SOLUTION: Hello, I need help with the following problem: A mathematician works for {{{ t }}} hours per day and solves {{{ p }}} problems per hour, where {{{ t }}} and {{{ p }}} are pos

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Question 1047022: Hello,
I need help with the following problem:
A mathematician works for +t+ hours per day and solves +p+ problems per hour, where +t+ and +p+ are positive integers and +1%3Cp%3C20+. One day, the mathematician drinks some coffee and discovers that he can now solve +3p%2B7+ problems per hour. In fact, he only works for +t-4+ hours that day, but he still solves twice as many problems as he would in a normal day. How many problems does he solve the day he drinks coffee?
Thank you,
Nicole
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
(3p+7)(t-4) = 2tp
===> t+=+4%2A%28%283p%2B7%29%2F%28p%2B7%29%29+=+4%2A%283+-+14%2F%28p%2B7%29%29.
Since both t and p are positive integers, 14%2F%28p%2B7%29 can only take on the values 0, 1, or 2. (Why?)
14%2F%28p%2B7%29+=+0 does not yield any value of p.
14%2F%28p%2B7%29+=+1 ===> p = 7.
14%2F%28p%2B7%29+=+2 ===> p = 0, which is not acceptable.
Therefore, p = 7,
===> t+=+4%2A%28%283%2A7%2B7%29%2F%287%2B7%29%29+=+8,
and the mathematician solved (3*7 + 7)(8-4) = 28*4 = 112 problems the day he drank coffee.