+2862 Chloe charged for admission to her play on three different nights. Each night, a different number of people were in attendance, but remarkably, Chloe collected $541 each night. If the admission charges for each child and each adult were $9 and $17, respectively, how many people in total came to the three showings?
I solved this tediously by guessing and checking half a year ago in the Mathcounts Competition. I finished just the problem in time with no time to spare for checking my other answers!
Please help, I need a faster strategy to solve this.
You posted the same question about 4 weeks ago. See here for the answers:
https://web2.0calc.com/questions/money-problem_5
+2862 Sorry I posted again I was confused by the explanation... I dont understand why it works.
It is locked now I couldnt reply to the answer today.
+128448 Here's the solution ....
Note that we should have this equation
9x + 17y = 541 which we can manipulate as follows
9x = 541 - 17y
x = 541 - 17y
________
9
x = 541 - (18 - 1)y
_____________
9
x = 541 + y 18y
_______ - ____
9 9
x = 540 + 1 + y 18y
___________ - ____
9 9
x = 540 1 + y 18y
___ + _____ - ___
9 9 9
Note that the first and last terms will always be divisible by 9
And.....the middle term will be divisible by 9 when y = 8, 17 or 26
So we have the following possibilities for x
x = 541 - 17(8) = 45
_________
9
x = 541 - 17*17 = 28
__________
9
x = 541 - 17*26 = 11
____________
9
Note if y = 35,we get the next multiple of 9...but x = -6 if this is so.... so....these are the only possible solutions
So....the solutions (x, y) = ( 11,26) (28, 17) (45, 8)
Just add these coordinates and you will get your answer, CU
+2862 thank you so much! that is a very beatiful solution I will make sure in the future to use the same strategy!
