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(1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?
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Updated on: 16 Sep 2018, 12:01
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[ Math Revolution GMAT math practice question] \(\frac{(1^2+2^2+3^2+…+20^2)}{(1+2+3+ …+20)}=?\) \(A. \frac{41}{3}\) \(B. \frac{41}{6}\) \(C. 41\) \(D. 210\) \(E. 420\)
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Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?
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13 Sep 2018, 08:51
MathRevolution wrote: [ Math Revolution GMAT math practice question] \(\frac{(1^2+2^2+3^2+…+20^2)}{(1+2+3+ …+20)}=?\) \(A. \frac{41}{3}\) \(B. \frac{41}{6}\) \(C. 41\) \(D. 210\) \(E. 420\) Since the answer choices are very spread out, we can BALLPARK  and we can be somewhat lax with our estimations. Numerator: If the 11 greatest terms were 10²=100, the sum of the 11 greatest terms = 11*100 = 1100. If the 11 greatest terms were 20²=400, the sum of the 11 greatest terms = 11*400 = 4400. Implication: The sum of the 11 greatest terms is around the average of 1100 and 4400: (1100+4400)/2 = 2750. The 9 smallest terms  the perfect squares between 1 and 81, inclusive  will increase the sum by another 200 or so. Thus, the numerator ≈ 2750 + 200 ≈ 3000. Denominator: The denominator is composed of the integers between 1 and 20, inclusive. For any evenly spaced set, sum = (number of terms)(average of first and last). Thus, the denominator = 20 * (1+20)/2 = 210. Approximate value of the given fraction: 3000/210 = a bit less than 15. Only A is viable: 41/3 = 13⅔.
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Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?
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13 Sep 2018, 00:51
Is there any formula that can be applied?
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Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?
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13 Sep 2018, 01:23
OA: ASum of the squares of first \(n\) natural numbers \(= \frac{n(n+1)(2n+1)}{6}\) ....(1) Sum of first \(n\) natural numbers \(=\frac{n(n+1)}{2}\) ....(2) Dividing \((1)\) by \((2)\), we get \(\frac{Sum \quad of \quad the \quad squares \quad of \quad first \quad n \quad natural \quad numbers}{Sum \quad \quad of \quad first \quad n \quad natural \quad numbers}=\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}=\frac{2n+1}{3}\) Putting \(n = 20\) , we get \(\frac{Sum \quad of \quad the \quad squares \quad of \quad first \quad 20 \quad natural \quad numbers}{Sum \quad \quad of \quad first \quad 20 \quad natural \quad numbers}=\frac{2*20+1}{3}=\frac{41}{3}\)
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Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?
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13 Sep 2018, 08:13
MathRevolution wrote: [ Math Revolution GMAT math practice question] \(\frac{(1^2+2^2+3^3+…+20^2)}{(1+2+3+ …+20)}=?\) \(A. \frac{41}{3}\) \(B. \frac{41}{6}\) \(C. 41\) \(D. 210\) \(E. 420\) The sum 1+2+3+...+20 is equal to 20*(1+20)/2 = 21*10 and it is certainly intheGMATscope. (We have used a"finite arithmetic progression" important result.) The sum of the squares of the first "n" (n=20) positive integers is outofGMAT´s scope, although a beautiful (and elementary) deduction for the formula: \({1^2} + {2^2} + \ldots + {n^2} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\) you may find here: https://mathschallenge.net/library/numb ... of_squares From the identity above, we have: \({1^2} + {2^2} + \ldots + {20^2} = \frac{{20 \cdot 21 \cdot 41}}{6} = 10 \cdot 7 \cdot 41\) Finally, \(? = \frac{{10 \cdot 7 \cdot 41}}{{21 \cdot 10}} = \frac{{41}}{3}\) The right answer is therefore (A) . This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. P.S.: perhaps there is a smart way of avoiding the "outofscope formula"... some "hidden pattern", for instance... Anyway, I could not find an alternate approach in approx. 5min (a generous "timelimit" in the GMAT).
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Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?
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13 Sep 2018, 09:10
GMATGuruNY wrote: MathRevolution wrote: [ Math Revolution GMAT math practice question] \(\frac{(1^2+2^2+3^2+…+20^2)}{(1+2+3+ …+20)}=?\) \(A. \frac{41}{3}\) \(B. \frac{41}{6}\) \(C. 41\) \(D. 210\) \(E. 420\) Since the answer choices are very spread out, we can BALLPARK  and we can be somewhat lax with our estimations. Numerator: If the 11 greatest terms were 10²=100, the sum of the 11 greatest terms = 11*100 = 1100. If the 11 greatest terms were 20²=400, the sum of the 11 greatest terms = 11*400 = 4400. Implication: The sum of the 11 greatest terms is around the average of 1100 and 4400: (1100+4400)/2 = 2750. The 9 smallest terms  the perfect squares between 1 and 81, inclusive  will increase the sum by another 200 or so. Thus, the numerator ≈ 2750 + 200 ≈ 3000. Hi, Mitch! You had a creative idea, no doubt, but... 01. You have made CONSIDERABLE modifications in the original parcels... the alternative choices are not that apart for these modifications... far from that! 02. We are dealing with sum of SQUARES, arithmetic means (averages) are dangerous (to say the least) in nonlinear situations. There was a "happy ending", I saw... but quoting the great Niels Bohr... "Prediction is very difficult, especially about the future." Regards, Fabio.
