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by shashank.ism » Wed Feb 17, 2010 6:50 am
abhi75 wrote:A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?

a) 1/20
b) 1/6
c) 1/5
d) 4/21
e) 5/21

I got this one correct but I am sure there are better ways to crack this than my method. Can you please suggest any alternative method to solve it more efficiently.
An efficient way could be this which will take just few seconds..Let S21 be the sum of 21 numbers..
A/Q, Average of other 20 numbers = (S21-n)/20
--> n= (4/5)x(S21-n)/20
--> n/S21 =[spoiler] 1/6 Ans B[/spoiler]
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by harrycool » Sat Apr 17, 2010 7:55 am
let x be sum of 21 numbers
let y be sum of 20 numbers

x = y + n

therefore

x/20 = y/20 + n/20

but 4 * (y/20) = n as per the question

therefore

x/20 = n/4 + n/20 = 6n/20

therefore

x/n = 6 and n/x = 1/6 which is B
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by ymach3 » Sat Apr 17, 2010 8:17 am
S(20)= sum of 20 numbers.

S(21)=sum of 21 numbers.


From the Qn,

n=4*S(20)/20 = S(20)/5 ---> S(20)=5n

S21=n+S(20)

-->S(21)=n+S(20)

-->S(21)=n+5n

-->n=S(21)/6
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