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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*}147 miles. If you use \begin{align*}\frac{22}{7}\end{align*}227 as an approximation for \begin{align*} \pi\end{align*}π, 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*}a=Fm. Suppose \begin{align*}F = 7\frac{1}{3}\end{align*}F=713 and \begin{align*}m= \frac{1}{5}\end{align*}m=15. Find \begin{align*}a\end{align*}a, 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*}a=223÷15.

\begin{align*}\frac{22}{3} \times \frac{5}{1} = \frac{110}{3} = 36 \frac{2}{3}\end{align*}223×51=1103=3623. Therefore, the acceleration is \begin{align*}36 \frac{2}{3} \ m/s^2.\end{align*}3623 m/s2.

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*}speed=distancetime.

\begin{align*}s = 1.5 \div \frac{1}{4}\end{align*}s=1.5÷14

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*}s=3241=4321=122=6 mi/hr.

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*}3÷12=3×21=3×2=6.

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*}2÷6=2×16=26=13.

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

Video Review

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*}a=Fm. Suppose \begin{align*}F = 5\frac{1}{2}\end{align*}F=512 and \begin{align*}m= \frac{2}{3}\end{align*}m=23. Find \begin{align*}a\end{align*}a, 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*}512=5×2+12=112

\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*}a=Fm=11223=112÷23=112×32=11×32×2=334=814

Therefore, the acceleration is \begin{align*}8\frac{1}{4}m/s^2\end{align*}814m/s2.

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

\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*}speed=distancetime=314÷12=134÷12=134×2=13×24.

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*}13×24=13×22×2=132=612.

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


In 1 – 3, evaluate the expression.

  1. \begin{align*}\frac{x}{y}\end{align*}xy for \begin{align*}x = \frac{3}{8}\end{align*}x=38 and \begin{align*}y= \frac{4}{3}\end{align*}y=43
  2. \begin{align*}4z \div u\end{align*}4z÷u for \begin{align*}u = 0.5\end{align*}u=0.5 and \begin{align*}z = 10\end{align*}z=10
  3. \begin{align*}\frac{-6}{m}\end{align*}6m for \begin{align*}m= \frac{2}{5}\end{align*}m=25
  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*}18-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*}38 mph. How long will it take to dig a trench \begin{align*}\frac{2}{3}\end{align*}23-mile long?
  6. A \begin{align*}\frac{2}{7}\end{align*}27 Newton force applied to a body of unknown mass produces an acceleration of \begin{align*}\frac{3}{10} \ m/s^2\end{align*}310 m/s2. Calculate the mass of the body. Note: \begin{align*}\text{Newton} = kg \ m/s^2\end{align*}Newton=kg m/s2
  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*}199(11)
  2. \begin{align*}-2.3 - (-3.1)\end{align*}2.3(3.1)
  3. \begin{align*}|16-84|\end{align*}|1684|
  4. \begin{align*}|\frac{-11}{4}|\end{align*}|114|
  5. \begin{align*}(4 \div 2 \times 6 + 10-5)^2\end{align*}(4÷2×6+105)2
  6. Evaluate \begin{align*}f(x)= \frac{1}{9} (x-3); f(21)\end{align*}f(x)=19(x3);f(21).
  7. Define range.

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

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