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A movie theater sold 120 tickets to the matinee showing of a popular

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A movie theater sold 120 tickets to the matinee showing of a popular [#permalink]

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A movie theater sold 120 tickets to the matinee showing of a popular children's movie, and 150 tickets to the evening showing. The theater sold the same number of adult tickets for each show, but for the evening show was just 20 children's tickets short of selling twice as many children's tickets as it did for the matinee. How many children's tickets were sold to the evening show?

A. 50
B. 60
C. 70
D. 80
E. 90
[Reveal] Spoiler: OA

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Re: A movie theater sold 120 tickets to the matinee showing of a popular [#permalink]

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New post 23 Feb 2017, 06:56
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MS = 120; ES = 150

Adult and MS = x; Adult and ES = x.
Child and MS = y; Child and ES = 2y - 20.

x + y = 120
x + 2y - 20 = 150 --> x + 2y = 170

Solving the two equations, we get x = 70; y = 50

We need the value of Child and ES = 2y - 20 = 80

Answer: D
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Re: A movie theater sold 120 tickets to the matinee showing of a popular [#permalink]

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Bunuel wrote:
A movie theater sold 120 tickets to the matinee showing of a popular children's movie, and 150 tickets to the evening showing. The theater sold the same number of adult tickets for each show, but for the evening show was just 20 children's tickets short of selling twice as many children's tickets as it did for the matinee. How many children's tickets were sold to the evening show?

A. 50
B. 60
C. 70
D. 80
E. 90


The theater sold the same number of adult tickets for each show
Let A = # of adult tickets sold for MATINEE show
So, A = # of adult tickets sold for EVENING show

The evening show was just 20 children's tickets short of selling twice as many children's tickets as it did for the matinee
In other words, if they had sold 20 extra children's tickets for the evening show, then the number of children's tickets for the evening show would have been TWICE the number of children's tickets for the matinee show
So, we can write: (# of EVENING show children's tickets) + 20 = 2(# of MATINEE show children's tickets)
Another way to write this is: (# of EVENING show children's tickets) = 2(# of MATINEE show children's tickets) - 20

Let C = # of children's tickets sold for MATINEE show
So, 2C - 20 = # of children's tickets sold for EVENING show

A movie theater sold 120 tickets to the matinee showing of a popular children's movie, and 150 tickets to the evening showing
We can write the following:
A + C = 120
A + (2C - 20) = 150

Rewrite as:
A + 2C = 170
A + C = 120

Subtract bottom equation from top equation to get: C = 50
So, 50 children's tickets were sold for the MATINEE show.

How many children's tickets were sold to the evening show?
We already know that 2C - 20 = # of children's tickets sold for EVENING show
Since C = 50, we can replace C with 50 to get: 2(50) - 20 = # of children's tickets sold for EVENING show
Evaluate: 80 = # of children's tickets sold for EVENING show

Answer: D

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Re: A movie theater sold 120 tickets to the matinee showing of a popular [#permalink]

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New post 27 Feb 2017, 11:47
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Bunuel wrote:
A movie theater sold 120 tickets to the matinee showing of a popular children's movie, and 150 tickets to the evening showing. The theater sold the same number of adult tickets for each show, but for the evening show was just 20 children's tickets short of selling twice as many children's tickets as it did for the matinee. How many children's tickets were sold to the evening show?

A. 50
B. 60
C. 70
D. 80
E. 90


We can let the number of adult tickets sold for each show = x, the number of children’s tickets sold to the matinee = y, and the number of children's tickets sold to the evening show = z.

Since 120 matinee tickets were sold:

x + y = 120

x = 120 - y

Since 150 evening tickets were sold:

x + z = 150

x = 150 - z

Since the evening show was just 20 children's tickets short of selling twice as many children's tickets as were sold for the matinee:

z + 20 = 2y

Since x = 120 - y and x = 150 - z, we have:

120 - y = 150 - z

z = 30 + y

Next, we can substitute 30 + y for z in the equation z + 20 = 2y and we have:

30 + y + 20 = 2y

50 = y

Thus, z = 30 + 50 = 80.

Answer: D
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Re: A movie theater sold 120 tickets to the matinee showing of a popular [#permalink]

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New post 24 Mar 2017, 11:28
we can solve the question with equation. But first what is the unswer? it's the number of children tickets sold for the evening show: 2y-20
So here are the equations:
x+y=120
x+2y-20=150
y=50
2*50-20=80 tickets were sold for the evening show for children
Answer is D
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Re: A movie theater sold 120 tickets to the matinee showing of a popular [#permalink]

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New post 04 Apr 2017, 17:28
Bunuel wrote:
A movie theater sold 120 tickets to the matinee showing of a popular children's movie, and 150 tickets to the evening showing. The theater sold the same number of adult tickets for each show, but for the evening show was just 20 children's tickets short of selling twice as many children's tickets as it did for the matinee. How many children's tickets were sold to the evening show?

A. 50
B. 60
C. 70
D. 80
E. 90


Solve for one variable only because second variable is same number in both scenarios.

C = # of children's tickets

120 - C = 150 - (2C - 20)
120 - C = 150 - 2C + 20

C = 50

Evening: 2C - 20 = 80
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Re: A movie theater sold 120 tickets to the matinee showing of a popular [#permalink]

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New post 09 Jul 2017, 00:33
Bunuel wrote:
A movie theater sold 120 tickets to the matinee showing of a popular children's movie, and 150 tickets to the evening showing. The theater sold the same number of adult tickets for each show, but for the evening show was just 20 children's tickets short of selling twice as many children's tickets as it did for the matinee. How many children's tickets were sold to the evening show?

A. 50
B. 60
C. 70
D. 80
E. 90


Let the number of children for matinee be \(- C\)

Let the number of Adults be \(= A\) (No. of adults remain same for both the shows)

Matinee Show \(- C + A = 120\) ===> 1

For number of children in the evening show we ca write: \(2C - 20\)

Evening Show \(- A + 2C - 20 = 150\)

\(2C + A = 170\) ===> 2

Solving (1) and (2)

\(C = 50\)

No. of children in matinee show = 50

No. if children in evening show \(= 2C - 20 = 2 * 50 - 20 = 100 - 20 = 80\)

Hence, Answer is D
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Re: A movie theater sold 120 tickets to the matinee showing of a popular [#permalink]

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New post 05 Dec 2017, 19:35
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A movie theater sold 120 tickets to the matinee showing of a popular children's movie, and 150 tickets to the evening showing. The theater sold the same number of adult tickets for each show, but for the evening show was just 20 children's tickets short of selling twice as many children's tickets as it did for the matinee. How many children's tickets were sold to the evening show?

A. 50
B. 60
C. 70
D. 80
E. 90


Quote:
Since the evening show was just 20 children's tickets short of selling twice as many children's tickets as were sold for the matinee:

z + 20 = 2y


Do we not have to account for adult tickets here?
As per me equation should be 150 = x + (2y-20) since we substitute for z ie no children in evening show.
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Re: A movie theater sold 120 tickets to the matinee showing of a popular   [#permalink] 05 Dec 2017, 19:35
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