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Adult tickets to a movie cost $9 each and children’s tickets cost $5 e
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14 Jun 2021, 10:15
14 Jun 2021, 10:15
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Adult tickets to a movie cost $9 each and children’s tickets cost $5 each. When 44 tickets were bought, the total cost was $312. How many of the tickets purchased were adult tickets?
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Show: :: OA
23
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Re: Adult tickets to a movie cost $9 each and children’s tickets cost $5 e
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14 Jun 2021, 19:27
14 Jun 2021, 19:27
1
Adult's = \($9\) & children's = \($5\)
Take no of adult's ticket = \(x\) & children's = \(y\)
\(x + y = 44\) __________________ (1)
\(9x + 5y = 312\) ____________________(2)
By solving both the equation, we will get
\(x = 23\) & \(y = 21\)
Adult's tickets = 23
Take no of adult's ticket = \(x\) & children's = \(y\)
\(x + y = 44\) __________________ (1)
\(9x + 5y = 312\) ____________________(2)
By solving both the equation, we will get
\(x = 23\) & \(y = 21\)
Adult's tickets = 23
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Re: Adult tickets to a movie cost $9 each and children’s tickets cost $5 e
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11 Nov 2021, 12:15
11 Nov 2021, 12:15
1
Carcass wrote:
Adult tickets to a movie cost $9 each and children’s tickets cost $5 each. When 44 tickets were bought, the total cost was $312. How many of the tickets purchased were adult tickets?
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Show: :: OA
23
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Here's an algebraic solution..
Let A = the number of adult tickets sold
Let C = the number of children tickets sold
A total of 44 tickets were bought.
So we can write: A + C = 44
The total cost was $312.
So we can write: 9A + 5C = 312
We now have the following system of equations:
9A + 5C = 312
A + C = 44
Take the second equation and multiply both sides by 5 to get the following equivalent equation:
9A + 5C = 312
5A + 5C = 220
At this point, we can subtract the bottom equation from the top equation to get: 4A = 92
Solve: A = 92/4 = 23
Answer: 23
