Example

As of March, 2016 the world's population was estimated at 7.4 billion. [footnote] "Current World Population." World Population Clock: 7.4 Billion People (2016). Accessed June 21, 2016. http://www.worldometers.info/world-population/. "Current World Population." World Population Clock: 7.4 Billion People (2016). Accessed June 21, 2016. http://www.worldometers.info/world-population/. "Current World Population." World Population Clock: 7.4 Billion People (2016). Accessed June 21, 2016. http://www.worldometers.info/world-population/.[/footnote].  According to water.org, 1 out of every 3 people on the planet lives without access to a toilet.  Find the number of people on the planet that do not have access to a toilet.

Answer: We can use a proportion to find the unknown number of people who live without a toilet since we are given that 1 in 3 don't have access, and we are given the population of the planet. We know that 1 out of every 3 people don't have access, so we can write that as a ratio (fraction)

[latex]\frac{1}{3}[/latex].

Let the number of people without access to a toilet be x. The ratio of people with and without toilets is then

[latex]\frac{x}{7.4\text{ billion }}[/latex]

Equate the two ratios since they are representing the same fractional amount of the population.

[latex]\frac{1}{3}=\frac{x}{7.4\text{ billion }}[/latex]

Solve:

[latex]\begin{array}{l}\frac{1}{3}=\frac{x}{7.4}\\\text{}\\7.4\cdot\frac{1}{3}=\frac{x}{7.4}\cdot{7.4}\\\text{}\\2.46=x\end{array}[/latex]

The original units were billions of people, so our answer is [latex]2.46[/latex] billion people don't have access to a toilet.  Wow, that's a lot of people.

Answer

2.46 billion people don't have access to a toilet.

Example

It has been shown that a person's height is proportional to the length of their femur [footnote]Obialor, Ambrose, Churchill Ihentuge, and Frank Akapuaka. "Determination of Height Using Femur Length in Adult Population of Oguta Local Government Area of Imo State Nigeria." Federation of American Societies for Experimental Biology, April 2015. Accessed June 22, 2016. http://www.fasebj.org/content/29/1_Supplement/LB19.short.[/footnote]. Given that a person who is 71 inches tall has a femur length of 17.75 inches, how tall is someone with a femur length of 16 inches?

Answer: Height and femur length are proportional for everyone, so we can define a ratio with the given height and femur length.  We can then use this to write a proportion to find the unknown height. Let x be the unknown height.  Define the ratio of femur length and height for both people using the given measurements.

Person 1:  [latex]\frac{\text{femur length}}{\text{height}}=\frac{17.75\text{inches}}{71\text{inches}}[/latex]

Person 2:  [latex]\frac{\text{femur length}}{\text{height}}=\frac{16\text{inches}}{x\text{inches}}[/latex]

Equate the ratios, since we are assuming height and femur length are proportional for everyone.

[latex]\frac{17.75\text{inches}}{71\text{inches}}=\frac{16\text{inches}}{x\text{inches}}[/latex]

 Solve by using the common denominator to clear fractions.  The common denominator is [latex]71x[/latex]

[latex]\begin{array}{c}\frac{17.75}{71}=\frac{16}{x}\\\\71x\cdot\frac{17.75}{71}=\frac{16}{x}\cdot{71x}\\\\17.75\cdot{x}=16\cdot{71}\\\\x=\frac{16\cdot{71}}{17.75}=64\end{array}[/latex]

The unknown height of person 2 is 64 inches. In general, we can reduce the fraction [latex]\frac{17.75}{71}=0.25=\frac{1}{4}[/latex] to find a general rule for everyone.  This would translate to sayinga person's height is 4 times the length of their femur.

Example

Given a scale factor of 1:557 on a map of the US, if the distance from Seattle, WA to San Jose, CA is 1.5 inches on the map,  define a proportion to find the actual distance between them.

