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The average of all the multiples of 5 between 199 and 706

The average of all the multiples of 10 between 199 and 706

A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.

Re: The average of all the multiples of 5 between 199 and 706 [#permalink]
21 Feb 2018, 02:18

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To take the average of evenly spaced numbers, we can determine it easily by taking the average of first and last numbers. Q.A The average of all the multiples of 5 between 199 and 706 is = (200+705)/2=452.5 Q.B The average of all the multiples of 10 between 199 and 706 is = (200+700)/2= 450

Re: The average of all the multiples of 5 between 199 and 706 [#permalink]
24 Feb 2018, 17:57

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A

If you don't know the trick above, as for people like me, below is the conventional way to solve it. Hope my solution is correct. Quantity A:- series -- 200, 205 .... 700, 705 (102 terms). Sum of all terms in series = [n (a + l)]/2 = 102(200+705)/2 = 905*51 Average = 905*51 / 102 = 452.5 Quantity B:- series -- 200, 210 .... 690, 700 (51 terms). Sum of all terms = 51(200+700)/2 = 51*450 Average = 51*450 / 51 = 450

Re: The average of all the multiples of 5 between 199 and 706 [#permalink]
13 Mar 2018, 15:59

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correct: A Sum of a sequence of numbers is (a1 +an) *n/2 and when we divide it by n, it will be the average of the numbers (a1+an)/2 The average of all the multiples of 5 between 199 and 706: 200, 205, ..., 700, 705 average = (a1 + an-1) / 2 = (705+200)/2 = 905/2 = 452.5 The average of all the multiples of 10 between 199 and 706: 200, 210, 220, ..., 700 average =(700 + 200)/ 2 = 900/2 = 450 _________________

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Re: The average of all the multiples of 5 between 199 and 706
[#permalink]
13 Mar 2018, 15:59