MathRevolution wrote:
[Math Revolution GMAT math practice question]
\(\frac{(1^2+2^2+3^2+…+20^2)}{(1+2+3+ …+20)}=?\)
\(A. \frac{41}{3}\)
\(B. \frac{41}{6}\)
\(C. 41\)
\(D. 210\)
\(E. 420\)
I guess I found a nice approximation solution inspired by Mitch´s clever idea.
(If he accepts the reasoning, we share the merits!)
\({1^2} + {2^2} + \ldots + {20^2} < 10 \cdot {10^2} + 10 \cdot {20^2}\)
\(? = \frac{{{1^2} + {2^2} + \ldots + {{20}^2}}}{{10 \cdot 21}} < \frac{{{{10}^2} + {{20}^2}}}{{21}} = \frac{{500}}{{21}}\,\,\mathop \cong \limits^{\left( * \right)} \,\,24\)
\(\left( * \right)\,\,\,\frac{{500}}{{21}} = \frac{{420 + 84 - 4}}{{21}} = 23\frac{{17}}{{21}}\)
\(\left( A \right)\,\,\frac{{41}}{3} = 13\frac{2}{3}\)
Note that (C), (D) and (E) are out, while (B) is half the value of (A)... "much much less" (?) than 24...
Obs.: we may divide the sum of squares into FOUR parcels (not TWO, as before) to MAJORATE (first edited) and also MINORATE (second edited) our numerator-focus with better precision:
\(\frac{{5\left( {{3^2} + {8^2} + {{13}^2} + {{18}^2}} \right)}}{{10 \cdot 21}}\,\,\, < \,\,\,?\,\,\, < \,\,\,\frac{{5\left( {{5^2} + {{10}^2} + {{15}^2} + {{20}^2}} \right)}}{{10 \cdot 21}}\)
\(13\frac{{10}}{{21}}\,\, = \,\,\frac{{283}}{{21}}\,\,\, < \,\,\,?\,\,\, < \,\,\,\frac{{125}}{7}\,\, = \,\,17\frac{6}{7}\)
Now we are sure (A) is the right answer... and this COULD be done in less than 5min!
Regards,
Fabio.