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If numbers N and K are added to set X{2,8,10,12}, its mean w [#permalink]
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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
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Re: If numbers N and K are added to set X{2,8,10,12}, its mean w [#permalink]
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20 Sep 2013, 00:22
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 ... eticmean/http://www.veritasprep.com/blog/2012/05 ... eviations/
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Re: Numbers added to a set [#permalink]
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09 Jan 2011, 11:19
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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) > \(n+k=6032=28\) > \(n^2+2nk+k^2=(n+k)^2=28^2=784\). Answer: D.
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Re: If numbers N and K are added to set X{2,8,10,12}, its mean w [#permalink]
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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 = 6032 = 28
The question is find the value of N^2+2NK+K^2 => (N+K)^2 = 28^2 = 784.



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If numbers N and K are added to set X [#permalink]
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19 Sep 2013, 03:50
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



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Re: If numbers N and K are added to set X [#permalink]
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19 Sep 2013, 04:49



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Re: If numbers N and K are added to set X{2,8,10,12}, its mean w [#permalink]
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20 Sep 2013, 02:59
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]
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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) > \(n+k=6032=28\) > \(n^2+2nk+k^2=(n+k)^2=28^2=784\). Answer: D. Magician!!! Why do my eyes escape these little details like the a+b whole square formula being expanded in the question ://



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