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# (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?

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Math Revolution GMAT Instructor
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Updated on: 16 Sep 2018, 13:01
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44% (01:03) correct 56% (01:46) wrong based on 61 sessions

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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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"Only $99 for 3 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 13 Sep 2018, 01:39. Last edited by MathRevolution on 16 Sep 2018, 13:01, edited 1 time in total.  Math Revolution Discount Codes Magoosh Discount Codes Veritas Prep GMAT Discount Codes ##### Most Helpful Expert Reply Senior Manager Joined: 04 Aug 2010 Posts: 294 Schools: Dartmouth College Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=? [#permalink] ### Show Tags 13 Sep 2018, 09:51 1 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. _________________ GMAT and GRE Tutor Over 1800 followers Click here to learn more GMATGuruNY@gmail.com New York, NY If you find one of my posts helpful, please take a moment to click on the "Kudos" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. ##### General Discussion Manager Joined: 09 Oct 2015 Posts: 234 Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=? [#permalink] ### Show Tags 13 Sep 2018, 01:51 Is there any formula that can be applied? Posted from my mobile device Senior Manager Joined: 22 Feb 2018 Posts: 357 Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=? [#permalink] ### Show Tags 13 Sep 2018, 02:23 OA: A Sum 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 $$\frac{Sum \quad of \quad the \quad squares \quad of \quad first \quad 20 \quad natural \quad numbers}{Sum \quad \quad of \quad first \quad 20 \quad natural \quad numbers}=\frac{2*20+1}{3}=\frac{41}{3}$$ _________________ Good, good Let the kudos flow through you GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 379 Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=? [#permalink] ### Show Tags 13 Sep 2018, 09:13 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 in-the-GMAT-scope. (We have used a"finite arithmetic progression" important result.) The sum of the squares of the first "n" (n=20) positive integers is out-of-GMAT´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 "out-of-scope formula"... some "hidden pattern", for instance... Anyway, I could not find an alternate approach in approx. 5min (a generous "time-limit" in the GMAT). _________________ Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT) Course release PROMO : finish our test drive till 31/Oct with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 60% discount! GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 379 Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=? [#permalink] ### Show Tags 13 Sep 2018, 10:10 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. _________________ Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT) Course release PROMO : finish our test drive till 31/Oct with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 60% discount! GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 379 Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=? [#permalink] ### Show Tags Updated on: 13 Sep 2018, 18:23 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 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. _________________ Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT) Course release PROMO : finish our test drive till 31/Oct with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 60% discount! Originally posted by fskilnik on 13 Sep 2018, 11:31. Last edited by fskilnik on 13 Sep 2018, 18:23, edited 2 times in total. Senior Manager Joined: 04 Aug 2010 Posts: 294 Schools: Dartmouth College Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=? [#permalink] ### Show Tags 13 Sep 2018, 12:01 1 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. _________________ GMAT and GRE Tutor Over 1800 followers Click here to learn more GMATGuruNY@gmail.com New York, NY If you find one of my posts helpful, please take a moment to click on the "Kudos" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6390 GMAT 1: 760 Q51 V42 GPA: 3.82 (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=? [#permalink] ### Show Tags 16 Sep 2018, 18:31 => 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 _________________ 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$99 for 3 month Online Course"
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Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?  [#permalink]

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21 Sep 2018, 03:30
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 [#permalink] 21 Sep 2018, 03:30
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