5x/6

Can somebody please tell me where am'I going wrong?
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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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23 Jul 2017, 08:16
NandishSS wrote:
Hi Bunuel, Vyshak,

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?

x=2y ==> y=x/2

x=4z ==> z=x/4

arithmetic mean of x, y and 3z = [x+(x/2)+(3x/4)]/3

=10x/12 ==>5x/6

Can somebody please tell me where am'I going wrong?

$$\frac{x+\frac{x}{2}+\frac{3x}{4}}{3}=\frac{\frac{4x+2x+3x}{4}}{3}=\frac{3x}{4}$$, not 5x/6.
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If x=2y=4z for positive integers x, y, and z, which of the following c [#permalink]

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23 Jul 2017, 10:55
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

y=x/2
z=x/4
x+x/2+3*x/4=9x/4
mean=(9x/4)/3=3x/4
only 15 is a multiple of 3
A

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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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24 Jul 2017, 06:58
Bunuel wrote:
NandishSS wrote:
Hi Bunuel, Vyshak,

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?

x=2y ==> y=x/2

x=4z ==> z=x/4

arithmetic mean of x, y and 3z = [x+(x/2)+(3x/4)]/3

=10x/12 ==>5x/6

Can somebody please tell me where am'I going wrong?

$$\frac{x+\frac{x}{2}+\frac{3x}{4}}{3}=\frac{\frac{4x+2x+3x}{4}}{3}=\frac{3x}{4}$$, not 5x/6.

Tagging VeritasPrepKarishma, Bunuel,

I'm Sorry for very silly mistake.But 3x/4, does ans should be multiple of 3. Can you please explain bit.
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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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07 Aug 2017, 10:12
NandishSS wrote:
Bunuel wrote:
NandishSS wrote:
Hi Bunuel, Vyshak,

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?

x=2y ==> y=x/2

x=4z ==> z=x/4

arithmetic mean of x, y and 3z = [x+(x/2)+(3x/4)]/3

=10x/12 ==>5x/6

Can somebody please tell me where am'I going wrong?

$$\frac{x+\frac{x}{2}+\frac{3x}{4}}{3}=\frac{\frac{4x+2x+3x}{4}}{3}=\frac{3x}{4}$$, not 5x/6.

Tagging VeritasPrepKarishma, Bunuel,

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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Re: If x=2y=4z for positive integers x, y, and z, which of the following c   [#permalink] 07 Aug 2017, 10:12
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