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At a certain theater, the cost of each adult's ticket is $5 and cost

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At a certain theater, the cost of each adult's ticket is $5 and cost  [#permalink]

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New post Updated on: 05 Jan 2015, 05:04
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At a certain theater, the cost of each adult's ticket is $5 and cost of each child's ticked is $2. What was the average (arithmetic mean) cost of all adult's and children's tickets sold at the theater yesterday.

(1) Yesterday the ratio of the number of children's tickets sold at the theater to the number of adult's tickets sold at the theater was 3 to 2.

(2) Yesterday 80 adult's tickets were sold at the theater.

Originally posted by prashi82 on 21 Aug 2008, 21:21.
Last edited by Bunuel on 05 Jan 2015, 05:04, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: At a certain theater, the cost of each adult's ticket is $5 and cost  [#permalink]

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New post 03 Jan 2015, 12:56
2
Hi All,

When you're dealing with "ratio data" in a DS question, it's actually really easy to prove if a pattern exists or not - just run through a few quick TESTs...

Here, we're told:
Adult tickets cost $5 each
Children's tickets cost $2 each

We're asked for the AVERAGE COST of all tickets sold yesterday.

Fact 1: The ratio of Children's tickets to Adult's tickets was 3:2 yesterday.

Some Test Takers can clearly see that this ratio IS enough information to say that Fact 1 is SUFFICIENT. Here's how you can quickly prove the consistency...

IF...
3 children
2 adults
3(2) + 2(5) = 16/5 = $3.20 average ticket price

6 children
4 adults
6(2) + 4(5) = 32/10 = $3.20 average ticket price

9 children
6 adults
9(2) + 6(5) = 48/15 = $3.20 average ticket price

With this ratio, the average is ALWAYS $3.20
Fact 1 is SUFFICIENT

Fact 2: 80 Adult tickets were sold

Here, we don't know the number of Children's tickets, so the average ticket price would change depending on THAT number. Again, here's the proof:

0 children
80 adults
0 + 80(5) = 400/80 = $5 average ticket price

80 children
80 adults
80(2) + 80(5) = 560/160 = $3.50 average ticket price
Fact 2 is INSUFFICIENT

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Re: At a certain theater, the cost of each adult's ticket is $5 and cost  [#permalink]

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New post 05 Jan 2015, 05:07
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1
At a certain theater, the cost of each adult's ticket is $5 and the cost of each child's ticket is $2. What was the average cost of all the adult's and children's tickets sold at the theater yesterday?

The average cost = (2*C+5*A)/(C+A)

(1) Yesterday ratio of # of children's ticket sold to the # of adult's tickets sold was 3 to 2 --> 3A =2CA = 2C/3 --> the average cost = C(2+5*2/3)/(C(1+2/3)) --> (2+5*2/3)/(1+2/3). Sufficient.

(2) Yesterday 80 adult's tickets were sold at the theater --> A = 80. We know nothing about C. Not sufficient.

Answer: A
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Re: At a certain theater, the cost of each adult's ticket is $5 and cost  [#permalink]

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New post 20 Oct 2016, 16:56
1
prashi82 wrote:
At a certain theater, the cost of each adult's ticket is $5 and cost of each child's ticked is $2. What was the average (arithmetic mean) cost of all adult's and children's tickets sold at the theater yesterday.

(1) Yesterday the ratio of the number of children's tickets sold at the theater to the number of adult's tickets sold at the theater was 3 to 2.

(2) Yesterday 80 adult's tickets were sold at the theater.


We are given that the cost of each adult's ticket is $5 and cost of each child's ticket is $2. We need to determine the average (arithmetic mean) cost of all adults’ and children's tickets sold at the theater yesterday. If we let a = the number of adults’ tickets sold and c = the number of children’s tickets sold, we can create the following average equation:

Average = (5a + 2c)/(a + c)

Statement One Alone:

Yesterday the ratio of the number of children's tickets sold at the theater to the number of adult's tickets sold at the theater was 3 to 2.

