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

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

Ratio.jpg [ 80.71 KiB | Viewed 34844 times ]

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

$$\frac{x + y + z }{ 3 }= 7x$$

$$x + y + z = 21x$$

$$y + z = 20x$$

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

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

$$\frac{x+y+z}{3}=7x$$

$$x+y+z=21x$$

$$y+z=20x$$

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

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

$$x + y + z = 21x$$

Or, $$y + z = 20x$$

Or, $$\frac{1}{20} = \frac{x}{( y + z )}$$ , Answer must be (B)
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If the average (arithmetic mean) of x, y, and z is 7x and x 0 [#permalink]
decotis74 wrote:
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.

wondering exactly the same!
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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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Thib33600 wrote:
decotis74 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

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.

wondering exactly the same!

We get $$y + z = 20x$$, and are asked to find $$\frac{x}{y+z}$$. Since $$y + z = 20x$$, then $$\frac{x}{y+z}=\frac{x}{20x}=\frac{1}{20}$$.

Hoppe it's clear.
Re: If the average (arithmetic mean) of x, y, and z is 7x and x 0 [#permalink]
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