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

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 7612
GMAT 1: 760 Q51 V42
GPA: 3.82

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Updated on: 02 May 2016, 22:44
00:00

Difficulty:

35% (medium)

Question Stats:

74% (02:05) correct 26% (02:50) wrong based on 68 sessions

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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.

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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" 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. Math Expert Joined: 02 Aug 2009 Posts: 7764 Re: (4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=? [#permalink] ### Show Tags 02 May 2016, 21:30 1 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 $$1^2+2^2+...n^2 = \frac{n(n+1)(2n+1)}{6}$$ _________________ Intern Joined: 05 Sep 2014 Posts: 2 Re: (4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=? [#permalink] ### Show Tags 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... Intern Joined: 06 May 2015 Posts: 23 Re: (4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=? [#permalink] ### Show Tags 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 . Math Expert Joined: 02 Aug 2009 Posts: 7764 Re: (4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=? [#permalink] ### Show Tags 02 May 2016, 22:28 1 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.. _________________ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7612 GMAT 1: 760 Q51 V42 GPA: 3.82 (4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=? [#permalink] ### Show Tags 05 May 2016, 22:35 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. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: (4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=?   [#permalink] 06 Mar 2019, 08:01
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