(In this note **a / b** represents the fraction whose numerator is “a” and whose denominator is “b” )

**A Mowing Problem**

Mary mows her rear lawn in three hours. With a comparable mower, her younger brother can complete the job in four hours. After the lawn is mowed, Mary and her brother plan to see the fireworks display in the park, a ten minutes walk from their house. Mother has granted them permission to go to the park after the lawn is mowed. If Mary and her brother work together, will they be able to see the start of the fireworks display? The firework display starts two hours after the mowing begins.

This discussion centers on the presentation and solution of this problem.

**What is needed?**

To solve this problem, one needs to know how to add fractions and how to find a particular equivalent fraction of a fraction.

**Counting and Addition**

Counting is one of the first operations a child learns. When she recites, for the first time, one, two, three, etc…, our faces light up with pride. Before long, she masters the counting procedure having learned to point to an object and assign a number from the sequence, 1, 2, 3, … ; stopping after she assigns to the last object a number. For small counting chores, this method is very effective. However, when counting a large number of objects or counting objects from two or more large piles, this method is time-consuming and often ineffective.

We accelerate the counting process with addition. Addition allows us to count without physically pairing the numbers 1, 2, 3, … etc. to each objects. We only need to know how many are in each pile in order to get the total.

**How is it done?**

Let’s take our lead from a child. A young boy describes to his friends the method he used to win his marbles. Boasting he says, ”I won 3 shooters and 12 marbles. Marbles are non-shooters. Shooters are special, so he counts them separately. What’s going on? The boy is counting the marbles that are alike in different piles. But sometimes we must group objects and count the contents of the group.

**Now consider this example.**

The proprietor of a small pest control company owns 12 light trucks for his technicians and 4 cars for his salesmen. Each truck has pest fighting equipment. When asked the size of his mobile inventory, he reclassifies the trucks and the cars in a common category with the answer, “I have 16 vehicles.” The word, vehicle, strips the trucks and the cars of their designated function and places them in a broader category as things that are alike — namely motor vehicles. Once they are reclassified into a common category, he counted them. That is, they can be added.

These two examples highlight two principles in fraction arithmetic.

**Principle 1**: We can only add things that are alike. And when things are alike, adding is simply counting.

**Principle 2**: When things are not alike and you wish to add them, first find representations of them as members of a common category. Then add (count) them as members of the common category.

**When Are Fractions Alike?**

Fractions are alike **when they have the same denominator**.

**Example**: 5 / 6 and 3 / 6 are alike.

**NOTATION:** We will use the plus sign to represent addition in this discussion.

**Example**: Explain 2 / 6 + 3 / 6.

2 / 6 is represented by a bar with two red cells.

and

3 / 6 is represented by a bar with three yellow cells.

The fractions are alike, so we count the number of parts in their representations.

There are (2+3) parts. Since each part is one-sixth, we have 5 one-sixths or 5 / 6.

This example displays the rule **to add fractions that have like denominators.** The numerator of the sum is the number of parts there are in the two fractions. The denominator of the sum is the common denominator of the two fractions.

Another way to express this rule is: “the numerator of the sum of the two fractions is the sum of the numerators of the two fractions. And the denominator of the sum is the common denominator.”

Now, how does one add fractions when the denominator are not alike?

Explain: 2 / 3 + 1 / 5 .

The bar diagram that represents 2 / 3 is:

The bar diagram that represents 1 / 5 is:

Since the fractions are not alike, we use Principle 2. So we must find a common category that conains a representative of each fraction then we add the representatives within that common category.

To find a common category that contains a representative for each fraction.

Draw a rectangle whose vertical side is divided into three equal parts: thirds. And its horizontal side is divided into five equal parts: fifths.

Shade two of the vertical subdivisions to get the representative of 2 / 3.

Shade one of the horizontal sundivisions to get the representative of 1 / 5.

Count the number of subdivisions in the representatives of the two fractions.

Shade that count in the commom category. This shading is the representative of the sum of the two fractions.

The bar diagrram that represent the sum:

Note:

- The red bar represents 2 / 3 of the the first rectangle.
- The yellow bar represents 1 / 5 of the second rectangle.
- 2 / 3 is equivalent to 10 / 15 in the common category, count the cells,
- 1 / 5 is equivalent to 3 / 15 in the common category.

Within this common category (the rectangle with 15 cells), (10+3) must be shaded. So the sum is 13 / 15 .

**Problem**: Explain 1 / 3 + 1 / 2.

Illustrations are powerful learning devices. They build mental maps that solidify what is being learned. Attempt to use them when teaching new concepts. Further illustrations are useful when solving problems, thereby giving more reverence to the phrase – a picture is worth a thousand words. Be advised, illustrations can mislead when improperly drawn or improperly interpreted.

For this reason formulas can proudly step forward as a preventor from error. For that reason, I summarize the addition procedure in a formula.

