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PS 13

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by magical cook » Thu Nov 29, 2007 10:46 pm
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

Hi, Answer seems to be wrong for thsi question, so can anyone confirm?
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by sujaysolanki » Thu Nov 29, 2007 11:03 pm
My bad ,,thnks for the correction guys ..
E it is
Last edited by sujaysolanki on Fri Nov 30, 2007 8:25 am, edited 1 time in total.
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Re: PS 13

by projase » Fri Nov 30, 2007 3:08 am
magical cook 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

Hi, Answer seems to be wrong for thsi question, so can anyone confirm?
(x+y+20)/3 = 10 + (x+y+20+30)/4
4x+4y+80 = 120 + 3x+3y+150
x+y=120+150-80=190
(x+y)/2= 95
Answer is E
Where there's a will, there's a way...
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by Suyog » Fri Nov 30, 2007 4:31 am
agree with Projase.

I wud solve this by substituting the ans values if i cudn't form the equation.
Here only 95 gives possible solution.

Since avg is 95, addition of x & y is 190.

avg of x, y & 20 = 70
and avg of x,y,20 & 30 = 60

Ans E
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by magical cook » Fri Nov 30, 2007 9:40 am
Hi, I though A as well but the answer is B (which I think is wrong...)

10 greater than should be expressed Aaverage - Baverage = 10 OR Aaverage = 10 + Baverage.....

Maybe gabriel will come figure it for us... :wink:
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