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 \(1^2+2^2+...n^2 = \frac{n(n+1)(2n+1)}{6}\)

_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372

2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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