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Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?
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Updated on: 13 Sep 2018, 17:23
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 numeratorfocus 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.
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Originally posted by fskilnik on 13 Sep 2018, 10:31.
Last edited by fskilnik on 13 Sep 2018, 17:23, edited 2 times in total.



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Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?
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13 Sep 2018, 11:01
fskilnik wrote: GMATGuruNY wrote: MathRevolution wrote: [ Math Revolution GMAT math practice question] \(\frac{(1^2+2^2+3^2+…+20^2)}{(1+2+3+ …+20)}=?\) \(A. \frac{41}{3}\) \(B. \frac{41}{6}\) \(C. 41\) \(D. 210\) \(E. 420\) Since the answer choices are very spread out, we can BALLPARK  and we can be somewhat lax with our estimations. Numerator: If the 11 greatest terms were 10²=100, the sum of the 11 greatest terms = 11*100 = 1100. If the 11 greatest terms were 20²=400, the sum of the 11 greatest terms = 11*400 = 4400. Implication: The sum of the 11 greatest terms is around the average of 1100 and 4400: (1100+4400)/2 = 2750. The 9 smallest terms  the perfect squares between 1 and 81, inclusive  will increase the sum by another 200 or so. Thus, the numerator ≈ 2750 + 200 ≈ 3000. Hi, Mitch! You had a creative idea, no doubt, but... 01. You have made CONSIDERABLE modifications in the original parcels... the alternative choices are not that apart for these modifications... far from that! 02. We are dealing with sum of SQUARES, arithmetic means (averages) are dangerous (to say the least) in nonlinear situations. There was a "happy ending", I saw... but quoting the great Niels Bohr... "Prediction is very difficult, especially about the future." Regards, Fabio. The answer choices represent the ratio between the numerator and the denominator of the given expression. Option A implies that the numerator is about 14 times the denominator Option B implies that the numerator is less than 7 times the denominator. Option C implies that the numerator is 41 times the denominator. Option D implies that the numerator is 210 times the denominator. Option E implies that the numerator is 420 times the denominator. The ratios implied by the answers choices are quite different. The estimations in my earlier post are sufficient to determine whether the numerator is less than 7 times the denominator, about 14 times the denominator, 41 times the denominator, 210 times the denominator, or 420 times the denominator. Consider the extremes: If the 11 greatest terms in the numerator were all 10²=100, the numerator = 11(100) + (sum of the 9 smallest terms) ≈ 1100 + 200 = 1300. In this case, numerator/denominator ≈ 1300/200 = 6.5. If the 11 greatest terms in the numerator were all 20²=400, the numerator = 11(400) + (sum of the 9 smallest terms) ≈ 4400 + 200 = 4600. In this case, numerator/denominator ≈ 4600/200 = 23. Thus, the correct ratio must be between 6.5 and 23  more specifically, somewhere in the MIDDLE of this range. Only option A is viable.
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(1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?
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16 Sep 2018, 17:31
=> Now, \(1^2 + 2^2 + 3^2 + … + n^2 = \frac{n(n+1)(2n+1)}{6}\) and \(1 + 2 + 3 + … + n = \frac{n(n+1)}{2}\). So, \(\frac{(1^2+2^2+3^2+…+20^2)}{(1+2+3+ …+20)} = ( 20*21*\frac{41}{6} ) / ( 20*\frac{21}{2} ) = (\frac{41}{6}) / (\frac{1}{2}) = \frac{41}{3}\) Therefore, the answer is A. Answer: A
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Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?
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21 Sep 2018, 02:30
A different approach would be to find a certain pattern. This one definitly takes more time, but it worked for me. \(\frac{1^2+2^2}{1+2}\) = \(\frac{5}{3}\) \(\frac{1^2+2^2+3^2}{1+2+3}\) = \(\frac{14}{6}\) = \(\frac{7}{3}\) \(\frac{1^2+2^2+3^2+4^2}{1+2+3+4}\) = \(\frac{30}{10}\) = \(3\) \(\frac{1^2+2^2+3^2+5^2}{1+2+3+4+5}\) = \(\frac{55}{15}\) = \(\frac{11}{3}\) One can see that the denominator of the simplified fractions follows a certain pattern > Number of terms in the denominator multiplied with 2 plus 1. One can see that the numerator of the simplified fractions follows a certain pattern > Always 3 As a result \(\frac{(1^2+2^2+3^2+…+20^2)}{(1+2+3+ …+20)}\) can be written as \(\frac{20*2+1}{3}\) = \(\frac{41}{3}\)
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Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=? &nbs
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