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

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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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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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New post 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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New post 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.

Hence Answer 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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New post 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.

Answer is A
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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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New post 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 \(3\).

Answer (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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New post 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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New post 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.

Therefore, the answer is A.
Answer: 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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New post 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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New post 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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New post 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
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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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New post 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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New post 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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