Video solution from Quant Reasoning:

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Ticket-sale reveue for:
Thursday night: 200*40=8,000
Friday night: 240*50=12,000
Saturday afternoon: 220*40=8,800
Saturday night: 300*50=15,000

Maximum: 15,000
Minimum: 8,000

As a result, the difference between the maximum and the minimum ticket-sale revenue is: 15,000-8,000=7,000

The correct answer is D.

Solution

Given:

• The data in the table shows the ticket sales from four performances of a certain play

To find:

• The difference between the maximum and the minimum ticket-sale revenue from a single performance

Approach and Working:
We can calculate the revenues for each show as follows:
Ticket-sale Revenue = ticket price * number of tickets sold

Therefore,

• Revenue on Thursday night = 40 * 200 = $8000
• Revenue on Friday night = 50 * 240 = $12000
• Revenue on Saturday afternoon = 40 * 220 = $8800
• Revenue on Saturday night = 50 * 300 = $15000

Maximum revenue = $15000 and minimum revenue = $8000

Hence, the difference between the maximum and the minimum ticket-sale revenue from a single performance = ($15000 - $8000) = $7000

Hence, the correct answer is option D.

Answer: D
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This question can be made simpler by considering that Thursday night's ticket price and number of tickets sold is the lowest of all four days, and vice versa for Saturday night.

So that means the minimum revenue has to be for Thursday night, i.e. $8,000, and highest for Saturday night, i.e. $15,000. Then calculations are as per everyone else'.

It's just good to be able to remove two potential calculations out of the equation early on, saves on time.
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The maximum revenue was 50 x 300 = $15,000.

The minimum revenue was 40 x 200 = $8,000.

So the difference is 15,000 - 8,000 = $7,000.

Answer: D
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The wording "single performance" causes confusion here.

Solution :

$40 * 200 = 8000 <- Min
$50 * 240 = 12000
$40 * 220 = 8800
$50 * 300 = 15000 <- Max

15,000 - 8000 = 7,000

Hence D.

Performance Time Ticket Price # of Tickets Sold Revenue Max Min Difference
Thursday 40 200 8000 15000 8000 7000
Friday 50 240 12000
Saturday afternoon 40 220 8800
Saturday night 50 300 15000

[quote="Bunuel"]

The table shows a summary of the ticket sales from four performances of a certain play. What is the difference between the maximum and the minimum ticket-sale revenue from a single performance?

A. $4,000

B. $5,100

C. $6,200

D. $7,000

E. $9,600

Min/MAx problem:

Max: 15000
Min: 8000

Thu: 8000
Fri: 12000
Sat-afternoon-8000
Sat-night-15000
Difference: 15000- 8000=7000
OA:D

Easy: 15000-8000= 7K

Thursday Night Revenue is 4*20 = 80 (Ignoring the zeros) { Max }
Friday Night Revenue is 5*24 = 120 (Ignoring the zeros)
Saturday Afternoon Revenue is 4*22 = 88 (Ignoring the zeros)
Saturday Night Revenue is 5*30 = 150 (Ignoring the zeros) { Max }

So, The difference between the maximum and the minimum ticket-sale revenue from a single performance is 150 - 80 = 70 , Answer must be (D)
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Answer: Option D

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Solution:

Minimum value should be b/w Thu and Sat afternoon with 40$/ticket whereas maximum price between Friday and Sat night with 50$/ticket
(as 220 corresponding to 40$,the bigger value b/w 200 and 220 is less than both 300 and 240 corresponding to 50$)

Between Thu and Sat the minimum value = 40*200 = 8000$

Between Fri and Sat night the maximum value =50*300= 15000$

Difference = 15000$-8000$

=7000$ (option d)

Devmitra Sen
GMAT SME
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