a museum sold admission tickets

[Needgmat]

for a certain art exhibit, a museum sold admission tickets to group of 30 people every 5 minutes from 9:00 in the morning to 5:55 in the afternoon, inclusive. The price of regular admission ticket was $10 and the price of a student ticket was $6. If on one day 3 times as many regular admission tickets were sold as student tickets, what was the total revenue from ticket sales that day?

A) $24960

B) $25920

C) $28080

D) $28500

E) $29160

OA E

[GMATGuruNY]

for a certain art exhibit, a museum sold admission tickets to group of 30 people every 5 minutes from 9:00 in the morning to 5:55 in the afternoon, inclusive. The price of regular admission ticket was $10 and the price of a student ticket was $6. If on one day 3 times as many regular admission tickets were sold as student tickets, what was the total revenue from ticket sales that day?

A) $24960

B) $25920

C) $28080

D) $28500

E) $29160

Of every 4 tickets sold, 3 are $10 regular tickets and 1 is a $6 student ticket:
3(10) + 6 = 36.

30 tickets are sold every 5 minutes.
Thus, in 60 minutes -- 12 times as long -- the number of tickets sold = 12*30 = 360.

Every 4 of these tickets brings in $36.
Thus, the amount earned every hour by 360 tickets = 360/4 * 36 = 90*36 = 3240.

Thus, in the 9 hours between 9 and 6pm, the total earnings = 9*3240 = 29,160.

The correct answer is E.

Please note that the last set of 30 tickets is sold at 5:55pm.
Thus, even in the last hour, a total of 360 tickets are sold.

[Needgmat]

Hi Sir ,

Can you please check and advise, whats wrong in my approach.

From 9 to 5:00 we have 8 hrs

so in minutes it will be 8*60 = 480 +55 = 535 right?

It is given that tickets sold in every 5 minutes so it will be 535/5 = 107.

now 107*30 = 3210.

Let the regular admission tickets be R and students tickets be S.

The equation will be R+S = 3210.

it is given that R=3S

3S+S = 3210

4S=3210

The result will not be an integer, but it has to be an integer.

Please advise..

Many thanks in advance.

Kavin

[regor60]

Hi Sir ,

Can you please check and advise, whats wrong in my approach.

From 9 to 5:00 we have 8 hrs

so in minutes it will be 8*60 = 480 +55 = 535 right?

It is given that tickets sold in every 5 minutes so it will be 535/5 = 107.

now 107*30 = 3210.

Let the regular admission tickets be R and students tickets be S.

The equation will be R+S = 3210.

it is given that R=3S

3S+S = 3210

4S=3210

The result will not be an integer, but it has to be an integer.

Please advise..

Many thanks in advance.

Kavin

You are missing the prompt in the question indicating that the ticket interval is inclusive of both the start time and the end time.

Your approach includes just the end time in the interval and misses the 30 tickets sold at 9:00.

For example, the question could read 1 pm to 3 pm inclusive. The naive approach is just to subtract 1 from 3 to equal 2.

Well, if you count two up from 1, your first number is 2 and the second is 3.

But you would have missed the activity that takes place at 1.

[[email protected]]

Hi All,

The math 'steps' behind this question can be done in a couple of different 'orders', depending on how you choose to think about the given information. Here's an approach that accounts for the difference in the price of the two types of tickets at the end of the calculation:

We're told that 30 tickets are sold every 5 minutes; there are 12 'groups' of 5 minutes in each hour, so...

(30)(12) = 360 tickets sold per hour (meaning from 9:00 to 9:55)

From 10:00 to 10:55, another 360 tickets are sold, etc. From 9:00am to 5:55pm, we have 9 hours worth of sales...

(9)(360) = 3240 total tickets

IF every ticket sold cost $10, then we'd have...

(3240)(10) = $32,400

HOWEVER, 1 out of every 4 tickets was actually a student ticket (which costs $6 instead of $10), so we have to subtract $4 for every student ticket in the total.

(3240 total tickets)(1/4) = 820 student tickets
(820 student tickets)($4) = $3280

Total revenue = $32,400 = $3280 = $29,160

Final Answer: E

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