Question

In a three hour examination of 350 questions there are 50 mathematical problems. If twice as much time should be allowed for each problem as for each of the other questions, how any minutes should be spent on the mathematical problems?

Answer 1

Start by solving in terms of two variables. Let x = time for math problems and y = time for nonmath problems. Solve for time in minutes.

350 problems total
50 math problems
300 nonmath problems
3 hours total = 180 minutes total

50x + 300y = 180

We know that math problems will take twice as long as nonmath problems so we let x = 2y and substitute back into the equation.

50(2y) + 300y = 180
100y + 300y = 180
400y = 180
y = .45

So,

y = time for nonmath problems = .45 min each
x = time for math problems = 2y = 2(.45) = .9 min each
total time for math problems = 50x = 50(.9) = 45 min