2.4: Rational Numbers in Applications
What if you sent out invitations to a birthday party to 64 of your friends? One week later \begin{align*}\frac{1}{2}\end{align*}
Watch This
Watch this video for help with the Examples above.
CK-12 Foundation: 0204S Solving Real-World Problems with Addition and Subtraction
Guidance
Let's use the skills we learned in the last concept to solve some real-world problems.
Example A
Peter is hoping to travel on a school trip to Europe. The ticket costs $2400. Peter has several relatives who have pledged to help him with the ticket cost. His parents have told him that they will cover half the cost. His grandma Zenoviea will pay one sixth, and his grandparents in Florida will send him one fourth of the cost. What fraction of the cost can Peter count on his relatives to provide?
The first thing we need to do is extract the relevant information. Peter’s parents will provide \begin{align*}\frac{1}{2}\end{align*}
To determine the sum, we first need to find the LCD. The LCM of 2, 6 and 4 is 12, so that’s our LCD. Now we can find equivalent fractions:
\begin{align*}\frac{1}{2} &= \frac{6 \cdot 1}{6 \cdot 2} = \frac{6}{12}\\
\frac{1}{6} &= \frac{2 \cdot 1}{2 \cdot 6} = \frac{2}{12}\\
\frac{1}{4} &= \frac{3 \cdot 1}{3 \cdot 4} = \frac{3}{12}\end{align*}
Putting them all together: \begin{align*}\frac{6}{12} + \frac{2}{12} + \frac{3}{12} = \frac{11}{12}\end{align*}
Peter will get \begin{align*}\frac{11}{12}\end{align*}
Example B
A property management firm is buying parcels of land in order to build a small community of condominiums. It has just bought three adjacent plots of land. The first is four-fifths of an acre, the second is five-twelfths of an acre, and the third is nineteen-twentieths of an acre. The firm knows that it must allow one-sixth of an acre for utilities and a small access road. How much of the remaining land is available for development?
The first thing we need to do is extract the relevant information. The plots of land measure \begin{align*}\frac{4}{5}, \frac{5}{12},\end{align*}
We can add and subtract multiple fractions at once just by finding a common denominator for all of them. The factors of 5, 9, 20, and 6 are as follows:
\begin{align*}&5 \ \qquad 5\\
&12 \qquad 2 \cdot 2 \cdot 3\\
&20 \qquad 2 \cdot 2 \cdot 5\\
&6 \ \qquad 2 \cdot 3\end{align*}
We need a 5, two 2’s, and a 3 in our LCD. \begin{align*}2 \cdot 2 \cdot 3 \cdot 5 = 60\end{align*}
\begin{align*}\frac{4}{5} &= \frac{12 \cdot 4}{12 \cdot 5} = \frac{48}{60}\\
\frac{5}{12} &= \frac{5 \cdot 5}{5 \cdot 12} = \frac{25}{60}\\
\frac{19}{20} &= \frac{3 \cdot 19}{3 \cdot 20} = \frac{57}{60}\\
\frac{1}{6} &= \frac{10 \cdot 1}{10 \cdot 6} = \frac{10}{60}\end{align*}
We can rewrite our sum as \begin{align*}\frac{48}{60} + \frac{25}{60} + \frac{57}{60} - \frac{10}{60} = \frac{48 + 25 + 57 - 10}{60} = \frac{120}{60}\end{align*}
Next, we need to reduce this fraction. We can see immediately that the numerator is twice the denominator, so this fraction reduces to \begin{align*}\frac{2}{1}\end{align*}
Solution
The property firm has two acres available for development.
Evaluate Change Using a Variable Expression
When we write algebraic expressions to represent a real quantity, the difference between two values is the change in that quantity.
Example C
The intensity of light hitting a detector when it is held a certain distance from a bulb is given by this equation:
\begin{align*}Intensity = \frac{3}{d^2} \end{align*}
where \begin{align*}d\end{align*}
We first find the values of the intensity at distances of two and three meters.
\begin{align*}\text{Intensity} \ (2) & = \frac{3}{(2)^2} = \frac{3}{4}\\
\text{Intensity} \ (3) & = \frac{3}{(3)^2} = \frac{3}{9} = \frac{1}{3}\end{align*}
The difference in the two values will give the change in the intensity. We move from two meters to three meters away.
\begin{align*}\text{Change} = \text{Intensity} \ (3) - \text{Intensity} \ (2) = \frac{1}{3} - \frac{3}{4}\end{align*}
To find the answer, we will need to write these fractions over a common denominator.