Using the information in statement one, we can create the following equation:

c/a = 3/2

2c = 3a

c = 1.5a

Since c = 1.5a, we can substitute 1.5a for c in our average equation and we have:

Average = (5a + 2 x 1.5a)/(a + 1.5a)

Average = (5a + 3a)/2.5a

Average = 8a/2.5a

Average = 3.2

Statement one alone is sufficient to answer the question. We can eliminate answer choices B, C, and E.

Statement Two Alone:

Yesterday 80 adult's tickets were sold at the theater.

Only knowing the number of adults’ tickets sold yesterday is not enough information to determine the average cost of all tickets sold. Statement two alone is not sufficient to answer the question.

Answer: A
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Re: At a certain theater, the cost of each adult's ticket is $5 and cost  [#permalink]

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New post 21 Aug 2008, 21:36
prashi82 wrote:
Que:- At a certain theater, the cost of each adult's ticket is $5 and cost of each child's ticked is $2. What was the average (arithmetic mean) cost of all adult's and children's tickets sold at the theater yesterday.

a) Yesterday the ratio of the number of children's tickets sold at the theater to the number of adult's tickets sold at the theater was 3 to 2.

b) Yesterday 80 adult's tickets were sold at the theater.

Can someone help me with this one?

Thanks
Prashi82


ratio of childre't ticket: adult tickets sold = 3:2
assume that 3 and 2 tickets are sold.

a) average = cost of adult tickets + cost of childrent tickets/total tickets sold=
( 5*2+ 2*3 )/5

sufficient

b) 80 adult tickets sold. we don't about children's tickets sold.
Insuffcieint


A.
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Re: At a certain theater, the cost of each adult's ticket is $5 and cost  [#permalink]

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New post 07 Feb 2010, 17:07
But the ratio provided by A doesn't hold true if you use different values: let's say that 120 children tickets were sold and 80 adults tickets were sold (same ratio as 3 to 2) then we have:

5(120)+2(80)/200 and the average price is not the same.

I choose C for this for that reason. I know it's wrong but can someone explain. Thanks.
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Re: At a certain theater, the cost of each adult's ticket is $5 and cost  [#permalink]

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New post 08 Feb 2010, 12:29
adamsmith2010 wrote:
But the ratio provided by A doesn't hold true if you use different values: let's say that 120 children tickets were sold and 80 adults tickets were sold (same ratio as 3 to 2) then we have:

5(120)+2(80)/200 and the average price is not the same.

I choose C for this for that reason. I know it's wrong but can someone explain. Thanks.


adamsith2010 - I think the ratio does hold true if you use other values. There is an error in your equation. If we use the 120 children tickets and 80 adult like you mentioned the equation looks as follows:

5(80)+2(120)/200 =====> 640/200 = 3.2 same average as if we used the original ration.

I hope this helps :)
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Re: At a certain theater, the cost of each adult's ticket is $5 and cost  [#permalink]

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New post Updated on: 08 Feb 2010, 20:07
If you take a conceptual approach to this problem, the ratio is all you need to get weighted average of the sold tickets.
X= the number of adult tickets sold
Y=number of the children ticket sold
Average = (5*X+2*Y)/(X+Y)
As you have a ratio of Y/X, you can express Y in terms of X (or vise versa). As a result Average expression will become independent of X and Y.

A is sufficient obviously

Originally posted by alexBLR on 08 Feb 2010, 15:23.
Last edited by alexBLR on 08 Feb 2010, 20:07, edited 1 time in total.
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Re: At a certain theater, the cost of each adult's ticket is $5 and cost  [#permalink]

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New post 08 Feb 2010, 19:07
prashi82 wrote:
Que:- At a certain theater, the cost of each adult's ticket is $5 and cost of each child's ticked is $2. What was the average (arithmetic mean) cost of all adult's and children's tickets sold at the theater yesterday.

a) Yesterday the ratio of the number of children's tickets sold at the theater to the number of adult's tickets sold at the theater was 3 to 2.

b) Yesterday 80 adult's tickets were sold at the theater.

Can someone help me with this one?