**The formula for adding fractions with unlike denominators**

The formula says that the sum of two fractions with unlike denominators is a fraction. Its numerator is the sum of two products. The first product is made the numerator of the first fraction (a) times the denominator of the second fraction (d). The second product is the denominator of the first fraction (b) times the numerator of the second fraction (c).

The denominator of the sum is the product of the denominators of the given fractions.

Graphically this rule is

The formula says the sum of two fractions with unlike denominators is a fraction. The numerator of the sum is a sum of two products. The first product is made with the numbers along the red arrow (ad). The second product is made with the numbers along the blue arrow (bc).

The denominator of the sum is made with the numbers along the fuscha arrow (bd).

Example: Solve 3/8 + 4/13

The numeratore of the the sum is: 3 x 13 + 8 x 4 = 39 + 32 = 71.

The denominator of the sum is 8 x 13 = 104.

So 3 / 8 + 4 / 13 = 71 / 104.

In order to understand how to solve the mowing, we must be able to find a certain fraction that is equivalent to a given fraction.

To find a fraction that is equivalent to a given fraction, we have two rules:

(m) Multiply the numerator and the denominator of the given fraction by the same number. The created fraction is equivalent to the given fraction

or

(d) Divide the numerator and the denominator of the given fraction by the same number. The created fraction is equivalent to the given fraction.

In either case, we cannot multiply or divide the numerator and denominator by ZERO.

Find the fraction that has ONE as its numerator that is equivalent to 7 /12.

Since we want ONE in the numerator, we divide the numerator of 7 / 12 by SEVEN to get the ONE. Rule (d) requires that we divide the denominator of 7 / 12 by SEVEN.

We get

Problem: What is the fraction with ONE as its numerator that is equivalent to 34 / 3 ?

**Answer:**

Pretty simple isn’t it.

**Can this be written as a fraction with a simple numerator and a simple denominator?**

Can this fraction be written as a fraction with a simple numerator and simple denominator?

Thankfully yes!

**To find the fraction with ONE in the numerator that is equivalent to a given fraction, write the given fraction upside down. **

**That is, the fraction you seek has the denominator of the given fraction as its numerator and it has the numerator ofthe given fraction as its denominator.**

Pretty simple isn’t it.

Find the fractions with ONE in the numerator that is equivalent to 12 / 7. It is

Can this fraction be written as a fraction with a simple numerator and simple denominator?

Thankfully yes!

It is 7 / 12

**To find the fraction that has one in the numerator that is equivalent to a given fraction, simply write the fraction upside down. That is, it will have the previous denominator as its numerator and it will have the previous numerator as its denominator.**

Now we can treat the problem.

Since Mary can mow the rear lawn in three hours, she does 1 / 3 of it in one hour.

If you can do something in 14 hours, you will do 1/ 14 of it in one hour.

If you can do something in 5 hours, you will do 1/5 of it in one hour.

Rule: If you can do something in n hours, you will do 1/n of it in one hour.

Note the relationship between the amount of time to do a task and how much of the task can be done in one hour. For either of these numbers, the other is its equivalent fraction having ONE in the numerator. (3, 1/3) (1/4. 4)

A similar reasoning holds for her brother. He mows 1 / 4 of the lawn in an hour.

If Mary starts on the left side of the lawn and mows in an up and down serpentine fashion and her brother, starting at the same time, begins on the lawn’s right side and mow in an up and down serpentine fashion, at the end of one hour they will have mowed 1 / 3 + 1 / 4 = 7 / 12 of the lawn.

Use the above rule in reverse. If 1/3 of a task can be done in 1 hour, then the task can be done in 3 hours.

Rule: If 1/n of a task can be done in 1 hour, then the task can be done in n hours.

A person who completely mows Mary’s rear lawn in 12 / 7 hours will mow 7/12 of Mary’s lawn in one hour. But Mary and her brother working together mow 7/12 of the lawn in one hour. So they can mow the lawn in 1 hours or 1 hour 42 minutes and 51 seconds also.

**Something To Think About ****Difficult1 1**: Jack Nerva had a backyard pool installed for his kids. To fill the pool, Jack plans to use his backyard hose and run it continuously. It will fill the pool in 26 hours. Jack’s next-door neighbor has access to a water source that can fill the pool in 18 hours. Jack convinces his neighbor to combine their water sources to fill the pool. How much time is needed to fill the pool?

**Difficult 2**: Assuming they work together, how much time will Mary and her brother need in order to complete the mowing and arrive at the park in time to see the start of the fireworks show? Give your answer as a fraction.

**Difficult 3:** When the spillway north of Coryville, USA is full and the water is not released, its waters will drains into an underground aquifer in 84 days. The State’s Corp of Engineers began a project to accelerate the drainage into the aquifer to 53 days but the construction funds were slashed severely and the construction ceased with only half of the project completed. So the water will drain from the spillway in 74 days. The spillway is full with winter rain. If it is not drained, how long will it take to empty the waters into the aquifer?

From these problems, do you see a pattern? Can you think of other situations where you might apply this analysis? Write your suggestion as a comment.