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# Mike makes 7 phone calls on the first day of a local fundraising drive

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Math Expert
Joined: 02 Sep 2009
Posts: 44617
Mike makes 7 phone calls on the first day of a local fundraising drive [#permalink]

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13 Apr 2018, 03:18
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(N/A)

Question Stats:

100% (01:32) correct 0% (00:00) wrong based on 15 sessions

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Mike makes 7 phone calls on the first day of a local fundraising drive and increases the number of calls he makes by 4 calls each day for the 10-day drive. What is the average number of calls he makes during the drive?

A. 25
B. 27
C. 28
D. 50
E. 54
[Reveal] Spoiler: OA

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Joined: 04 Apr 2018
Posts: 21
Re: Mike makes 7 phone calls on the first day of a local fundraising drive [#permalink]

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13 Apr 2018, 03:34
Since the increase is constant and the total number of days = 10 days which is even.
Hence the average can simply be calculated by taking mean of 5th and 6th day which is 7+(4*4) & 7+(4*6)= (23,37)
So the average = (23+27)/2 = 25. Option A. 25
SC Moderator
Joined: 22 May 2016
Posts: 1551
Mike makes 7 phone calls on the first day of a local fundraising drive [#permalink]

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14 Apr 2018, 18:01
Bunuel wrote:
Mike makes 7 phone calls on the first day of a local fundraising drive and increases the number of calls he makes by 4 calls each day for the 10-day drive. What is the average number of calls he makes during the drive?

A. 25
B. 27
C. 28
D. 50
E. 54

After the first day of 7 calls, he increases each day's number of calls by 4 for 9 days. It's an arithmetic progression, whose average =

$$\frac{FirstTerm+LastTerm}{2}$$

First term = 7
Last term* = (7 + (9*4)) = (7 + 36) = 43
Or just list terms: 7,11,15,19,23,27,31,35,39,43

$$\frac{7+43}{2}=25$$

$$A_{n} = A_1 + (n-1)d$$
$$A_{10} = 7 + (9)*(4) = 43$$

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Mike makes 7 phone calls on the first day of a local fundraising drive   [#permalink] 14 Apr 2018, 18:01
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