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2.11: Applications of Reciprocals

Difficulty Level: Basic Created by: CK-12
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Suppose that a car did one lap around a circular race track with a circumference of \begin{align*} 1 \frac{4}{7}\end{align*} miles. If you use \begin{align*}\frac{22}{7}\end{align*} as an approximation for \begin{align*} \pi\end{align*}, could you find the diameter of the race track? After completing this Concept, you'll be able to solve real-world problems such as this by using reciprocals.


Using Reciprocals to Solve Real-World Problems

The need to divide rational numbers is necessary for solving problems in physics, chemistry, and manufacturing. The following example illustrates the need to divide fractions in physics.

Example A

Newton’s Second Law relates acceleration to the force of an object and its mass: \begin{align*}a = \frac{F}{m}\end{align*}. Suppose \begin{align*}F = 7\frac{1}{3}\end{align*} and \begin{align*}m= \frac{1}{5}\end{align*}. Find \begin{align*}a\end{align*}, the acceleration.

Solution: Before beginning the division, the mixed number of force must be rewritten as an improper fraction.

Replace the fraction bar with a division symbol and simplify: \begin{align*}a = \frac{22}{3} \div \frac{1}{5}.\end{align*}

\begin{align*}\frac{22}{3} \times \frac{5}{1} = \frac{110}{3} = 36 \frac{2}{3}\end{align*}. Therefore, the acceleration is \begin{align*}36 \frac{2}{3} \ m/s^2.\end{align*}

Example B

Anne runs a mile and a half in one-quarter hour. What is her speed in miles per hour?

Solution: Use the formula \begin{align*}speed = \frac{distance}{time}\end{align*}.

\begin{align*}s = 1.5 \div \frac{1}{4}\end{align*}

Rewrite the expression and simplify: \begin{align*}s = \frac{3}{2} \cdot \frac{4}{1} = \frac{4 \cdot 3} {2 \cdot 1} = \frac{12}{2} = 6 \ mi/hr.\end{align*}

Example C

For a certain recipe of cookies, you need 3 cups of flour for every 2 cups of sugar. If Logan has 1/2 cup flour, how many cups of sugar will he need to use to make a smaller batch?

Solution: First we need to figure out how many times biger 3 is than 1/2, by dividing 3 by 1/2:

\begin{align*}3 \div \frac{1}{2}=3 \times \frac{2}{1}=3\times 2=6.\end{align*}

Since 1/2 goes into 3 six times, then we need to divide the 2 cups of sugar by 6:

\begin{align*}2 \div 6=2\times \frac{1}{6}=\frac{2}{6}=\frac{1}{3}.\end{align*}

Logan needs 1/3 cup of sugar to make a smaller batch with 1/2 cup flour.

Guided Practice

1. Newton’s Second Law relates acceleration to the force of an object and its mass: \begin{align*}a = \frac{F}{m}\end{align*}. Suppose \begin{align*}F = 5\frac{1}{2}\end{align*} and \begin{align*}m= \frac{2}{3}\end{align*}. Find \begin{align*}a\end{align*}, the acceleration.

2. Mayra runs 3 and a quarter miles in one-half hour. What is her speed in miles per hour?


1. Before we substitute the values into the formula, we must turn the mixed fraction into an improper fraction:

\begin{align*} 5\frac{1}{2}=\frac{5\times 2+1}{2}=\frac{11}{2}\end{align*}

\begin{align*}&a = \frac{F}{m}=\frac{\frac{11}{2}}{\frac{2}{3}}=\\ &\frac{11}{2}\div \frac{2}{3}=\frac{11}{2}\times \frac{3}{2}=\\ & \frac{11\times 3}{2\times 2}=\frac{33}{4}=8\frac{1}{4}\end{align*}

Therefore, the acceleration is \begin{align*}8\frac{1}{4}m/s^2\end{align*}.

2. Use the formula \begin{align*}speed = \frac{distance}{time}\end{align*}:

\begin{align*}&speed = \frac{distance}{time}=3\frac{1}{4}\div \frac{1}{2}=\\ & \frac{13}{4}\div \frac{1}{2}= \frac{13}{4}\times 2=\\ &\frac{13\times 2}{4}.\end{align*}

Before we continue, we will simplify the fraction:

\begin{align*} &\frac{13\times 2}{4}=\frac{13\times 2}{2\times 2}=\frac{13}{2}=6\frac{1}{2}.\end{align*}

Mayra can run 6-and-a-half miles per hour.


Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Division of Rational Numbers (8:20)

In 1 – 3, evaluate the expression.

  1. \begin{align*}\frac{x}{y}\end{align*} for \begin{align*}x = \frac{3}{8}\end{align*} and \begin{align*}y= \frac{4}{3}\end{align*}
  2. \begin{align*}4z \div u\end{align*} for \begin{align*}u = 0.5\end{align*} and \begin{align*}z = 10\end{align*}
  3. \begin{align*}\frac{-6}{m}\end{align*} for \begin{align*}m= \frac{2}{5}\end{align*}
  4. The label on a can of paint states that it will cover 50 square feet per pint. If I buy a \begin{align*}\frac{1}{8}\end{align*}-pint sample, it will cover a square two feet long by three feet high. Is the coverage I get more, less, or the same as that stated on the label?
  5. The world’s largest trench digger, “Bagger 288,” moves at \begin{align*}\frac{3}{8}\end{align*} mph. How long will it take to dig a trench \begin{align*}\frac{2}{3}\end{align*}-mile long?
  6. A \begin{align*}\frac{2}{7}\end{align*} Newton force applied to a body of unknown mass produces an acceleration of \begin{align*}\frac{3}{10} \ m/s^2\end{align*}. Calculate the mass of the body. Note: \begin{align*}\text{Newton} = kg \ m/s^2\end{align*}
  7. Explain why the reciprocal of a nonzero rational number is not the same as the opposite of that number.
  8. Explain why zero does not have a reciprocal.

Mixed Review


  1. \begin{align*}199 - (-11)\end{align*}
  2. \begin{align*}-2.3 - (-3.1)\end{align*}
  3. \begin{align*}|16-84|\end{align*}
  4. \begin{align*}|\frac{-11}{4}|\end{align*}
  5. \begin{align*}(4 \div 2 \times 6 + 10-5)^2\end{align*}
  6. Evaluate \begin{align*}f(x)= \frac{1}{9} (x-3); f(21)\end{align*}.
  7. Define range.

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reciprocal The reciprocal of a nonzero rational number \frac{a}{b} is \frac{b}{a}.

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