A mathematician works for t hours per day and solves p problems per hour, where and are positive integers. One day, the mathematician drinks some coffee and discovers that he can now solve 4p+5 problems per hour. In fact, he only works for t-3 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?
Let's assume that the mathematician works for x hours a day and can solve y problems per hour. Also, the mathematician drinks some coffee and discovers that he can now solve z problems per hour. So, the mathematician works for n hours that day. We are given that:x*y = number of problems solved in a dayz * n = number of problems solved on the day he drank coffee
Then, we can write the equations:x*y = n * 2*z (he still solves twice as many problems as he would in a normal day)andx = n (he only works for n hours that day)Now, we need to simplify these equations to solve for the number of problems solved on the day he drank coffee. Here is how to do it:\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)Since x, y, n, and z are all positive integers, we can say that the expression 2*n*z/x is also a positive integer. Therefore, we can write:\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)where k is a positive integer.
Finally, the number of problems solved on the day he drank coffee is: y = 2k Therefore, the answer is that the mathematician solved 2k problems on the day he drank coffee.
Answer: 2000 problems
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Sorry, my first response was wrong because i read the question incorrectly. Here's the corrected version:
Let's break down the problem step by step:
1. On a normal day, the mathematician solves t×p problems.
2. On the day he drinks coffee, he solves (t−3)×(4p+5) problems.
3. According to the problem, he solves twice as many problems as he would on a normal day: (t−3)×(4p+5)=2×(t×p)
4. Now, we can solve the equations to find the value of t and p:
\((t−3)×(4p+5)=2×t×p\)
\(4tp−12p+5t−15=2tp\)
\( 2tp−5t−12p+15=0 \)
5. The solution to this equation will give us the values of t and p, which we can then use to calculate how many problems the mathematician solves on the day he drinks coffee.