In a certain batch of guests in a museum, there are 50 guests; each gu

Question

In a certain batch of guests in a museum, there are 50 guests; each guest buys either a $40 ticket or a $60 ticket, with at least one guest of each ticket type. The average (arithmetic mean) value of ticket-receipts from the batch is more than $50. If the average value of ticket-receipts is to be reduced to less than $50 by including few new guests with $40 tickets, what could definitely NOT be the number of new guests with $40 tickets that could be included?

I. 1
II. 2
III. 3

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

Show Answer

Official Answer

D

I. 1
II. 2
III. 3

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

Capakki · Answer

I am not sure if my approach is correct. Giving a try.
if the split between $40 and $60 tickets is equal then (40x25)+(60x25)=2500. Avg=2500/50=50.
Now, to keep the avg above 50, number of people with $60 tickets needs to be increased.
New split: 24x40$ and 26x60$= 2520. Avg=2520/50=50.xy>50.0. Since its given that avg is more than 50, so we will solve further using this split, i.e. 24 tickets for 40$ and 26 tickets for $60.

To bring the avg below 50, lets add one to $40 ticket count
Split: 25x40$+26x60$=2560. Avg=2560/51=50.xy>50.00. Hence '1' can definitely not be included. 
Adding one more to $40 count
Split: 26x40+ 26x60=2600. Avg=2600/52=50.00. Not less than 50. Hence '2' can definitely not be included.
Adding one more:
Split: 27x40 + 26x60=2640. Avg=2640/53=49.xy<50.00. Hence '3' can be there atleast in one of the possible case.

Therefore, option D is the answer.

It took me some time. A different approach might save some time.

rocky620 · Answer

If the average is exactly was exactly $50 than equal number of guests must have bought $40 & $60 tickets (25 each). But, since the average is greater than $50, more than 25 guests bought $60 tickets. The minimum will be 26 guests, as per this 24 guests bought $40 tickets.

To bring the average below $50 we need to add minimum 3 guests who will buy $40 ticket.

So 1 & 2 cannot be the value.

Option D

To bring the average below $50 we need to add minimum 3 guests who will buy $40 ticket.

So 1 & 2 cannot be the value.

Option D

dhruv23320 · Answer

Let the original batch have k guests with $60 tickets and 50 − k with $40 tickets.

Total revenue = 40(50 − k) + 60k = 2000 + 20k.
Given the average is more than $50:
(2000 + 20k) / 50 > 50
2000 + 20k > 2500
20k > 500 → k > 25

The minimum possible k is 26, which gives the smallest total revenue above $50 average:
Total revenue = 2000 + 20(26) = 2520.

Now add n new guests with $40 tickets. New average must be less than $50:
(2520 + 40n) / (50 + n) < 50
2520 + 40n < 2500 + 50n
20 < 10n
n > 2

So at least 3 new $40 guests are required.

Thus:
1 and 2 are impossible, 3 is possible.

Answer: D (I and II only)

In a certain batch of guests in a museum, there are 50 guests; each gu

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In a certain batch of guests in a museum, there are 50 guests; each gu

Prep Toolkit

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