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Y+Z

by sk8ternite » Wed Aug 26, 2009 1:33 pm
The average of x, y, and z is one-third greater than x. The sum of y+z is what percent greater than x?

a. 50
b. 100
c. 200
d. 300
e. 400
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by vani_13in » Wed Aug 26, 2009 1:55 pm
Is the answer - 200? Please let me know. Here is how I have calculated:

Let A be the average.
A= x+y+z/3

Also, A= x+x/3
A=4x/3
so x+y+z/3 = 4x/3
x+y+z = 4x
y+z = 3x
Percentage increase = 3x-x= 2 i.e. 200%
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by sk8ternite » Wed Aug 26, 2009 2:00 pm
vani_13in wrote:Is the answer - 200? Please let me know. Here is how I have calculated:

Let A be the average.
A= x+y+z/3

Also, A= x+x/3
A=4x/3
so x+y+z/3 = 4x/3
x+y+z = 4x
y+z = 3x
Percentage increase = 3x-x= 2 i.e. 200%
Vani, how come I cant set the equation to be..
(x+y+z/3)(2/3) = x. It says the average is 1/3 greater, so 2/3 of the average should equal x. Am I doing something incorrectly. I will post the answer shortly
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by vani_13in » Wed Aug 26, 2009 2:15 pm
As I have understood
" The average of x, y, and z is one-third greater than x "

i.e. A= x+x/3 = 4x/3 (where A is Average)
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by vani_13in » Thu Aug 27, 2009 8:03 am
Please provide the answer.
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by sk8ternite » Thu Aug 27, 2009 9:15 am
OA is C
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by heshamelaziry » Thu Aug 27, 2009 10:41 pm
I just substituted numbers for the values:

x+y+z/3=A ------->90+135+135/3=120 , Since Y+Z=270, (270/90)*100 = 300
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