SOLUTION: The refreshment stand at the little league field sells beverages for 1.25 each and sandwiches for 3.00 each. If 134.50 was earned on the sale of beverages and sandwiches at one gam

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Question 158222: The refreshment stand at the little league field sells beverages for 1.25 each and sandwiches for 3.00 each. If 134.50 was earned on the sale of beverages and sandwiches at one game, how many beverages were sold?
Found 2 solutions by helpwithhomework, gonzo:
Answer by helpwithhomework(4) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = number of beverages sold and y = number of sandwiches.
So
1.25 x + 3 y = 134.5
Multiply both sides by 4
5 x + 12 y = 538
You might think that this equation has infinite number of solutions (because it is one equation with two unknowns) but in fact, that is not the case since x and y have to be integers. This type of equation is called a Diophantine equation.
5 x = 538 - 12 y
RHS is even so LHS is even. 5 x is even means x is even. Let x = 2 w.
10 w = 538 - 12 y
which means 538 - 12 y has to be divisible by 10. Also, both x and y are >= 0 so we only have to consider all multiples of 12 that make (538 - 12 y) divisible by 10 and are <= 538. Those are multiples of 12 that end in an 8, which are of the form 48 + 60 v where v = 1, 2, ..., 8. So substitute 12 y = 48 + 60 v or y = 4 + 5 v. Now
10 w = 538 - (48 + 60 v) = 538 - 48 - 60 v = 490 - 60 v, dividing by 10
w = 49 - 6 v as v goes from 1 to 8
Substitute each value,
v = 1 => w = 43 => x = 2 w = 86 and y = 4 + 5 v = 9
Check: 86 * 1.25 + 9 * 3 = 134.5
Similarly
v = 2 => w = 37 => x = 74 and y = 14
Check: 74 * 1.25 + 14 * 3 = 134.5.
So the problem has 8 answers, all corresponding to v = 1, 2, 3, ..., 8.

Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
you don't have enough information to solve this for one unique solution.
you have multiple possible solutions based on the fact that a common multiple of $1.25 and $3.00 is $15.00
at $15.00, you can sell 12 beverages at $1.25 or you can sell 5 sandwiches at $3.00
some examples of possible solutions are:
solution 1:
44 sandwiches and 2 beverages = 44*$3.00 + 2*1.25 = $132.00 + $2.50 = $134.5
to get other solutions, all you have to do is subtract 5 sandwiches and add 12 beverages each time.
solution 2:
39 sandwiches and 14 beverages = $117 + $17.5 = $134.5
solution 3:
34 sandwiches and 26 beverages = $102 + $32.5 = $134.5
etc................................
normally this type of problem says:
the number of beverages and sandwiches sold in total is 60 (for example)
each beverage costs $1.25 and each sandwich costs $3.00
now you have enough information.
you would then set up your formula to say
s + b = 60 (first equation)
s*3 + b*1.25 = 134.5 (second equation)
to solve you would find s in terms of b, or b in terms of s.
solving in the first equation for s, we get s = 60-b
substituting 60-b for s in the second equation, we get
3*(60-b) + 1.25*b = 134.5
solving for b, this equation becomes
180 - 3*b + 1.25 * b = 134.5 which becomes
180 - 134.5 = 3*b - 1.25*b which becomes
45.5 = 1.75*b which becomes
b = 45.5/1.75 = 26
if this looks familiar it's because i took the sum of sandwiches sold and beverages sold from solution 3 above.
if s + b = 60, then s = 60 - 26 = 34, so
s = 34, and
b = 26
solving in the second equation, we get
34*3 + 26*1.25 = $102 + $32.5 = $134.5
this confirms the answer as correct.