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There are 15 children going to a movie. Tickets cost 5 dollars each

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There are 15 children going to a movie. Tickets cost 5 dollars each  [#permalink]

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New post 04 May 2017, 01:52
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Difficulty:

  55% (hard)

Question Stats:

75% (02:24) correct 25% (02:25) wrong based on 111 sessions

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There are 15 children going to a movie. Tickets cost 5 dollars each, popcorn costs $3 per serving, and candy costs $2.50 per serving. If each child gets one snack, popcorn or candy, and $21 are spent for popcorn, what is the average (arithmetic mean) cost of the trip per child to the nearest cent?

A. $2.73
B. $3.86
C. $5.96
D. $7.70
E. $7.73

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There are 15 children going to a movie. Tickets cost 5 dollars each  [#permalink]

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New post Updated on: 04 May 2017, 05:17
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There are 15 children going to the movie, and each must have a snack

Given data :
Money spent on popcorn : 21$
Cost of a popcorn : 3$

We can infer the people eating popcorn is \(\frac{21}{3}\) or 7
Hence people eating candy is 8, money spent is 20$(8 * $2.5)

After we figure this out, without making any calculations, we can infer the cost of food to be between 2.5 and 3.
The reason for that is there are almost the same number of people eating each snack
Hence the price of food is in that range,2.73
Price spent per child = Price of ticket+Price of food = 5+2.73 = 7.73(Option E)

Thanks for pointing out the error. Please find corrected solution
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Originally posted by pushpitkc on 04 May 2017, 02:23.
Last edited by pushpitkc on 04 May 2017, 05:17, edited 1 time in total.
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Re: There are 15 children going to a movie. Tickets cost 5 dollars each  [#permalink]

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New post 04 May 2017, 04:02
pushpitkc wrote:
There are 15 children going to the movie, and each must have a snack

Given data :
Money spent on popcorn : 21$
Cost of a popcorn : 3$

We can infer the people eating popcorn is \(\frac{21}{3}\) or 7
Hence people eating candy is 8, money spent is 20$(8 * $2.5)

After we figure this out, without making any calculations, we can infer the answer to be between 2.5 and 3.
The reason for that is there are almost the same number of people eating each snack
Only one answer choice is in that range,2.73(Option A)



Buddy I think you missed the price of the ticket per person ..
So 2.73 + 5 = 7.73
Answer : E
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Re: There are 15 children going to a movie. Tickets cost 5 dollars each  [#permalink]

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New post 04 May 2017, 05:17
mihir0710 wrote:
pushpitkc wrote:
There are 15 children going to the movie, and each must have a snack

Given data :
Money spent on popcorn : 21$
Cost of a popcorn : 3$

We can infer the people eating popcorn is \(\frac{21}{3}\) or 7
Hence people eating candy is 8, money spent is 20$(8 * $2.5)

After we figure this out, without making any calculations, we can infer the answer to be between 2.5 and 3.
The reason for that is there are almost the same number of people eating each snack
Only one answer choice is in that range,2.73(Option A)



Buddy I think you missed the price of the ticket per person ..
So 2.73 + 5 = 7.73
Answer : E



Thanks, made the change to the solution!
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Re: There are 15 children going to a movie. Tickets cost 5 dollars each  [#permalink]

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New post 06 May 2017, 17:59
Bunuel wrote:
There are 15 children going to a movie. Tickets cost 5 dollars each, popcorn costs $3 per serving, and candy costs $2.50 per serving. If each child gets one snack, popcorn or candy, and $21 are spent for popcorn, what is the average (arithmetic mean) cost of the trip per child to the nearest cent?

A. $2.73
B. $3.86
C. $5.96
D. $7.70
E. $7.73


We are given that 15 children are going to a movie and that tickets costs $5 each, popcorn costs $3 per serving, and candy costs $2.50 per serving. Each child gets 1 snack and $21 is spent on popcorn. We need to determine the average cost per child for the trip.

Since 21 dollars is spent on popcorn, 21/3 = 7 children purchased popcorn. Thus, 8 children purchased candy, which means 8 x 2.5 = $20 was spent on candy. We also know that 15 x 5 = $75 dollars was spent on tickets. Let’s now determine the average amount of money spent per child:

(21 + 20 + 75)/15 = 116/15

≈ 7.73

Answer: E
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Re: There are 15 children going to a movie. Tickets cost 5 dollars each  [#permalink]

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New post 07 May 2017, 05:48
Bunuel wrote:
There are 15 children going to a movie. Tickets cost 5 dollars each, popcorn costs $3 per serving, and candy costs $2.50 per serving. If each child gets one snack, popcorn or candy, and $21 are spent for popcorn, what is the average (arithmetic mean) cost of the trip per child to the nearest cent?

A. $2.73
B. $3.86
C. $5.96
D. $7.70
E. $7.73


$21 for popcorn = 21/3 = 7 servings of popcorn

i.e. remaining 8 children got one candy each = 8*2.5 = $20

Ticket cost = 15*5 = $75

Average cost = total cost / total children = (75+21+20)/15 = $7.73


Answer: Option E
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Re: There are 15 children going to a movie. Tickets cost 5 dollars each  [#permalink]

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New post 18 Jul 2017, 12:09
The long division of 116/15 took a while only because there was another option (D) that was close to the answer. I always thought GMAT spaces answer choices fairly widely, no?
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Re: There are 15 children going to a movie. Tickets cost 5 dollars each   [#permalink] 18 Jul 2017, 12:09
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