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

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Current Student
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Joined: 09 Jan 2016
Posts: 117
Location: Hong Kong
GMAT 1: 750 Q50 V41
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If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0  [#permalink]

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26 May 2016, 21:17
2
14
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5% (low)

Question Stats:

82% (00:54) correct 18% (01:01) wrong based on 559 sessions

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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
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Joined: 18 Jan 2010
Posts: 246
Re: If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0  [#permalink]

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26 May 2016, 22:32
3
1
$$\frac{x+y+z}{3}$$ = 7x

y+z= 20x.

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

$$\frac{x}{y+z}$$ = $$\frac{1}{20}$$

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

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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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17 Mar 2017, 20:27
1
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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Joined: 06 Dec 2016
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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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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

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Joined: 21 Jun 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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22 Aug 2017, 12:09
1
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.
Attachments

Ratio.jpg [ 80.71 KiB | Viewed 7534 times ]

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Joined: 09 Mar 2016
Posts: 1230
If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0  [#permalink]

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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 ?
Intern
Joined: 15 Oct 2016
Posts: 28
Re: If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0  [#permalink]

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04 Mar 2018, 10:58
1
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.
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Posts: 13386
Re: If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0  [#permalink]

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17 Mar 2019, 10:17
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Re: If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0   [#permalink] 17 Mar 2019, 10:17
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