Thanks
Prashi82

Answer: A

(5x+2y)/(x+y) is the average. However from 1) we know that x/y = 3/2 ie 2x = 3y. use this to solve for first equation which will be 3.2 Hence Sufficent
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Re: At a certain theater, the cost of each adult's ticket is $5 and cost  [#permalink]

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New post 09 Feb 2010, 17:03
I picked A because it gave the same average cost per ticket for different numbers i picked.

algebrically when you form the equation it is like :

( 5x+2y ) / x+y , ST 1 says that y = 3/2x , when you substitute you get one value which is 16/5 or 3.2
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Re: At a certain theater, the cost of each adult's ticket is $5 and cost  [#permalink]

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New post 18 Oct 2016, 10:26
This is a double Matrix Question :

We know Adult price is $5 and Child Price is $2.

Option A:

The ratio of C:A is 3:2

Number Price Total
Adult 2x 5 10x
Child 3x 2 6x
Total 5x 16x

Average can be found as we have the total and individual components.

Option B: Individual component of Adult number of tickets are given, giving no clue to the number of tickets sold for the children.

Answer : A.

Hope it helps.
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Re: At a certain theater, the cost of each adult's ticket is $5 and cost  [#permalink]

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New post 08 Dec 2016, 14:50
prashi82 wrote:
At a certain theater, the cost of each adult's ticket is $5 and cost of each child's ticked is $2. What was the average (arithmetic mean) cost of all adult's and children's tickets sold at the theater yesterday.

(1) Yesterday the ratio of the number of children's tickets sold at the theater to the number of adult's tickets sold at the theater was 3 to 2.

(2) Yesterday 80 adult's tickets were sold at the theater.


Statement 1 - let number of children's ticket and adult's ticket sold = 3x and 2x
So average cost = (2x *5 + 3x*2)/3x+2x
= 16x/5x = 16/5 ----sufficient

Statement 2 - Number of adult tickets sold = 80 but no info about number of children's ticket. --not sufficient.

Answer - A
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Re: At a certain theater, the cost of each adult's ticket is $5 and cost  [#permalink]

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New post 16 Feb 2017, 18:58
Here is another approach.

Assume that a adults and c children tickets were sold.
The Question can be modified as follows
(5a +2c)/(a+c) = ?

stmt says c/a = 3/2
ie c = 3a/2 ----------(1)

substitute this ie the question.
(5a + 3a)/(a +3a/2)
simplifying,
16a/5a = 16/5 .. a definite answer
So sufficient.

Stmt 2 is not sufficient.

hence A
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Re: At a certain theater, the cost of each adult's ticket is $5 and cost  [#permalink]

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New post 01 Sep 2018, 09:22
prashi82 wrote:
At a certain theater, the cost of each adult's ticket is $5 and cost of each child's ticked is $2. What was the average (arithmetic mean) cost of all adult's and children's tickets sold at the theater yesterday.

(1) Yesterday the ratio of the number of children's tickets sold at the theater to the number of adult's tickets sold at the theater was 3 to 2.

(2) Yesterday 80 adult's tickets were sold at the theater.


OA: A
Let the Adult's ticket price be \(T_{Adult}\) and Child's ticket price be \(T_{Child}\)

\(T_{Adult}=$5 \qquad T_{Child}=$2\)

Let the number of Adults be \(A\) and the number of children be \(C\).

We have to find out the average cost of all tickets sold at the theater i.e We have to find \(\frac{A*T_{Adult}+C*T_{Child}}{A+C}......(1)\)

1) Yesterday the ratio of the number of children's tickets sold at the theater to the number of adult's tickets sold at the theater was 3 to 2.

\(\frac{C}{A}=\frac{3}{2}\) ;\(C = \frac{{3A}}{2}......(2)\)

putting (2) and \(T_{Adult}=$5 ; T_{Child}=$2\) in (1), we get

Average cost of all tickets sold \(= \frac{A*T_{Adult}+C*T_{Child}}{A+C} =\frac{A*5+\frac{3A}{2}*2}{A+\frac{3A}{2}} =\frac{8A}{\frac{5A}{2}}=\frac{16}{5}\)

Statement 1 alone is sufficient.

2)Yesterday 80 adult's tickets were sold at the theater.

We know about the number of Adult's ticket sold but are not aware about the number of children ticket sold.
Statement 2 alone is not sufficient
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Re: At a certain theater, the cost of each adults ticket is $5  [#permalink]

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Re: At a certain theater, the cost of each adults ticket is $5   [#permalink] 22 Oct 2019, 08:36
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