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Rational Numbers in Applications

Story problems including fractions

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Rational Numbers in Applications

Rational Numbers in Applications

Let's use the skills we learned in the last concept to solve some real-world problems.

Real-World Application: School Trip

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*} the cost; his grandma Zenoviea will provide \begin{align*}\frac{1}{6}\end{align*}; and his grandparents in Florida \begin{align*}\frac{1}{4}\end{align*}. We need to find the sum of those numbers, or \begin{align*}\frac{1}{2} + \frac{1}{6} + \frac{1}{4}\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*} the cost of the trip, or2200 out of \$2400, from his family.

Real-World Application: Property Management

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*} and \begin{align*}\frac{19}{20}\end{align*} acres, and the firm can use all of that land except for \begin{align*}\frac{1}{6}\end{align*} of an acre. The total amount of land the firm can use is therefore \begin{align*}\frac{4}{5} + \frac{5}{12} + \frac{19}{20} - \frac{1}{6}\end{align*} acres.

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*}, so that’s our common denominator. Now to convert the fractions:

\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*} or simply 2. One is sometimes called the invisible denominator, because every whole number can be thought of as a rational number whose denominator is one.

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.

License: CC BY-NC 3.0

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*} is the distance measured in meters, and intensity is measured in lumens. Calculate the change in intensity when the detector is moved from two meters to three meters away.

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*}. The negative value means that the intensity decreases as we move from 2 to 3 meters away.

When moving the detector from two meters to three meters, the intensity falls by \begin{align*}\frac{5}{12}\end{align*} lumens.

Guided Practice

Example 1

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?

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.

Review

Which property of addition does each situation involve?

1. Whichever order your groceries are scanned at the store, the total will be the same.
2. However many shovel-loads it takes to move 1 ton of gravel, the number of rocks moved is the same.
3. 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.

1. \begin{align*}\frac{7}{15} + \frac{2}{9}\end{align*}
2. \begin{align*}\frac{5}{19} + \frac{2}{27}\end{align*}
3. \begin{align*}\frac{5}{12} - \frac{9}{18}\end{align*}
4. \begin{align*}\frac{2}{3} - \frac{1}{4}\end{align*}
5. 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?
6. 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?
7. 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.

Review (Answers)

To view the Review answers, open this PDF file and look for section 2.4.

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Vocabulary Language: English

Equivalent Fractions

Equivalent fractions are fractions that can each be simplified to the same fraction. An equivalent fraction is created by multiplying both the numerator and denominator of the original fraction by the same number.

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1. [1]^ License: CC BY-NC 3.0

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