Tickets sold - 2 different price ranges

Question

A school made $887 from selling 159 tickets. Ticket price for students is $5. Ticket price for non-students is $8. How many of each type of tickets were sold?

Answer 1

Hey there!

 

I saw this and tried solving the system with elimination but instead of rounding up completely I did this:

 

X + Y = 159 --> -5X - 5Y = -795

5X + 8Y = 887 -->5X + 8y = 887

_______________________________

 

3Y = 92 --> Y = 30.7 ∴ X = 128.3

 

and when we verify this result we get:

 

(128.3*5) + (30.7*8) = 887.1 ≈ 887

 

I know its a little odd but its much closer to 887 then 888. Just a thought :)

 

Answer 2

Let s = number of student tickets

Let n= number of non student tickets

 

1). s+n= 159   total tickets sold

2). 5s+8n = 887  total money made

 

Solve by elimination or substitution.

For elimination multiply equation 1 by

-5 and add the equations

 

-5s - 5n = -795.  (equation 1)(-5)

 5s + 8n = 887

-------------------

0s + 3n = 92

3n = 92

n = 92/3

n = 30 2/3

 

As can be seen the given numbers do not come

out evenly.  The total money made should

probably have been listed as $888 for the

problem to come out correctly.  However,

we can finish with what we have.

 

s + n = 159

s + 30 2/3 = 159

s = 159 - 30 2/3

s = 158 3/3 - 30 2/3

s = 128 1/3

 

With the information given there were 

128 1/3 student tickets sold and

30 2/3 nonstudent tickets.

Answer 3

This does not come out to an even number:

 

X + Y = 159 (# of tickets sold)

5X + 8Y = 887 ($ taken in)

 

Y = 159 - X

 

5X + 8(159 - X) = 887

5X + 1272 - 8X = 887

-3X  = -385

X = 128.333 (rounded to 128)

 

128 + Y = 159

Y = 31

 

128 * 5 = $640

+ 31 * 8 = $248

 

 

Again this does not equal $887 (You get $888).