Question

Maria is a hard-working college sophomore. One Sunday, she decides to work nonstop until she has answered 100 practice problems for her math course. She starts work at 8:00 AM and uses a table to keep track of her progress throughout the day. She notices that as she gets tired, it takes her longer to solve each problem.
Time Total Problems Answered
8:00 AM 0
9:00 AM 40
10:00 AM 70
11:00 AM 90
Noon 100

Use the table to answer the following questions.
The marginal, or additional, gain from Maria’s first hour of work, from 8:00 AM to 9:00 AM, is _______ problems.
The marginal gain from Maria’s third hour of work, from 10:00 AM to 11:00 AM, is _______ problems.
Later, the teaching assistant in Maria’s math course gives her some advice. "Based on past experience," the teaching assistant says, "working on 25 problems raises a student’s exam score by about the same amount as reading the textbook for 1 hour." For simplicity, assume students always cover the same number of pages during each hour they spend reading.
Given this information, in order to use her 4 hours of study time to get the best exam score possible, how many hours should she have spent working on problems, and how many should she have spent reading?

a) 0 hours working on problems, 4 hours reading
b) 1 hour working on problems, 3 hours reading
c) 2 hours working on problems, 2 hours reading
d) 3 hours working on problems, 1 hour reading

Answer 1

Answer:

The correct answers are:

i. The marginal, or additional, gain from Maria’s first hour of work, from 8:00 AM to 9:00 AM, is  40   problems.

ii. The marginal gain from Maria’s third hour of work, from 10:00 AM to 11:00 AM, is   20   problems

iii. 2 hours working on problems, 2 hours reading (c)

Explanation:

i. The marginal gain from 8:00 AM to 9:00 AM is calculated thus:

work done at 8:00 AM = 0 problems

Work done at 9:00 AM = 40 problems

Marginal gain = (Work done at 9:00 AM) - (work done at 8:00) = 40 - 0 = 40 problems

ii. work done at 10:00 AM = 70 problems

work done at 11:00 AM = 90 problems

Marginal gain = 90 - 70 = 20 problems

iii. To get the best exam score, the most number of problems possible in total must be answered within the 4-hour period, and this is determined as follows:

we are told that Maria gets tired as time progresses and the number of questions answered reduces we notice the following numbers of questions answered in each hour:

8:00 AM to 9:00 AM = 40 - 0 = 40 problems (hour 1)

9:00 AM to 10:00 AM = 70 - 40 = 30 problems (hour 2)

10:00 AM to 11:00 AM = 90 - 70 = 20 problems (hour 3)

11:00 AM to Noon = 100 - 90 = 10 problems (hour 4)

Also, we are told that;

1 hour of reading = 25 problems

Comparing the reading and solving problems, we see that the number of problems solved in hour-1 and hour-2  (40 and 30 respectively) are more than the equivalent number of problems solved in reading for 1 hour, while hour-3 and hour-4 (20 and 10 respectively), have lesser equivalent number of problems each, than reading for one hour. Therefore to make the most out of the 4 hours, Maria will spend the first 2 hours solving problems and the next two hours reading, giving a total of: 40 + 30 + 25 + 25 = 120 problems solved.