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If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0

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If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0 [#permalink]

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New post 26 May 2016, 21:17
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If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0, what is the ratio of x to the sum of y and z?

(A) 1:21
(B) 1:20
(C) 1:6
(D) 6:1
(E) 20:1
[Reveal] Spoiler: OA
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Re: If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0 [#permalink]

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New post 26 May 2016, 22:32
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\(\frac{x+y+z}{3}\) = 7x

y+z= 20x.

Important: ratio of x to the sum of y and z, has been asked

\(\frac{x}{y+z}\) = \(\frac{1}{20}\)

Answer: B
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If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0 [#permalink]

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New post 17 Dec 2016, 19:41
Nice Official Question.
Here is my Solution to this one =>

Using \(Mean = \frac{Sum}{#}\)

\(x+y+z=21x\)
Hence \(y+z=20x\)
\(\frac{x}{y+z}=1/20\)
Hence the Required Ratio=1:20

Hence B

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Re: If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0 [#permalink]

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New post 17 Mar 2017, 20:27
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I feel a little stupid for asking, but how do we know to divide both sides by 20(y+z)? I can see how we get the answer from this method, but don't understand how we get there.
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Re: If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0 [#permalink]

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New post 18 Mar 2017, 12:09
decotis74

(x+y+z)/3 = 7x
x + y + z = 21x
y + z = 20 x

y + z is twenty times x

Therefore the ratio is
x - 1
y + z - 20

1:20 which is answer B
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If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0 [#permalink]

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New post 22 Aug 2017, 12:09
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nalinnair wrote:
If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0, what is the ratio of x to the sum of y and z?

(A) 1:21
(B) 1:20
(C) 1:6
(D) 6:1
(E) 20:1


1. (X + Y + Z) / 3 = 7x
2. You want to isolate (X + Y + Z); therefore, multiple both sides of the equation by 3. Multiplying 1/3 by 3 neutralizes the denominator.
3. (X + Y + Z) = 21x
"what is the ratio of x to the sum of y and z?" given this, we want to isolate Y + Z; therefore, we subtract X from both sides of the equation.
4. Y + Z = 21X - 1X (remember the coefficients when add/subtracting variables)
4.1 Y + Z = 20x
5. divide by X: doing this will help you visualize the ratio as fraction --> Y+Z / X = 20/1 or see Attached:


Meaning, for every sum of Y and Z there is 20, and for every sum of Y and Z there is exactly 1 X

Therefore, The Answer is (B) 1:20

As for a roundabout method, Let X = 1, Y = 14, Z = 6

(1) + (14) + (6) / 3 = 7(1)
21 / 3 = 7

The five steps I posted first is a little quicker.

This Question is Q 100 on OG 2017.
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If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0 [#permalink]

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New post 04 Mar 2018, 06:13
nalinnair wrote:
If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0, what is the ratio of x to the sum of y and z?

(A) 1:21
(B) 1:20
(C) 1:6
(D) 6:1
(E) 20:1



\(\frac{x+y+z}{3} = 7x\)

\(x+y+z = 21x\)

\(20x = y+z\)

\(x = \frac{y+z}{20}\)

can anyone tell me if my solution is correct ? :-)
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Re: If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0 [#permalink]

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New post 04 Mar 2018, 10:58
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dave13 wrote:
nalinnair wrote:
If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0, what is the ratio of x to the sum of y and z?

(A) 1:21
(B) 1:20
(C) 1:6
(D) 6:1
(E) 20:1



\(\frac{x+y+z}{3} = 7x\)

\(x+y+z = 21x\)

\(20x = y+z\)

\(x = \frac{y+z}{20}\)

can anyone tell me if my solution is correct ? :-)


Yes Dave - It is correct.
Re: If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0   [#permalink] 04 Mar 2018, 10:58
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