Answer: We need to define a proportion to solve for the unknown distance between Seattle and San Jose.  The scale factor is 1:557, and we will call the unknown distance x. The ratio of inches to miles is [latex]\frac{1}{557}[/latex]. We know inches between the two cities, but we don't know miles, so the ratio that describes the distance between them is [latex]\frac{1.5}{x}[/latex]. The proportion that will help us solve this problem is [latex]\frac{1}{557}=\frac{1.5}{x}[/latex]. Solve using the common denominator [latex]557x[/latex] to clear fractions. [latex-display]\begin{array}{ccc}\frac{1}{557}=\frac{1.5}{x}\\557x\cdot\frac{1}{557}=\frac{1.5}{x}\cdot{557x}\\x=1.5\cdot{557}=835.5\end{array}[/latex-display] We used the scale factor 1:557 to find an unknown distance between Seattle and San Jose. We also checked our answer of 835.5 miles with Google maps, and found that the distance is 839.9 miles, so we did pretty well!

Example

Two cities are 2.5 inches apart on a map.  Their actual distance from each other is 325 miles.  Write a proportion to represent and solve for the scale factor for one inch of the map.

Answer: We know that for each 2.5 inches on the map, it represents 325 actual miles. We are looking for the scale factor for one inch of the map. The ratio we want is [latex]\frac{1}{x}[/latex] where x is the actual distance represented by one inch on the map.  We know that for every 2.5 inches, there are 325 actual miles, so we can define that relationship as [latex]\frac{2.5}{325}[/latex] We can use a proportion to equate the two ratios and solve for the unknown distance.

[latex]\begin{array}{ccc}\frac{1}{x}=\frac{2.5}{325}\\325x\cdot\frac{1}{x}=\frac{2.5}{325}\cdot{325x}\\325=2.5x\\x=130\end{array}[/latex]

[latex]\begin{array}{ccc}\frac{1}{x}=\frac{2.5}{325}\\325x\cdot\frac{1}{x}=\frac{2.5}{325}\cdot{325x}\\325=2.5x\\x=130\end{array}[/latex]

The scale factor for one inch on the map is 1:130, or for every inch of map there are 130 actual miles.

ExAMPLE

Greg went to a conference in a city 120 miles away. On the way back, due to road construction he had to drive 10 mph slower which resulted in the return trip taking 2 hours longer. How fast did he drive on the way to the conference?

