Skills Needed for Mathematical Problem Solving (Part 4)

Q: What is the difference between concepts and facts in problem solving?

Facts are discrete pieces of information (e.g., formulas, definitions, theorems), while concepts are broader understandings that connect multiple facts and allow transfer to new situations. Problem solving requires both, but deep conceptual understanding is what enables students to tackle unfamiliar problems.

This section examines how concepts and facts are applied and taught in mathematical problem solving. Using a series of classroom case studies, we show how the concept of rate of work was reinforced through progressive problem assignments, and how a population growth problem can serve as a natural entry point to introduce exponential functions.

Table of Contents

3.1 Concepts and Facts in Problem Solving

Learners need a deep understanding of algebraic, geometric, and statistical concepts and facts to solve challenging mathematical problems. Here I will explain how I used a set of problems assigned as homework with my Level 2 advanced mathematics classes to reinforce the concept of rate of work that was necessary to solve the problem.

Distinguishing Concepts from Facts

Facts are discrete pieces of information: a formula, a definition, or a theorem. For example, the quadratic formula is a fact.
Concepts are broader, interconnected understandings that allow the transfer of knowledge to new situations. For example, "rate" as a general idea applies to work, speed, flow, and growth.
Problem solving requires both, but deep conceptual understanding is what enables students to tackle problems they have never seen before.

Case Study 1: Teaching Rate of Work Through Progressive Problems

Step 1: The concept of rate of work is first presented, then discussed. The whole class then discussed the solution to the following problem.

Problem 1: It takes Carla 1 hour longer to mow the lawn than it takes Sharon to mow the lawn. If they can mow the lawn in 5 hours working together, then how long would it take each girl by herself?

Step 2: Students were then asked to solve the following problem as homework, where the same concept as Problem 1 is needed.

Problem 2: John takes 3 hours longer than Andrew to peel 500 pounds (lb) of apples. If together they can peel 500 lb of apples in 8 hours, then how long would it take each one working alone?

Although the above problem was discussed in class before they attempted to solve it, students had difficulties in solving it. I decided that the concept of rate of work had to be discussed again. A few days later, a few students managed to solve the given problem, and the solution to Problem 2 was discussed with the whole class.

Step 3: In order to assess students' understanding of the concept of rate of work and the process of problem solving, I assigned the following problem as homework.

Problem 3: It takes pump A 2 hours less time than pump B to empty a certain swimming pool. Pump A is started at 8:00 A.M., and pump B is started at 11:00 A.M. If the pool is still half full at 5:00 P.M., then how long would it take pump A working alone?

Problem 3 also needs a deep understanding of the concept of rate of work. Except for a few students, most found the problem very challenging and could not solve it. I carefully examined the solutions generated by students and understood that the concept of rate of work was their main difficulty. I decided to give a full one-hour lesson on the rate of work with many examples, and I made sure that not only did they understand the concept but also how to use it to formulate problems. I requested that they look again at the problem. A few days later, more than half the class solved Problem 3 correctly.

Step 4: I then assigned the following problem in a quiz.

Problem 4: It takes pump B 2 hours more time than pump A to fill a swimming pool. Both pumps are started at 7 a.m. At 10 a.m. pump A breaks down. It took 1 hour to repair it and then it was restarted again. At 3 p.m. 80% of the swimming pool was filled with water. How long would it take each pump working alone to fill the swimming pool?

About half the class solved the problem correctly, and a quarter of the class had solutions with minor mistakes. I carefully examined the solutions generated by students, and it was clear that the students had a better understanding of the concept of rate of work and, more importantly, they knew how to apply it to solve problems.

Students fail to solve problems involving concepts that are not thoroughly understood. Moreover, mathematical problems may be used as teaching methodologies not only to introduce concepts but also to help students gain a deeper understanding of these concepts. In fact, some concepts cannot be thoroughly understood unless they are used in problem solving or any other activity where critical thinking and reasoning are involved.

Worked Solutions to Rate-of-Work Problems

To illustrate the underlying concept, here are complete solutions to Problems 1 and 3.

Solution to Problem 1 (Carla and Sharon)

Show Solution

Let \( t \) be the number of hours Sharon takes to mow the lawn alone. Then Carla takes \( t + 1 \) hours.

Their work rates are:

Sharon: \( \dfrac{1}{t} \) of the lawn per hour
Carla: \( \dfrac{1}{t+1} \) of the lawn per hour

Working together, they take 5 hours, so their combined rate is \( \dfrac{1}{5} \) of the lawn per hour:

\[ \frac{1}{t} + \frac{1}{t+1} = \frac{1}{5} \]

Multiply through by \( 5t(t+1) \):

\[ 5(t+1) + 5t = t(t+1) \] \[ 5t + 5 + 5t = t^2 + t \] \[ t^2 - 9t - 5 = 0 \]

Using the quadratic formula:

\[ t = \frac{9 \pm \sqrt{81 + 20}}{2} = \frac{9 \pm \sqrt{101}}{2} \]

Taking the positive root:

\[ t \approx \frac{9 + 10.05}{2} \approx 9.53 \text{ hours} \]

Conclusion: Sharon takes about 9.53 hours alone, and Carla takes about 10.53 hours alone.

