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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

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) --> \(n+k=60-32=28\) --> \(n^2+2nk+k^2=(n+k)^2=28^2=784\).

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

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

Re: If numbers N and K are added to set X{2,8,10,12}, its mean w [#permalink]

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08 Sep 2014, 23:09

Bunuel wrote:

gmatpapa wrote:

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

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.

Magician!!! Why do my eyes escape these little details like the a+b whole square formula being expanded in the question ://

Re: If numbers N and K are added to set X{2,8,10,12}, its mean w [#permalink]

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09 Oct 2015, 02:31

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