SOLUTION: A grade 11 class, on a field trip to Montreal, had lunch in a restaurant. The bill came to $239.25. Four students had birthdays that day, and it was agreed that these four should n

Algebra ->  Finance -> SOLUTION: A grade 11 class, on a field trip to Montreal, had lunch in a restaurant. The bill came to $239.25. Four students had birthdays that day, and it was agreed that these four should n      Log On


   

Question 255012: A grade 11 class, on a field trip to Montreal, had lunch in a restaurant. The bill came to $239.25. Four students had birthdays that day, and it was agreed that these four should not have to pay for lunch. The other students had to pay $1 more than if all students had paid. How many students had lunch?
I know that the answer is 33 children ate on the trip and that 29 paid. The 29 that paid, had to pay 8.25 each. If all the children paid then each would have paid 7.25. I found this answer through trial and error, but can not find what the equation should be to get this answer. If you could help I would be very greatful. Thanks so much.
Found 2 solutions by Theo, ankor@dixie-net.com:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
bill = $239.25

let x = number of students.

let y = amount each student paid.

if all students paid the bill, then:

239.25+%2F+x+=+y (first equation)

if all but 4 paid the bill, then:

239.25+%2F+%28x-4%29+=+y%2B1 (second equation)

substitute for y in second equation by the value for y in first equation to get:

239.25+%2F+%28x-4%29+=+%28239.25%2Fx%29+%2B+1

you had one equation in 2 unknowns. the substitution allows you to have 1 equation in 1 unknown.

multiply both sides of this equation by (x-4) to get:

239.25+=+%28x-4%29+%2A+%28%28239.25%2Fx%29+%2B+1%29

multiply the factors to get:

239.25+=+239.25+%2B+x+-+4%2A%28239.25%2Fx%29+-+4

subtract 239.25 from both sides of the equation to get:

0+=+x+-+4%2A%28239.25%2Fx%29+-+4

remove parentheses to get:

0+=+x+-+%28957%29%2Fx+-+4

multiply both sides of equation by x to get:

0+=+x%5E2+-+957+-+4%2Ax

rearrange terms to get:

x%5E2+-+4%2Ax+-+957+=+0

use quadratic formula to solve for x to get:

x = %28-%28-4%29+%2B-+sqrt%2816%5E2+-+%284%2A1%2A%28-957%29%29%29%29%2F%282%29

this becomes:

x = %284+%2B-+sqrt%2816%5E2+%2B+3828%29%29%2F%282%29 which becomes:

x = %284+%2B-+sqrt%283844%29%29%2F2 which becomes:

x = %284+%2B-+62%29%2F2 which becomes:

x+=+%284%2B62%29%2F2+=+66%2F2+=+33+

or:

x+=+%284-62%29%2F2+=+-58%2F2+=+-29

since x has to be positive, then the solution narrows down to:

x+=+33.

when 33 students pay the bill, the bill is $239.25 / 33 = $7.25 apiece.

when 29 students pay the bill (33-4 = 29), the bill is $239.25 / 29 = $8.25.

















Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The bill came to $239.25. Four students had birthdays that day, and it was agreed that these four should not have to pay for lunch.
The other students had to pay $1 more than if all students had paid. How many students had lunch?
:
Let x = no. of students that had lunch
then
(x-4) = no. that actually paid
:
239.25%2Fx = cost per student if all paid
239.25%2F%28%28x-4%29%29%29 = cost if 4 did not pay
:
The equation
:
239.25%2F%28%28x-4%29%29 - 239.25%2Fx= $1
:
Multiply by x(x-4), results:
239.25x - 239.25(x-4) = x(x-4)
:
239.25x - 239.25x + 957 = x^2 - 4x
:
a quadratic equation from this
x^2 - 4x - 957 = 0
:
Factors to
(x-33)(x+29) = 0
:
positive solution
x = 33 students had lunch
and
29 students paid; 239.25/29 = $8.25 as you found