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Mathematics 16 Online
OpenStudy (dls):

At a museum, ticket prices are $13 for adults and $11 for children. If a tour group paid $179 for their tickets, how many people were in the tour group?

OpenStudy (anonymous):

7 adults 8 kids

OpenStudy (dls):

arent there many combinations possible?

OpenStudy (anonymous):

(7x13) + (8x11) = 179

OpenStudy (dls):

that is the ONLY combination?

terenzreignz (terenzreignz):

Well, you can certainly try many possible numbers of adults, from 1 to 14, and see which work.

OpenStudy (dls):

so no formula?hit and trial?

terenzreignz (terenzreignz):

Diophantine problem. Makes me dizzy. I opened this page. http://www.wikihow.com/Solve-a-Linear-Diophantine-Equation

hartnn (hartnn):

13a+11c=179 yup Diophantine equation....

OpenStudy (dls):

Diophantine equation....:O

hartnn (hartnn):

http://openstudy.com/users/ParthKohli#/updates/50e43ad3e4b0e36e35145cc6 i have listed 2 different methods here.

hartnn (hartnn):

one involves modular arithmetic and other involves finding the GCD and its reverse process.

OpenStudy (zpupster):

Let A denote the # of adult tickets Let C denote the # of child tickets Therefore... $13 x A + $11 x C = $179 13A + 11C = 179 So from this point forward, I would ask the following. What # must you subtract from $179 in order for it be divisible by 11. That number must also be a factor of 13. Essentially what I'm saying is 11C = 179-13A The answer to that would be the following: 13 x 7 = 91 179- 91 = 88 88/11 - 8 Therefore, 7 adults and 8 kids. PS - You could do it the other way around where you could try to find the number divisible by 179 which is also a factor of 11 but I thought it would be easier to do it the other way around since we're more familiar with our 11 times tables. from here http://www.jiskha.com/display.cgi?id=1362991103

Parth (parthkohli):

No.

OpenStudy (dls):

?

Parth (parthkohli):

Just kidding. I meant yes.

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