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

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

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.

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

Hi,
Understanding averages conceptually requires some imagination.


Say we have group A and their combined strength is x
we have two persons B and C with different strength Let the strength of B> C,

Now if strength of person B is same as strength of Group A , and the person B is added to group A then then the overall strength of group A along with B does not change

However if we added Person C to the group A in place of B, then the overall strength of group would not increase for sure .

So u see how we can reject choices A, C, D ,and the two choices we are having at hand are B and E.

So in situation of time crunch, where we guess the answer choices, we need to increase our odds of getting it right. Choices A,C,D if marked show conceptual gap in understanding averages.

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.

let's consider the fraction before the number we want to switch a constant X. Because this fraction is intact when changing the 66 by the 51, we will have the following formula:

X+66/21-66/21+52/21=T+52/21-66/21

this will be X+ 52/21=T-2/3

so it is the answer B

Just want to understand this further, what happens if we do the opposite and add a film that is 14 minutes longer?

The answer then would be t + 14/21 = t + 2/3 correct?

So basically the main idea for these types of questions is

We either add or subtract | x - y | / average to the original average (x is the original term, y is the new term).

For me personally, this seems really tricky for a sub-600 question. Any experts want to comment if this is something to expect in the 500-600 range?

The average (arithmetic mean) length per film for a group of 21 films is t minutes = (\(21*t\))

If a film that runs for 66 minutes is removed from the group and replaced by one that runs for 52 minutes = (66-52)

what is the average length per film, in minutes, for the new group of films, in terms of t?

New Film =\(\frac{ [21t - (66-52)] }{ 21}\) = \(\frac{21t}{21}\) - \(\frac{14}{21}\) = \(t\) - \(\frac{2}{3}\) = Option B

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