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The average of all the multiples of 5 between 199 and 706
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15 Sep 2017, 07:38
15 Sep 2017, 07:38
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Question Stats:
79% (01:22) correct
20% (01:38) wrong based on 128 sessions
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Quantity A |
Quantity B |
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
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21 Feb 2018, 02:18
21 Feb 2018, 02:18
5
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
So, A is the answer.
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
So, A is the answer.
General Discussion
Re: The average of all the multiples of 5 between 199 and 706
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24 Feb 2018, 17:57
24 Feb 2018, 17:57
1
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
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
