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# The average (arithmetic mean) length per film for a group of 21 films

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The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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05 May 2016, 19:27
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Difficulty:

15% (low)

Question Stats:

78% (01:49) correct 22% (02:06) wrong based on 622 sessions

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The average (arithmetic mean) length per film for a group of 21 films is t minutes. If a film that runs for 66 minutes is removed from the group and replaced by one that runs for 52 minutes, what is the average length per film, in minutes, for the new group of films, in terms of t?

A) $$t + \frac{2}{3}$$

B) $$t - \frac{2}{3}$$

C) $$21t+14$$

D) $$t + \frac{3}{2}$$

E) $$t - \frac{3}{2}$$

Spoiler: :: Solution
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Re: The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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05 May 2016, 20:09
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mcelroytutoring wrote:
The average (arithmetic mean) length per film for a group of 21 films is t minutes. If a film that runs for 66 minutes is removed from the group and replaced by one that runs for 52 minutes, what is the average length per film, in minutes, for the new group of films, in terms of t?

A) $$t + \frac{2}{3}$$

B) $$t - \frac{2}{3}$$

C) $$21t+14$$

D) $$t + \frac{3}{2}$$

E) $$t - \frac{3}{2}$$

hi,
you do not have to get into any calculations if you understand and take the Q logically..

avg time for 21 is t min..
you add 52 and remove 66.. so basically you are removing 14 min from total..
effect on each or average =$$\frac{14}{21}= \frac{2}{3}$$..

so final average = $$t-\frac{2}{3}$$...
B
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Re: The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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Updated on: 06 May 2016, 12:30
1
Logic tells us that the average length per film will go down, so we can eliminate all answers that add to the original average of t. That leaves either B or E. When executing the algebra, the key step is being able to quickly and easily turn one fraction with two numerators $$(\frac{21t-14}{21})$$ into two fractions with one numerator each $$(\frac{21t}{21}-\frac{14}{21})$$.

Above is a visual that should help.
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Originally posted by mcelroytutoring on 05 May 2016, 19:28.
Last edited by mcelroytutoring on 06 May 2016, 12:30, edited 6 times in total.
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Re: The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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15 Dec 2016, 14:05
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Great Solutions above.
Here is my solution =>
Average length = t minutes
Sum(21) = 21t
New sum = 21t-66+52 = 21t-14
Hence new mean = 21t-14 / 21 = t-2/3

Hence B

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Re: The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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20 Jul 2017, 15:08
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average = sum of numbers/total number
t = s/21
s = 21t
Now as per question
s =21t -66 +52
s = 21t - 14
New average
t1 = (21t - 14) / 21
=> t - 2/3
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Re: The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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06 Mar 2018, 09:16
stonecold wrote:
Great Solutions above.
Here is my solution =>
Average length = t minutes
Sum(21) = 21t
New sum = 21t-66+52 = 21t-14
Hence new mean = 21t-14 / 21 = t-2/3

Hence B

stonecold how from this 21t-14 / 21 you got t-2/3 , I understand you divided by 7 , but there are two 21 could you write in detail
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Re: The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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06 Mar 2018, 11:18
1
dave13 wrote:
stonecold wrote:
Great Solutions above.
Here is my solution =>
Average length = t minutes
Sum(21) = 21t
New sum = 21t-66+52 = 21t-14
Hence new mean = 21t-14 / 21 = t-2/3

Hence B

stonecold how from this 21t-14 / 21 you got t-2/3 , I understand you divided by 7 , but there are two 21 could you write in detail

Hey dave13

$$\frac{21t-14}{21} = \frac{21t}{21} - \frac{14}{21} = t - \frac{2}{3}$$

Hope that helps
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Re: The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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15 Aug 2018, 16:09
Hi,
We can use deviation technique to solve such type of questions. The answer would be in less than 30 Secs

refer: https://gmatclub.com/forum/a-student-s- ... l#p2113194

So new average would be
t-(14/21)
t-2/3
Hence B.
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Re: The average (arithmetic mean) length per film for a group of 21 films   [#permalink] 15 Aug 2018, 16:09
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