Adult tickets cost $4 and children tickets cost $1. 285 tickets are sold. And $765 is collected. How many adult tickets were sold.?Use substitution or elimination to solve the problem

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bullgatortail's profile pic

bullgatortail | High School Teacher | (Level 1) Distinguished Educator

Posted on

If adult tickets cost $4, and children's tickets cost $1, and 285 tickets total were sold, and the total amount collected is $765, there can only be one possible answer. I believe you will find that 160 adult tickets were sold (total of $640) and 125 children's tickets were sold (total of $125). Combining these two amounts will give you 285 tickets and a total of $765. Any difference in tickets sold by either adults or children would result in a higher or lower total amount.

pohnpei397's profile pic

pohnpei397 | College Teacher | (Level 3) Distinguished Educator

Posted on

The answer to this is that 160 adult tickets were sold.  This means that 125 children's tickets were sold.  Here is how to solve this problem:

Let's call adult tickets A and children's tickets C.

We know that A + C = 285 because that's how many total tickets were sold.

We know that 4A + C = 765.  That's because each adult ticket sold cost $4 so 4 times the number of adult tickets, plus the number of kids tickets ($1 each) make the total amount collected.

Let's use the first equation to find for C.  C = 285 - A.

Now just substitute that into the other equation and you have

4A + 285 - A = 765.

3A = 480

A = 160

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

In the question the need is to determine the number of adult and child tickets  tickets sold out. So one of the unknown  is x.

We presume  x number of tickets for sdults are sold. The child tickets is automatically must be 285-x.

The collection of revenue from x adult tickets =  number of tickets* rate of tickets = 4x

The collection of revenue from 285-x child tickets = number*rate = (285*x)*1 =285-x.

The total collection  = 4x+285-x algebraically.......(1)

The actual collection = $765..............................(2)

Therefore the required equation of the problem:

Algebraic collection as at (1)  = actual collection as at (2). So,

$(4x+285-x) = %765. Or

4x-x +285 = 765. Or

3x- 765 - 285 = 480. Or

3x = 480. Or

3x/3 = 480/3. So

x = 160  is the number of adult tickets sold.

285 - x = 285 - 160 = 125 is the number of child tickets sold.

Check: 160+125 = 285 and revenue $(160*4+125) =$ (640+125) = $765

 

 

 

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