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

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Kudos [?]: 162 [0], given: 75

If numbers N and K are added to set X{2,8,10,12}, its mean w [#permalink]  09 Jan 2011, 09:43
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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
[Reveal] Spoiler: OA

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Kudos [?]: 7352 [2] , given: 186

Re: If numbers N and K are added to set X{2,8,10,12}, its mean w [#permalink]  19 Sep 2013, 23:22
2
KUDOS
Expert's post
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

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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Joined: 02 Sep 2009
Posts: 28331
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Kudos [?]: 45373 [1] , given: 6759

1
KUDOS
Expert's post
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) --> $$n+k=60-32=28$$ --> $$n^2+2nk+k^2=(n+k)^2=28^2=784$$.

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Kudos [?]: 10 [1] , given: 8

Re: If numbers N and K are added to set X{2,8,10,12}, its mean w [#permalink]  19 Sep 2013, 21:47
1
KUDOS
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.
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Kudos [?]: 7 [0], given: 4

If numbers N and K are added to set X [#permalink]  19 Sep 2013, 02:50
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
N2 + 2NK + K2 ?

(A) 28 (B) 32 (C) 64 (D) 784 (E) 3600
Math Expert
Joined: 02 Sep 2009
Posts: 28331
Followers: 4483

Kudos [?]: 45373 [0], given: 6759

Re: If numbers N and K are added to set X [#permalink]  19 Sep 2013, 03:49
Expert's post
christykarunya 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
N2 + 2NK + K2 ?

(A) 28 (B) 32 (C) 64 (D) 784 (E) 3600

Merging similar topics. Please refer to the solution above.
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Kudos [?]: 7 [0], given: 4

Re: If numbers N and K are added to set X{2,8,10,12}, its mean w [#permalink]  20 Sep 2013, 01:59
Thanks all for helping me on this
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Re: If numbers N and K are added to set X{2,8,10,12}, its mean w [#permalink]  08 Sep 2014, 22:09
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) --> $$n+k=60-32=28$$ --> $$n^2+2nk+k^2=(n+k)^2=28^2=784$$.

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] 08 Sep 2014, 22:09
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