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If the average (arithmetic mean) of x, 25, y, and 30 is x + y, which

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If the average (arithmetic mean) of x, 25, y, and 30 is x + y, which [#permalink]

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27 Dec 2015, 08:09
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If the average (arithmetic mean) of x, 25, y, and 30 is x + y, which of the following equals the value of x?

A. $$18\frac{1}{3}y$$
B. $$18\frac{1}{3}+y$$
C. $$18-\frac{3}{y}$$
D. $$y-18\frac{1}{3}y$$
E. $$18\frac{1}{3}-y$$
[Reveal] Spoiler: OA

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Re: If the average (arithmetic mean) of x, 25, y, and 30 is x + y, which [#permalink]

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01 Jan 2016, 09:45
If the average (arithmetic mean) of x, 25, y, and 30 is x + y, which of the following equals the value of x?

x + 5 + y + 30 / 4= x + y
x + y = 55/3

x= $$18\frac{1}{3}-y$$[/quote]

E. $$18\frac{1}{3}-y$$[/quote]

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Re: If the average (arithmetic mean) of x, 25, y, and 30 is x + y, which [#permalink]

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15 Oct 2016, 08:33
E is correct. Here's why:

(x+y+55)/4 = x+y --> manipulate to give (55-3y)/3 = x

3 x 18 = 54 --> Thus answer can be made into a proper fraction (i.e. 18 1/3 - y = x)

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Re: If the average (arithmetic mean) of x, 25, y, and 30 is x + y, which [#permalink]

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15 Oct 2016, 10:44
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Bunuel wrote:
If the average (arithmetic mean) of x, 25, y, and 30 is x + y, which of the following equals the value of x?

A. $$18\frac{1}{3}y$$
B. $$18\frac{1}{3}+y$$
C. $$18-\frac{3}{y}$$
D. $$y-18\frac{1}{3}y$$
E. $$18\frac{1}{3}-y$$

$$\frac{(x + 25 + y + 30)}{4} = ( x + y )$$

So, $$(x + 25 + y + 30) = 4x + 4y$$

Or, $$x + y + 55 = 4x + 4y$$

Or, $$3x = 55 - 3y$$

Or, $$x = 18\frac{1}{3} - y$$

Hence correct answer is (E) $$18\frac{1}{3}-y$$
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Re: If the average (arithmetic mean) of x, 25, y, and 30 is x + y, which [#permalink]

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17 Dec 2016, 16:22
Here is my solution to this one =>

Using Mean= Sum/#

x+y+55 = 4(x+4)
x=55/3 -y

Hence E

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Re: If the average (arithmetic mean) of x, 25, y, and 30 is x + y, which [#permalink]

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20 Dec 2017, 09:19
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Re: If the average (arithmetic mean) of x, 25, y, and 30 is x + y, which   [#permalink] 20 Dec 2017, 09:19
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