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09 Jan 2011, 10:43
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If numbers N and K are added to set X{2,8,10,12}, its mean will increase by 25%. What is the value of \(N^2+2NK+K^2\) ?
A. 28
B. 32
C. 64
D. 784
E. 3600
A. 28
B. 32
C. 64
D. 784
E. 3600
09 Jan 2011, 11:19
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gmatpapa
If numbers N and K are added to set X{2,8,10,12}, its mean will increase by 25%. What is the value of \(N^2+2NK+K^2\) ?
A. 28
B. 32
C. 64
D. 784
E. 3600
A. 28
B. 32
C. 64
D. 784
E. 3600
Mean of {2,8,10,12} is \(mean=\frac{2+8+10+12}{4}=\frac{32}{4}=8\) --> new mean thus should equal to \(8*1.25=10\), so \(\frac{2+8+10+12+n+k}{6}=10\) (note that now we have the set of 6 terms not 4) --> \(n+k=60-32=28\) --> \(n^2+2nk+k^2=(n+k)^2=28^2=784\).
Answer: D.
General Discussion
19 Sep 2013, 22:47
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If numbers N and K are added to set X{2,8,10,12}, its mean will increase by 25%. What is the value of N^2+2NK+K^2 ?
A. 28
B. 32
C. 64
D. 784
E. 3600
Before adding N & K, mean of X = 32/4 = 8.
After adding N & K, the new mean of X = 125/100*8 = 10.
i,e (32 + N+K )/6 = 10 => N+K = 60-32 = 28
The question is find the value of N^2+2NK+K^2 => (N+K)^2 = 28^2 = 784.
A. 28
B. 32
C. 64
D. 784
E. 3600
Before adding N & K, mean of X = 32/4 = 8.
After adding N & K, the new mean of X = 125/100*8 = 10.
i,e (32 + N+K )/6 = 10 => N+K = 60-32 = 28
The question is find the value of N^2+2NK+K^2 => (N+K)^2 = 28^2 = 784.
