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My friend and I both have the same math homework one day. I work at a rate of $p$ problems per hour and it takes me $t$ hours to finish my homework. My friend works at a rate of $2p-4$ problems per hour and it only takes him $t-2$ hours to finish his homework. Given that $p$ and $t$ are positive whole numbers and I do more than $10$ problems per hour, how many problems did I do?

Respuesta :

If you do [tex] p [/tex] problems per hour, after [tex] t [/tex] hours you do [tex] pt [/tex] problems.

Similarly, your friend does [tex] 2p-4 [/tex] problems per hour, so after [tex] t-2 [/tex] he has completed [tex](2p-4)(t-2) = 8 - 4p - 4t + 2pt [/tex] problems.

Since you have the same number of problems, we deduce

[tex] pt = 8 - 4p - 4t + 2pt [/tex]

which we can rewrite as

[tex] 8 - 4p - 4t + pt = 0 [/tex]

If we solve this equation for one of the two variables, for example t, we have

[tex] t = \cfrac{4(p-2)}{p-4} [/tex]

Since p and t are positive integers, p must be at least 4.

Finding integers solutions require a bit of trial and error, and you can figure out that the only positive and integer solution where p > 10 is

[tex] p=12,\ t=5 [/tex]

why are you booing him? he's right

12 * 5 = 60

ok?

60

smh

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