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(4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=?

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(4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=? [#permalink]

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New post Updated on: 02 May 2016, 22:44
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\(\frac{(4*1^2+4*2^2+4*3^2+…+4*10^2)}{(3*1+3*2+…+3*10)}=?\)

A. 71/9
B. 79/9
C. 28/3
D. 93/9
E. 125/9

* A solution will be posted in two days.
[Reveal] Spoiler: OA

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Originally posted by MathRevolution on 02 May 2016, 21:16.
Last edited by Bunuel on 02 May 2016, 22:44, edited 3 times in total.
Moved to PS forum and edited the tags.
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Re: (4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=? [#permalink]

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New post 02 May 2016, 21:30
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MathRevolution wrote:
(4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=?

A. 71/9
B. 79/9
C. 28/3
D. 93/9
E. 125/9

* A solution will be posted in two days.


Hi,
two points..
1) it is not a MIXTURE problem..
2) (4*12+4*22+4*32+…+4*102) is not likely to give you 11 on numerator..
You are missing out on some signs.. as 12 may be 1^2.. pl recheck

The Q is supposed to be --
(4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)= \(\frac{(4*1^2+4*2^2+4*3^2+…+4*10^2)}{(3*1+3*2+…+3*10)}\)

\(\frac{4(1^2+2^2+...+10^2)}{3(1+2+3...+10)} = \frac{4*10*(10+1)(2*10+1)}{6}/ \frac{3*10*11}{2}\)

=> \(\frac{4*10*11*21*2}{3*10*11*6} = \frac{28}{3}\)
C..

Ofcourse you require to know that \(1^2+2^2+...n^2 = \frac{n(n+1)(2n+1)}{6}\)
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Re: (4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=? [#permalink]

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New post 02 May 2016, 21:37
According to my understanding, the answer does not match with any of the options as per my solution which is as follows:-

4(12+22+32....+102)/3(1+2+3+....10).
Now, both the numerator and denominator form an A.P. The sum of an A.p. is = n/2(first Term + Last Term)
4(570)/3(55) = 152/11.

Please let me know if i have made any mistake in the solution...
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Re: (4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=? [#permalink]

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New post 02 May 2016, 22:21
Hello Chetan2u.
however hit the right option but is the sum can be solved as-
we know that 4/3 is common numerator & denominator & if we look the option only c is the multiple of 4/3.
so no calculation is required.
correct if approach is wrong .
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Re: (4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=? [#permalink]

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New post 02 May 2016, 22:28
shravan2025 wrote:
Hello Chetan2u.
however hit the right option but is the sum can be solved as-
we know that 4/3 is common numerator & denominator & if we look the option only c is the multiple of 4/3.
so no calculation is required.
correct if approach is wrong .



Yes , there are many Qs which can be solved by this approach..
BUT be careful, we do not know what comes out of (1+2+3....+10) because if a multiple of 4 comes out of it the above 4 in numerator can get cancelled..
But a bit of THINKING--- that in 10 numbers 5 are ODD and 5 are EVEN--- the SUM 1+2+..+10 will be ODD..

Now you can USE your approach since the Denominator is ODD*ODD so the NUMERATOR has to be EVEN because of 4..
as Observed by you , ONLY C fits in..
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Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


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(4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=? [#permalink]

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New post 05 May 2016, 22:35
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We know that 1^2+2^2+….+n^2=n(n+1)(2n+1)/6 and 1+2+….+n=n(n+1)/2. Then, from the question, the numerator becomes 4(1^2+2^2+..10^2)=4(10)(11)(21)/6 and the denominator becomes 3(1+2+….+10)=3(10)(11)/2. Hence, (4*1^2+4*2^2+4*3^2+…+4*10^2)/(3*1+3*2+…+3*10) =[4(10)(11)(21)/6]/[3(10)(11)/2]=28/3. Hence, the correct answer is C.
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Re: (4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=? [#permalink]

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Re: (4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=?   [#permalink] 29 Jan 2018, 04:41
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