All these work and rate problems revolve around Volume(Quantity) proportional to rate * time. This means in two different situations, these 3 variables will be in direct or inverse proportion. So you don't have to think of them as 3 different variables and there would be no need to build 3 equations and solve.

Here, the questions asks for time in one situation. Since you know the time in some situation(first situation) put it down first and then multiply it with ratios of the remaining variables in 2 situations. See below

4 * 3000/5000 * 6/3 = 24/5

Explanation for the 3 parts.

1) 4 - time in the situation that you know

2) 3000/5000 - (Ratio of papers in 2 situations) In the second situation, there less number of papers are printed, therefore lesser time would be required. Hence the lesser number 3000 would go to the numerator.

3) 6/3 - (Ratio of presses in 2 situaions) In the second situation, there are less number of presses, so more time would be required. Hence the higer number would go to the numerator.

Supposing there was another variable in the problem, say number of hours that the presses work. Say 6 in the fist case and 8 in the second case, we would have had a fourt part

4) 6/8

get the drift...?

vivek_dj wrote:

This problem has three variables...how do we solve such problems.

Thanks your responses would be highly appreciated.

Vivek