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A movie theater sold 120 tickets to the matinee showing of a popular
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23 Feb 2017, 06:40

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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?

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23 Feb 2017, 10:10

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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

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

Re: A movie theater sold 120 tickets to the matinee showing of a popular
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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
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14 Apr 2017, 08:32

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ziyuen 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

I have difficulty to answer this question. Does someone could help to interpret the sentence in blue font?

Hi,

At a straight look, children tickets in matinée show is 30+20=50.. And evening becomes 2*50-20=80..

Now how n why? Difference in tickets is 150-120=30 and this difference is ONLY in children tickets. If in morning it is C, in evening it is C+30.. 2C-20=C+30....C=50 Evening=2*50-20=80 D
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Re: A movie theater sold 120 tickets to the matinee showing of a popular
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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
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12 Jul 2017, 06:36

hazelnut 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

I have difficulty to answer this question. Does someone could help to interpret the sentence in blue font?

Let A be the number of adult ticket sold in each show (The no. of adult ticket sold in each show is same ) Let C be the number of children's ticket sold in matinee show.

So A + C = 120 ..............................(i) Also considering evening show A+2C-20 = 150 A+ 2C = 170 ...................................(ii)

So from (i) and (ii) C = 50 So, no. of children's ticket sold in evening show = 2C-20 = 2*50 -20 = 80

Answer D
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Re: A movie theater sold 120 tickets to the matinee showing of a popular
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16 Jul 2017, 17:16

hazelnut 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 a = the number of adult tickets sold for each show, x = the number of children’s tickets sold for the evening show, and y = the number of tickets sold for the matinee.

The number of children’s tickets for the evening show was just 20 tickets short of selling twice as many children's tickets as were sold for the matinee, which we can express as:

x = 2y - 20

We also have that:

a + x = 150

Substituting (2y - 20) for x, we have:

a + 2y - 20 = 150

a + 2y = 170

We also have the following equation for the matinee ticket sales:

a + y = 120

Subtracting a + y = 120 from a + 2y = 170, we have:

y = 50

So x = 2(50) - 20 = 80 children’s tickets were sold for the evening show.

Answer: D
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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 &nbs
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