Answer: [latex]\begin{eqnarray*} \begin{array}{|c|c|c|c|} \hline & \,\text{rate}\,& \,\text{time}\,& \,\text{distance}\\ \hline \,\text{There}\,& r & t & 120\\ \hline \,\text{Back}\,& & & \\ \hline \end{array}\,& & \begin{array}{l} \,\text{We}\,\,\text{do}\,\,\text{not}\,\,\text{know}\,\,\text{rate}, r, \,\text{or} \,\text{time}, t \,\text{he}\,\,\text{traveled}\\ \,\text{on}\,\,\text{the}\,\,\text{way}\,\,\text{to}\,\,\text{the}\,\,\text{conference}\,. \,\text{But}\,\,\text{we}\,\,\text{do}\,\,\text{know}\\ \,\text{the}\,\,\text{distance}\,\,\text{was}\,120 \,\text{miles}\,. \end{array}\\ & & \\ \begin{array}{|c|c|c|c|} \hline & \,\text{rate}\,& \,\text{time}\,& \,\text{distance}\\ \hline \,\text{There}\,& r & t & 120\\ \hline \,\text{Back}\,& r - 10 & t + 2 & 120\\ \hline \end{array}& & \begin{array}{l} \,\text{Coming}\,\,\text{back}\,\,\text{he}\,\,\text{drove}\,10 \,\text{mph} \,\text{slower}\,(r - 10)\\ \,\text{and}\,\,\text{took}\,2 \,\text{hours}\,\,\text{longer}\,(t + 2) . \,\text{The} \,\text{distance}\\ \,\text{was}\,\,\text{still}\,120 \,\text{miles}\,. \end{array}\\ & & \\ r t = 120 & & \,\text{Equations}\,\,\text{are}\,\,\text{product}\,\,\text{of} \,\text{rate}\,\,\text{and}\,\,\text{time}\\ (r - 10) (t + 2) = 120 & & \,\text{We}\,\,\text{have}\,\,\text{simultaneous} \,\text{product}\,\,\text{equations}\\ & & \\ t = \frac{120}{r}\,\,\text{and}\,t + 2 = \frac{120}{r - 10} & & \,\text{Solving}\,\,\text{for}\,\,\text{rate}, \,\text{divide}\,\,\text{by}\,r \,\text{and}\,r - 10\\ & & \\ \frac{120}{r}\,+ 2 = \frac{120}{r - 10}& & \,\text{Substitute} \frac{120}{r} \,\text{for}\,t \,\text{in}\,\,\text{the}\,\,\text{second} \,\text{equation}\\ & & \\ \frac{120 r (r - 10)}{r} + 2 r (r - 10) = \frac{120 r (r - 10)}{r - 10} & & \,\text{Multiply}\,\,\text{each}\,\,\text{term}\,\,\text{by}\,\,\text{LCD}\,: r (r - 10)\\ & & \\ 120 (r - 10) + 2 r^2 - 20 r = 120 r & & \,\text{Reduce}\,\,\text{each} \,\text{fraction}\\ 120 r - 1200 + 2 r^2 - 20 r = 120 r & & \,\text{Distribute}\\ 2 r^2 + 100 r - 1200 = 120 r & & \,\text{Combine}\,\,\text{like} \,\text{terms}\\ \underline{- 120 r - 120 r} & & \,\text{Make}\,\,\text{equation}\,\,\text{equal} \,\text{to}\,\,\text{zero}\\ 2 r^2 - 20 r - 1200 = 0 & & \,\text{Divide}\,\,\text{each}\,\,\text{term} \,\text{by}\,2\\ r^2 - 10 r - 600 = 0 & & \,\text{Factor}\\ (r - 30) (r + 20) = 0 & & \,\text{Set}\,\,\text{each}\,\,\text{factor} \,\text{equal}\,\,\text{to}\,\,\text{zero}\\ r - 30 = 0 \,\text{and}\,r + 20 = 0 & & \,\text{Solve}\,\,\text{each} \,\text{equation}\\ \underline{+ 30 + 30} \underline{- 20 - 20} & & \\ r = 30 \,\text{and}\,r = - 20 & & \,\text{Can}' t \,\text{have}\,a \,\text{negative}\,\,\text{rate}\\ 30 \,\text{mph}\,& & \,\text{Our}\,\,\text{Solution} \end{eqnarray*}[/latex]          

Exercises

A man rows down stream for 30 miles then turns around and returns to his original location, the total trip took 8 hours. If the current flows at 2 miles per hour, how fast would the man row in still water?