The LCM of 3 and 4 is 12, so we need to rewrite each fraction with a denominator of 12:
\begin{align*}\frac{1}{3} &= \frac{4 \cdot 1}{4 \cdot 3} = \frac{4}{12}\\
\frac{3}{4} &= \frac{3 \cdot 3}{3 \cdot 4} = \frac{9}{12}\end{align*}
So we can rewrite our equation as \begin{align*}\frac{4}{12} - \frac{9}{12} = -\frac{5}{12}\end{align*}
Solution
When moving the detector from two meters to three meters, the intensity falls by \begin{align*}\frac{5}{12}\end{align*}
Watch this video for help with the Examples above.
CK-12 Foundation: Solving Real-World Problems Using Addition and Subtraction
Vocabulary
- Subtracting a number is the same as adding the opposite (or additive inverse) of the number.
- To add fractions, rewrite them over the lowest common denominator (LCD). The lowest common denominator is the lowest (or least) common multiple (LCM) of the two denominators.
- When adding fractions: \begin{align*}\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}\end{align*}
ac+bc=a+bc - When subtracting fractions: \begin{align*}\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}\end{align*}
ac−bc=a−bc - Commutative property: the sum of two numbers is the same even if the order of the items to be added changes.
- Associative Property: When three or more numbers are added, the sum is the same regardless of how they are grouped.
- Additive Identity Property: The sum of any number and zero is the original number.
- The number one is sometimes called the invisible denominator, as every whole number can be thought of as a rational number whose denominator is one.
- The difference between two values is the change in that quantity.
Guided Practice
Elsa baked a small cake for her family. First her sister ate one quarter and her mom ate one third. How much was left for Elsa?
Solution:
The whole cake is represented by 1. To solve this problem, we subtract the fraction that each person ate.
\begin{align*}1-\frac{1}{4}-\frac{1}{3}\end{align*}
To complete this problem, we must give the terms common denominators. Since the denominators do not share any factors, we simply multiply them together: \begin{align*} 4\cdot 3=12\end{align*}.
\begin{align*}&1-\frac{1}{4}-\frac{1}{3} \qquad \text{Start with the original expression.}\\ =&1\cdot\frac{12}{12}-\frac{1}{4}\cdot \frac{3}{3}-\frac{1}{3}\cdot \frac{4}{4} \qquad \text{Give each term a common denominator.} \\ =&\frac{12}{12}-\frac{3}{12}-\frac{4}{12} \qquad \text{Simplify.} \\ = & \frac{12-3-4}{12}=\frac{5}{12} \end{align*}
There is \begin{align*}\frac{5}{12}\end{align*} of the original cake left for Elsa.
Practice
Which property of addition does each situation involve?
- Whichever order your groceries are scanned at the store, the total will be the same.
- However many shovel-loads it takes to move 1 ton of gravel, the number of rocks moved is the same.
- If Julia has no money, then Mark and Julia together have just as much money as Mark by himself has.
In 4-7, practice your addition and subtraction skills.
- \begin{align*}\frac{7}{15} + \frac{2}{9}\end{align*}
- \begin{align*}\frac{5}{19} + \frac{2}{27}\end{align*}
- \begin{align*}\frac{5}{12} - \frac{9}{18}\end{align*}
- \begin{align*}\frac{2}{3} - \frac{1}{4}\end{align*}
- Ilana buys two identically sized cakes for a party. She cuts the chocolate cake into 24 pieces and the vanilla cake into 20 pieces, and lets the guests serve themselves. Martin takes three pieces of chocolate cake and one of vanilla, and Sheena takes one piece of chocolate and two of vanilla. Which of them gets more cake?
- Nadia, Peter and Ian are pooling their money to buy a gallon of ice cream. Nadia is the oldest and gets the greatest allowance. She contributes half of the cost. Ian is next oldest and contributes one third of the cost. Peter, the youngest, gets the smallest allowance and contributes one fourth of the cost. They figure that this will be enough money. When they get to the check-out, they realize that they forgot about sales tax and worry there will not be enough money. Amazingly, they have exactly the right amount of money. What fraction of the cost of the ice cream was added as tax?
- The time taken to commute from San Diego to Los Angeles is given by the equation \begin{align*}time = \frac{120}{speed}\end{align*} where time is measured in hours and speed is measured in miles per hour (mph). Calculate the change in time that a rush hour commuter would see when switching from traveling by bus to traveling by train, if the bus averages 40 mph and the train averages 90 mph.
Image Attributions
Here you'll apply the properties of addition and subtraction to solve real-world problems involving rational numbers.