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

Question

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) \(t - \frac{3}{2}\)

chetan2u · Accepted Answer

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

stonecold · Answer

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

mcelroytutoring · Answer

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.

Above is a visual that should help.

devctg · Answer

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

dave13 · Answer

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

pushpitkc · Answer

Hey  dave13

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

Hope that helps

Probus · Answer

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.

Hama95 · Answer

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

CEdward · Answer

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).

KarlLindoff · Answer

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?

100mitra · Answer

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{   }{ 21}  =  \frac{21t}{21}  -  \frac{14}{21}  =  t  -  \frac{2}{3}  =  Option B

A_Nishith · Answer

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
 Answer: B

shivani1351 · Answer

Given:
The average (arithmetic mean) length per film for a group of 21 films is t minutes. 
Hence, the total length of all the 21 films = 21t (As Average = Sum of all values/Number of all values)

Now, on replacing the film that runs for 66 mins with the one that runs for 52 mins, the total length of the films becomes = 21t-66+52 = 21t-14

Thus, the new average becomes: (21t-14)/21 (Note: Denominator = 21 as we are simply replacing one film with another, and not adding/subtracting any film as such)

Hence, the new average in terms of t = t-2/3 (Option B) (Answer)

sujoykrdatta · Answer

Since 66 minutes is replaced by 52 minutes, there is a reduction of 14 minutes from the total. 
Since there are 21 films, the average reduction is simply 14/21 = 2/3 minutes

=> New average = old average - 2/3 = t - 2/3

Answer B

PGTLrowanhand · Answer

Here's my video solution:  https://www.youtube.com/watch?v=YkyRxCLx7vI

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

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