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.
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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.

You can also use the conceptual and deviation based methods to solve this.

Mean of set X{2,8,10,12} = 8 (Note that if we assume 8 to be the mean, 2 is 6 less than 8 and 10 and 12 are together 6 more than 8. Hence 8 must be the actual mean - the deviation approach)

When we add two numbers, the mean increases by 25% i.e. by 2. Had the two numbers been both 8, the mean would have stayed 8. But they have 2 extra for each of the 6 numbers so they must be a total of 8 + 8 + 6*2 = 28
(N + K)^2 = 28^2 = 784

For more on these, check:
http://www.veritasprep.com/blog/2012/04 ... etic-mean/
http://www.veritasprep.com/blog/2012/05 ... eviations/
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Magician!!! Why do my eyes escape these little details like the a+b whole square formula being expanded in the question ://

The average before adding N and K is 8 which is come from (32/4)
So with N and K the average will be 10 which is come from (8 + 8/4)

we can conclude that N+K is equal to 28

By using factorisation we can reformulate the equation to get (N+K)(N+K) so (28)(28).

To simplify the computation and avoid a lot of calculs, we just need to look for the unit digit of the problem : 8*8 which is 4

Based on that we know that the unit digit will be 4 and from the number we also know that the number is bigger than 8*8 so we can conclude that it is 784

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