Word Problem: Cakes and Tea at a Café

[geekily]

Word Problem: Cakes and Tea at a Café

This word problem is driving me nuts because I know I know how to get the answer, but I just can't think of it right now: A café sold tea at 30 cents a cup and cakes at 50 cents each. Everyone in a group had the same number of cakes and the same number of cups of tea. (Note: This is not to say that the number of cakes is the same as the number of teas.) The bill came to $13.30. How many cups of tea did each have?

I set this up as .30x + .50y = 13.30, but I don't know where to go from there because I don't see a second equation that I can set up. We've done similar problems, but there are always 2 equations to be set up, so that then I can find out one equation or do the quadratic formula or something. I can move .50y over to the other side, so I get .30x = 13.30 - .50y, but 13.30 divided by .3 is 44.3 repeating. I don't know where to go from here - could anyone give me a hint? Thanks so much!

 

[skeeter]

one person in the group has x cups of tea and y cakes

total money spent by 1 person is .30x + .50y

if there are "n" people in the group, then the total spent is

n(.30x + .50y) = 13.30

multiply both sides by 10 to clear the decimals ...

n(3x + 5y) = 133

divide both sides by n ...

3x + 5y = 133/n

since "n" is a whole number > 0, it has to divide 133 evenly

133 will only factor as 7 times 19, so n has to be either 7 or 19 ... there is no other possible choice.

if n = 7, then ...

3x + 5y = 19 ... since x and y have to be whole numbers, the only integral solution is x = 3 and y = 2.

if n = 19, then ...

3x + 5y = 7 ... this equation has no integral solutions for x and y.

 

[geekily]

Thank you so much, Skeeter, I really appreciate it! Just one question: Say the numbers weren't as simple as 3x and 5y, where you can just look and say, well, 3 time 3 is nine and 5 times 2 is ten. How would you then figure out what x and y equal? Or, if instead of 19, we were trying to get to a number that can be equaled by more than one combination of 5 and 3, or if the numbers we can factor by (19 and 7 in this case) both gave us solutions of whole numbers > 0? Ok, I guess that's more than one question. I just want to make sure I know how to do the problem in different situations. Thanks so much!

 

[skeeter]

I really can't speak about all possible situations such as thsi, but I can say it is obvious this problem was "engineered" to give a single unique solution ... oft times, there may be more than one set of solutions for this type of problem, and if the problem gives further restrictions on the solution, then that number of solutions may be reduced to a possible single solution ... then again, maybe not.