Solution to Problem 3 (Two Pumps)

Show Solution

Let \( t \) be the number of hours pump A takes to empty the pool alone. Then pump B takes \( t + 2 \) hours.

Their rates are \( \frac{1}{t} \) and \( \frac{1}{t+2} \) of the pool per hour.

Time analysis: Pump A runs from 8:00 A.M. to 5:00 P.M. = 9 hours. Pump B runs from 11:00 A.M. to 5:00 P.M. = 6 hours.

The total work done is \( \frac{1}{2} \) of the pool (the pool is half full at 5 P.M.):

\[ \frac{9}{t} + \frac{6}{t+2} = \frac{1}{2} \]

Multiply through by \( 2t(t+2) \):

\[ 18(t+2) + 12t = t(t+2) \] \[ 18t + 36 + 12t = t^2 + 2t \] \[ t^2 - 28t - 36 = 0 \]

Using the quadratic formula:

\[ t = \frac{28 \pm \sqrt{784 + 144}}{2} = \frac{28 \pm \sqrt{928}}{2} = \frac{28 \pm 4\sqrt{58}}{2} = 14 \pm 2\sqrt{58} \]

Taking the positive root:

\[ t \approx 14 + 15.23 \approx 29.23 \text{ hours} \]

Conclusion: Pump A takes about 29.23 hours alone, and pump B takes about 31.23 hours alone.

Case Study 2: Introducing Exponential Functions Through a Population Problem

Mathematical problem solving may also be used to introduce a new concept. An example of a mathematical problem that may be used to introduce a new concept is now presented.

Problem 5: The present population of the UAE is 4.5 million. If we assume that the population grows at an annual rate \( r = 3\% \) for the next 15 years, what will be the population \( P \) of the UAE in \( t \) years? (Assume \( t \) is smaller than 15.)

Students can easily be guided to use percentages and come up with the following result:

\[ P(t) = 4.5(1 + 0.03)^t \quad \text{(in millions)} \]

Or equivalently, \( P(t) = 4.5(1.03)^t \) million.

At this point, the concept of exponential functions can easily be introduced using the result obtained. Students will have understood that exponential functions may be used to solve population problems and make a connection between the concept and its possible application. Experience shows that students are more motivated when solving problems related to their daily life.

Sample Values of the Population Model

\( t \) (years)	\( P(t) \) (millions)
0	4.50
5	\( 4.5(1.03)^5 \approx 5.22 \)
10	\( 4.5(1.03)^{10} \approx 6.05 \)
15	\( 4.5(1.03)^{15} \approx 7.01 \)

Note: After 15 years, the population is projected to be approximately 7.01 million — a growth of about 56%.

Pedagogical Implications

Key Findings from This Study

Concepts must be taught thoroughly. Students fail to solve problems when the underlying concept is not well understood, even if they have memorized the relevant formula.
Progressive difficulty helps. Assigning a sequence of problems with gradually increasing complexity — and revisiting the same concept multiple times — significantly improves mastery.
Problem solving can introduce new concepts. A well-chosen problem (such as the population model) can naturally lead students to discover a new concept (exponential functions).
Real-world context improves motivation. Students are more engaged when the problem connects to their daily life or familiar context.
Assessment reveals conceptual gaps. Careful analysis of student solutions helps the teacher identify exactly where understanding breaks down.

As illustrated by the rate-of-work case study, meaningful mastery of a concept is not achieved by a single exposure but through repeated, varied application. The teacher's role is to provide multiple opportunities for students to encounter the concept in different contexts, deepening their understanding each time.

Frequently Asked Questions

What is the rate of work concept in mathematics?

The rate of work concept describes how quickly a task can be completed. If a worker completes a job in \( t \) hours, their work rate is \( 1/t \) of the job per hour. When multiple workers cooperate, their rates add: \( \frac{1}{t_1} + \frac{1}{t_2} = \frac{1}{T} \), where \( T \) is the combined time.

How can mathematical problem solving be used to teach new concepts?

Mathematical problems can introduce new concepts by guiding students to discover patterns. For example, a population growth problem using percentages naturally leads students to discover the exponential function \( P(t) = P_0(1 + r)^t \). This discovery-based approach improves motivation and understanding.

Why do students struggle with rate of work problems?

Students often struggle because the concept of rate of work is not thoroughly understood. They may know the formula \( \frac{1}{t_1} + \frac{1}{t_2} = \frac{1}{T} \) but cannot apply it flexibly to non-standard scenarios, such as staggered start times or partial work followed by interruption.

What is the difference between concepts and facts in problem solving?

Facts are discrete pieces of information (e.g., formulas, definitions, theorems), while concepts are broader understandings that connect multiple facts and allow transfer to new situations. Problem solving requires both, but deep conceptual understanding is what enables students to tackle unfamiliar problems.

How does exponential growth appear in real-world problems?

Exponential growth appears in population growth, compound interest, radioactive decay, and the spread of diseases. The general model is \( P(t) = P_0(1 + r)^t \) for discrete growth or \( P(t) = P_0 e^{rt} \) for continuous growth, where \( P_0 \) is the initial amount, \( r \) is the growth rate, and \( t \) is time.