# Word Problems

## Introduction

Word problems, and problem solving exemplify why non mathematicians care about mathematics. The general process for solving word problems is: translate the word problem into some number of equations, solve the equations, and retranslate so we have the solution to the original question. Some of the common word problems are mixture, uniform motion (rate ${\displaystyle \times }$ time = distance), and work allocation problems.

## Mixture Problems

Mixture problems usually involving having two different types of product, for example two types of coffee beans, and selling a mixture of the two products at a price between the price of each product individually.

Example: Ellen's coffee shop has a very popular grade A coffee, and a poorly selling grade C coffee. In order to sell the rest of their grade C coffee, Ellen will mix some grade A coffee with some grade C coffee and sell it as an economic grade B coffee. Since the grade A coffee usually sells for $10 per pound, and the grade C coffee sells for$5 per pound, Ellen will sell the grade B coffee for \$7 per pound. In order to test how well the grade B coffee sells, Ellen will make 100 pounds of the grade B coffee at no difference in revenue, compared to selling the pounds of grade A and grade C coffee individually. How many pounds of grade A coffee is required to make this 100 pound mixture?

Solution: Since we do not know how many pounds of grade A coffee Ellen needs, well let the variable A represent this value. Since the mixture will weigh 100 pounds, we can conclude Ellen will need 100 - A pounds of grade C coffee. The 100 pounds of grade B coffee will earn Ellen 100 * 7 = 700 dollars. The amount of money earned from selling the grade A coffee alone is 10 * A dollars, while the amount of money earned from selling the grade C coffee alone is (100 - A) * 5 dollars. Since there is no difference in revenue we can conclude that 10A + (100 - A)5 = 700.

${\displaystyle {\begin{array}{rcl}10A+(100-A)5&=&700\\10A+500-5A&=&700\\5A+500&=&700\\5A&=&200\\A&=&40\end{array}}}$

Thus, Ellen needs 40 pounds of grade A coffee, and 100 - 40 = 60 pound of grade C coffee.

## Uniform Motion

Uniform motion problems involve an object moving at a fixed speed. It is important to note that this is unrealistic since very few objects move at a fixed rate. However, it is a basic version that helps us figure out what to do in more complicated problems.

Example: A boat travels 24 up a river against the current, and back again. The trip takes 6 hours and the current of the river is a constant 3 mi/h. Assuming that the boat travels at a constant speed relative to the water, what is the speed of the boat in still water?

Solution: Since we are looking for the speed of the boat in still water, let r represent the speed of the boat. While moving upstream the boat was moving at a constant speed of r - 3 mi/h, and the boat traveled at a constant speed of r + 3 mi/h while traveling downstream. Since ${\displaystyle rate\cdot time=distance}$, we know the boat took ${\displaystyle {\frac {24}{r-3}}}$ hours to travel up stream, and ${\displaystyle {\frac {24}{r+3}}}$ hours to travel downstream. Adding these two times together we get the total travel time, 6 hours. So ${\displaystyle {\frac {24}{r-3}}+{\frac {24}{r+3}}=6}$. Now we have to solve for r.

${\displaystyle {\begin{array}{rcll}{\frac {24}{r-3}}+{\frac {24}{r+3}}&=&6&{\text{Given equation}}\\\\{\frac {24(r+3)+24(r-3)}{(r-3)(r+3)}}&=&6&{\text{Finding the GCD and adding the fractions together}}\\\\{\frac {48r}{r^{2}-9}}&=&6&{\text{Simplifying using FOIL.}}\\\\48r&=&6(r^{2}-9)&{\text{Multiply both sides by }}(r^{2}-9)\\8r&=&r^{2}-9&{\text{Divide both sides by 6}}\\r^{2}-8r-9&=&0&\\(r-9)(r+1)&=&0&{\text{Factor}}\\r=9&{\text{ or }}&r=-1&{\text{Zero Product Property( If the product of two numbers is zero, one of them must be zero.)}}\end{array}}}$

Now we know r = 9 or r = -1. However, we defined r to be the speed of the boat in still water, in miles per hour. Having a boat moving -1 mi/h in still water is absurd, so we remove that as a possible solution.

Thus, the speed of the boat in still water is 9 mi/h.

## Work Allocation Problems

Work allocation problems involve two people/machines that could complete a job on their own, but we they can save time if they work together. Just like the uniform motion problems, we assume that everyone is able to work at a constant rate.

Example: Patrice, by herself, can paint a room in 10 hours. If she asks April to help, they can complete the job in 6 hours. How long would it take April to do the job herself?

Solution: Since it will take Patrice 10 hours to complete the job alone, she can complete 1/10th of the job in an hour, and since we do not know what fraction of the job April can complete in one hour, we will let the amount of work she can complete in one hour as 1/x. Since they can take 6 hours when working together, we have ${\displaystyle {\frac {6}{10}}+{\frac {6}{x}}=1}$. We have 1 on the right hand side since they completed the whole job. Now we want to solve for x.

${\displaystyle {\begin{array}{rcll}{\frac {6}{10}}+{\frac {6}{x}}&=&1&{\text{Given equation}}\\\\{\frac {6x+60}{10x}}&=&1&{\text{Finding a GCD and adding the fractions together.}}\\6x+60&=&10x&{\text{Clearing the denominator}}\\60&=&4x&{\text{Simplify}}\\x&=&15&\end{array}}}$

Now we know x is the number of hours it will take Alice to complete the job alone. Thus, the answer is 15 hours.