+586 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?
What do you mean by " and I do more than 10 problems per hour"? Your stated rate is p problems per hour! Unless you mean your rate is "p + 10". You have to clarify it. There are many different solutions, but not a unique one as the problem is stated.
+118609 I work at p problems per hour.
Say there are x problems altogether. It takes me x/p hours to finish and this equals t
It will take my friend x/(2p-4) hours to finish and this is t-2 hours.
p>10
so
\(\frac{x}{p}=t \qquad and \qquad \frac{x}{2p-4}=t-2\\ so\\ \frac{x}{2p-4}=\frac{x}{p}-2\\ 2=\frac{x}{p}-\frac{x}{2p-4}\\ 2=x(\frac{1}{p}-\frac{1}{2p-4})\\ 2=x(\frac{2p-4}{{p(2p-4)}}-\frac{p}{p(2p-4)})\\ 2=x(\frac{2p-4-p}{{p(2p-4)}})\\ 2=x(\frac{p-4}{{p(2p-4)}})\\ 2*(\frac{{p(2p-4)}}{p-4})=x\\ x=\frac{4p(p-2)}{p-4} \)
I decided to use a graphing program to help me.
so I had to swap the p for an y.
p>10 so y>10
Here is the full graph
https://www.desmos.com/calculator/e7xt74l0mi
Here is the extract
It looks like there is more than one answer.
If p = 12, x=60 and t = 60/12=5 So that is a good possiblilty
If p=20 then x=90 but t = 4.5 which is not an integer so that is not a possiblilty.
So the smallest answer that works appars to be p=12, x=60 and t =5
So there are 60 questions.
I answer 12 an hour so it takes me 5 hours.
My friend answers 20 an hour so it takes hime 3 hours.
So it takes him 2 hours less than me which is how it should be.
So the smallest and most sensible answer is that there are 60 questions.
