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Of 30 theater tickets sold, 20 tickets were sold at prices b

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Of 30 theater tickets sold, 20 tickets were sold at prices b [#permalink] New post 23 Feb 2017, 09:10
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Question Stats:

70% (00:46) correct 29% (00:23) wrong based on 117 sessions


Of 30 theater tickets sold, 20 tickets were sold at prices between $10 and $30 each and 10 tickets were sold at prices between $40 and $60 each.

Quantity A
Quantity B
The average (arithmetic mean) of the prices of the 30 tickets
$50


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: Of 30 theater tickets sold, 20 tickets were sold at prices b [#permalink] New post 28 Feb 2017, 16:06
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Explanation

Here we have two types of tickets

Type A: 20 tickets @ $10 to $30
Type B: 10 tickets @ $40 to $60


Now for the average of tickets to be maximum the tickets need to be sold at the highest possible price. For type A it is $30 and type B is $60.

Average price of tickets = \(\frac{20 \times $30 + 10 \times $60}{20 +10}= $40\).

Hence Quantity B is greater.


If we select the lower most prices the average price of tickets = \(\frac{20 \times $10 + 10 \times $40}{20 +10}= $20\).

Quantity B is still greater (as expected).

Hence option B is correct.
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Re: Of 30 theater tickets sold, 20 tickets were sold at prices b [#permalink] New post 14 Nov 2018, 10:15
sandy wrote:
Explanation

Here we have two types of tickets

Type A: 20 tickets @ $10 to $30
Type B: 10 tickets @ $40 to $60


Now for the average of tickets to be maximum the tickets need to be sold at the highest possible price. For type A it is $30 and type B is $60.

Average price of tickets = \(\frac{20 \times $30 + 10 \times $60}{20 +10}= $40\).

Hence Quantity B is greater.


If we select the lower most prices the average price of tickets = \(\frac{20 \times $10 + 10 \times $40}{20 +10}= $20\).

Quantity B is still greater (as expected).

Hence option B is correct.


Is it necessary to check for lower range?
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Re: Of 30 theater tickets sold, 20 tickets were sold at prices b [#permalink] New post 14 Nov 2018, 11:50
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As I can see, sandy did.
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Re: Of 30 theater tickets sold, 20 tickets were sold at prices b [#permalink] New post 14 Nov 2018, 13:04
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AE wrote:

Is it necessary to check for lower range?


Not necessary in this case but its a good practice if you are fast enough. It's, what we call, a sanity check. On the test you might commit silly calculation errors, this is a good way to mitigate it.

Also a good practice while in non test situations.
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Re: Of 30 theater tickets sold, 20 tickets were sold at prices b [#permalink] New post 13 Dec 2018, 11:46
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Answer: B

30 sold tickets
20 of them: priced between 10$ and 30$ each
10 of them: priced between 40$ and 60$ each

A: The average of prices of the 30 tickets,
We don’t have any accurate data about the exact price of the tickets, the only thing we know is their prices ranges. But we can construe their min and max possible values:

20 of them: priced between 10$ and 30$ each
For having the max: at least one ticket is cost 10$ and the others 30$

10 of them: priced between 40$ and 60$ each
For having the max: at least one ticket is cost 40$ and the others 60$

But in this case will the average of the ticket prices be 50 or more?
We assumed having 9 tickets with 60$, 1 with 40$, 19 with 30$ and 1 with 10$ to have the maximum possible average:
But it is obvious that their mean is less than 50$ because +10$ in 60$ tickets will compensate for just some of the lower prices than 50, and the others will be still less than 50$ So the mean is less than 50$ and B is bigger than A.
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Re: Of 30 theater tickets sold, 20 tickets were sold at prices b [#permalink] New post 18 Mar 2019, 09:38
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Carcass wrote:


Of 30 theater tickets sold, 20 tickets were sold at prices between $10 and $30 each and 10 tickets were sold at prices between $40 and $60 each.

Quantity A
Quantity B
The average (arithmetic mean) of the prices of the 30 tickets
$50


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


Let's examine the EXTREME CASES

Case i: the MINIMUM possible average price of the 30 tickets
To MINIMIZE the average price, we'll consider:
20 tickets sold at $10 each, and 10 tickets sold at $40 each
Cost of all 30 tickets = (20 x $10) + (10 x $40) = $200 + $400 = $600
So, average price of the 30 tickets = $600/30 = $20
We get:
QUANTITY A: $20
QUANTITY B: $50
In this case, Quantity B is greater

Case ii: the MAXIMUM possible average price of the 30 tickets
To MAXIMIZE the average price, we'll consider:
20 tickets sold at $30 each, and 10 tickets sold at $60 each
Cost of all 30 tickets = (20 x $30) + (10 x $60) = $600 + $600 = $1200
So, average price of the 30 tickets = $1200/30 = $40
We get:
QUANTITY A: $40
QUANTITY B: $50
In this case, Quantity B is greater

Since we've examined the two EXTREME cases, we can be certain that Quantity B is greater

Answer: B

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Re: Of 30 theater tickets sold, 20 tickets were sold at prices b [#permalink] New post 22 Mar 2019, 01:13
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Answer: B
Total number of tickets: 30
Between 10 and 30$: 20
Between 40 and 60$: 10

The most possible value would be if 19 out of 20 were sold by 30 and 1 by 10 dollars.
And for the other 10, 9 in price 60 and 1 in price 40.
Just create a sequence like :
10, 30,30,30,..,30, 40, 60, 60, …, 60
There are 9 ones in 60 which are 10 more than B (50) and they will never compensate that 19 tickets with price 30 (each 20 less than 50) or the two others (40,10). So the average is definitely less than 50 even in the maximum case.
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Re: Of 30 theater tickets sold, 20 tickets were sold at prices b   [#permalink] 22 Mar 2019, 01:13
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