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If the average (arithmetic mean) of x, y, and 20 is 10

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If the average (arithmetic mean) of x, y, and 20 is 10 [#permalink] New post 01 Dec 2005, 13:31
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If the average (arithmetic mean) of x, y, and 20 is 10 greater than the average of x, y, 20, and 30, what is the average of x and y?

A. 40
B. 45
C. 60
D. 75
E. 95
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 [#permalink] New post 01 Dec 2005, 14:10
E. 95

(x+y+20)/3 = 10 + (x + y + 50)/4

solve for (x+y)

find out : (x+y)/2
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 [#permalink] New post 01 Dec 2005, 18:38
(x+y+20)/3 -10 = (x+y+20+30)/4
(x+y)/3 + 20/3 -10 = (x+y)/4 + 50/4
(x+y)/12 = 95/6
(x+y)/2 = 95
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 [#permalink] New post 02 Dec 2005, 01:10
(x+y+20)/3=a and (x+y+20+30)/4=a-10

3a=x+y+20 and 4a-40=x+y+20+30

a=70

(x+y+20)/3=70; x+y=95
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Re: one more PS [#permalink] New post 02 Dec 2005, 02:49
sem wrote:
If the average (arithmetic mean) of x, y, and 20 is 10 greater than the average of x, y, 20, and 30, what is the average of x and y?

A. 40
B. 45
C. 60
D. 75
E. 95


E - 95

(X+Y+20)/3 - (x+y+20+30)/4 = 10
Gives X+y = 190
or (X+Y)/2= 95
Re: one more PS   [#permalink] 02 Dec 2005, 02:49
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