Answer: \begin{eqnarray*} \begin{array}{|c|c|c|c|} \hline & \,\text{rate}\,& \,\text{time}\,& \,\text{distance}\\ \hline \,\text{down}\,& & t & 30\\ \hline \,\text{up}\,& & 8 - t & 30\\ \hline \end{array} & & \begin{array}{l} \,\text{We}\,\,\text{know}\,\,\text{the}\,\,\text{distance}\,\,\text{up}\,\,\text{and}\, \,\text{down}\,\,\text{is}\,30.\\ \,\text{Put}\,t \,\text{for}\,\,\text{time}\,\,\text{downstream}\,. \,\text{Subtracting}\\ 8 - t \,\text{becomes}\,\,\text{time}\,\,\text{upstream}\, \end{array}\\ & & \\ \begin{array}{|c|c|c|c|} \hline & \,\text{rate}\,& \,\text{time}\,& \,\text{distance}\\ \hline \,\text{down}\,& r + 2 & t & 30\\ \hline \,\text{up}\,& r - 2 & 8 - t & 30\\ \hline \end{array}& & \begin{array}{l} \,\text{Downstream}\,\,\text{the}\,\,\text{current}\,\,\text{of}\,2 \,\text{mph} \,\text{pushes}\,\\ \,\text{the}\,\,\text{boat} (r + 2) \,\text{and}\,\,\text{upstream}\,\,\text{the}\, \,\text{current}\,\\ \,\text{pulls}\,\,\text{the}\,\,\text{boat}\,(r - 2) \end{array}\\ & & \\ (r + 2) t = 30 & & \,\text{Multiply}\,\,\text{rate}\,\,\text{by}\,\,\text{time} \,\text{to}\,\,\text{get}\,\,\text{equations}\,\\ (r - 2) (8 - t) = 30 & & \,\text{We}\,\,\text{have}\,a \,\text{simultaneous} \,\text{product}\\ & & \\ t = \frac{30}{r + 2} \,\text{and}\,8 - t = \frac{30}{r - 2} & & \,\text{Solving}\,\,\text{for}\,\,\text{rate}, \,\text{divide}\,\,\text{by}\,r + 2 \,\text{or}\,r - 2\\ & & \\ 8 - \frac{30}{r + 2}\,= \frac{30}{r - 2} & & \,\text{Substitute}\,\frac{30}{r + 2} \,\text{for}\,t \,\text{in}\,\,\text{second}\,\,\text{equation}\,\\ & & \\ 8 (r + 2) (r - 2) - \frac{30 (r + 2) (r - 2)}{r + 2} = \frac{30 (r + 2) (r - 2)}{r - 2} & & \,\text{Multiply}\,\,\text{each}\,\,\text{term}\,\,\text{by} \,\text{LCD}\,: (r + 2) (r - 2)\\ & & \\ 8 (r + 2) (r - 2) - 30 (r - 2) = 30 (r + 2) & & \,\text{Reduce} \,\text{fractions}\,\\ 8 r^2 - 32 - 30 r + 60 = 30 r + 60 & & \,\text{Multiply} \,\text{and} \,\text{distribute}\\ 8 r^2 - 30 r + 28 = 30 r + 60 & & \,\text{Make} \,\text{equation} \,\text{equal}\,\,\text{zero}\,\\ \underline{- 30 r - 60 - 30 r - 60} & & \\ 8 r^2 - 60 r - 32 = 0 & & \,\text{Divide}\,\,\text{each}\,\,\text{term}\,\,\text{by} 4\\ 2 r^2 - 15 r - 8 = 0 & & \,\text{Factor}\\ (2 r + 1) (r - 8) = 0 & & \,\text{Set}\,\,\text{each}\,\,\text{factor} \,\text{equal}\,\,\text{to}\,\,\text{zero}\,\\ 2 r + 1 = 0 \,\text{or}\,r - 8 = 0 & & \,\text{Solve}\,\,\text{each}\, \,\text{equation}\\ \underline{- 1 - 1} \underline{+ 8 + 8} & & \\ 2 r = - 1 \,\text{or}\,r = 8 & & \\ \overline{2} \overline{2} & & \\ r = - \frac{1}{2} \,\text{or}\, r = 8 & & \,\text{Can}' t \,\text{have}\,a \,\text{negative}\,\,\text{rate}\,\\ 8 \,\text{mph}\,& & \,\text{Our}\,\,\text{Solution} \end{eqnarray*}

Example

A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning?

Answer:

Let t be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the tank is a linear equation, as is the amount of sugar in the tank. We can write an equation independently for each:

The concentration, C, will be the ratio of pounds of sugar to gallons of water

The concentration after 12 minutes is given by evaluating [latex]C\left(t\right)\\[/latex] at [latex]t=\text{ }12\\[/latex].

This means the concentration is 17 pounds of sugar to 220 gallons of water.

At the beginning, the concentration is

Since [latex]\frac{17}{220}\approx 0.08>\frac{1}{20}=0.05\\[/latex], the concentration is greater after 12 minutes than at the beginning.