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Last week a certain comedian had an audience of 120 people for each of

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Bunuel
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Last week a certain comedian had an audience of 120 people for each of [#permalink] New post   05 Aug 2023, 10:00
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Last week a certain comedian had an audience of 120 people for each of the afternoon performances and 195 people for each of the evening performances. What was the average (arithmetic mean) number of people in an audience if the comedian gave only afternoon and evening performances last week?

(1) Last week the comedian gave 3 more evening performances than afternoon performances.

(2) Last week the comedian gave twice as many evening performances as afternoon performances.
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Re: Last week a certain comedian had an audience of 120 people for each of [#permalink] New post   05 Aug 2023, 22:33
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Bunuel wrote:
Last week a certain comedian had an audience of 120 people for each of the afternoon performances and 195 people for each of the evening performances. What was the average (arithmetic mean) number of people in an audience if the comedian gave only afternoon and evening performances last week?

(1) Last week the comedian gave 3 more evening performances than afternoon performances.

(2) Last week the comedian gave twice as many evening performances as afternoon performances.


  • Number of afternoon performance shows = \(x\)
  • Number of evening performance shows = \(y\)

Average number of people = \(\frac{120x + 195y}{(x+y)}\)

Statement 1

(1) Last week the comedian gave 3 more evening performances than afternoon performances.

y = x + 3

Average number of people = \(\frac{120x + 195y}{(x+y)}\)

Average number of people = \(\frac{120x + 195(x+3)}{(x+ x + 3)}\)

Average number of people = \(\frac{120x + 195x+(195*3)}{2x+3}\)

Average number of people = \(\frac{315x+(195*3)}{2x+3}\)

We have a variable in x. As the value of x is not known to us, we cannot find the average.

The statement alone is not sufficient, and we can eliminate A and D.

Statement 2

(2) Last week the comedian gave twice as many evening performances as afternoon performances.

y = 2x

Average number of people = \(\frac{120x + 195(2x)}{(x+2x)}\)

Average number of people = \(\frac{x(120 + 195*2)}{3x}\)

The variable x gets canceled out from the numerator and from the denominator. We have a definite answer to the question.

This statement is sufficient to answer the question.

Option B
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Re: Last week a certain comedian had an audience of 120 people for each of [#permalink] New post   03 Sep 2023, 16:47
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Bunuel wrote:
Last week a certain comedian had an audience of 120 people for each of the afternoon performances and 195 people for each of the evening performances. What was the average (arithmetic mean) number of people in an audience if the comedian gave only afternoon and evening performances last week?

(1) Last week the comedian gave 3 more evening performances than afternoon performances.

(2) Last week the comedian gave twice as many evening performances as afternoon performances.


If a is the number of afternoon performances, and e is the number of evening performances, then we need to answer the question:

(120a + 195e)/(a + e) = ?

Statement One Alone:

=> Last week the comedian gave 3 more evening performances than afternoon performances.

e = a + 3

Without knowing the ratio of the weights in the above weighted average, we can’t answer the question.

Statement one is not sufficient. Eliminate answer choices A and D.

Statement Two Alone:

=> Last week the comedian gave twice as many evening performances as afternoon performances.

e = 2a

Since we know the ratio of the weights, we could determine the value of the weighted average in question.

Statement two is sufficient.

Answer: B
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Last week a certain comedian had an audience of 120 people for each of [#permalink] New post   31 Mar 2024, 20:51
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Bunuel wrote:
Last week a certain comedian had an audience of 120 people for each of the afternoon performances and 195 people for each of the evening performances. What was the average (arithmetic mean) number of people in an audience if the comedian gave only afternoon and evening performances last week?

(1) Last week the comedian gave 3 more evening performances than afternoon performances.

(2) Last week the comedian gave twice as many evening performances as afternoon performances.

If we really understand the concept of the Teeter Totter for weighted averages, ­we can recognize that (1) is Insufficent and (2) is Sufficient within seconds.

Whenever we have the endpoints and the ratio of the weights, we can find the weighted average. (2) gives us a 2:1 ratio for evening to afternoon performances. (1) does not give us the ratio of the weights; for example, we could have a 4:1 ratio, or a 103:100 ratio of evening to afternoon performances.

Or, as a real-life example, if we know the profit margins for 2 products, and the weighting of how much revenue we have for each product, we could find the weighted average for the overall profit margin. 

If we're comfortable with this, we may not need to draw the diagram, but here's what it would look like if we needed to solve:




Here's a playlist with 4 basic examples going over the process for this Teeter Totter method: https://www.youtube.com/playlist?list=PL2exXfCUscn8Hvafet5-IPH1eNNLSjQBP


 ­
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Last week a certain comedian had an audience of 120 people for each of [#permalink]
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