Inverse Variation Formula. Practice Problems

Problem 1

If $$ y $$ varies inversely as $$ x $$ and $$ y = 14 $$ when $$ x = 20 $$. What is the value of $$ y $$ when $$ x $$ is 5?

Step 1

Write down the variation equation.

$$ y = \frac k x $$

Step 2

Use $$ \red{y = 14} $$ and $$ \blue{x = 20} $$ to find the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{14} & = \frac k {\blue{20}}\\[6pt] \red{14}(\blue{20}) & = \frac k {\blue{20}}\cdot \blue {20}\\[6pt] 280 & = \frac k {\cancelred{20}}\cdot \cancelred{20}\\[6pt] k & = 280 \end{align*} $$

The updated variation equation is $$ y = \frac{280} x $$

Step 3

Determine the value of $$ y $$ when $$ \blue{x = 5} $$.

$$ \begin{align*} y & = \frac{280}{\blue x}\\[6pt] y & = \frac{280}{\blue 5} = 56 \end{align*} $$

Answer

$$ y = 56 $$ when $$ x = 5 $$.

Problem 2

Suppose $$ s $$ varies indirectly as $$ t $$, and $$ s(1.5)=25 $$. Determine the value of $$ s(2.25) $$.

Step 1

Write down the variation equation.

$$ s = \frac k t $$

Step 2

Use $$ \red{s = 25} $$ and $$ \blue{t = 1.5} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{25} & = \frac k {\blue{1.5}}\\[6pt] \red{25}(\blue{1.5}) & = \frac k {\blue{1.5}}\cdot \blue{1.5}\\[6pt] 27.5 & = \frac k {\cancelred{1.5}}\cdot \cancelred{1.5}\\[6pt] k & = 37.5 \end{align*} $$

The updated variation equation is $$ s = \frac{37.5} t $$

Step 3

Determine the value of $$ s $$ when $$ \blue{t = 2.25} $$.

$$ \begin{align*} s & = \frac{37.5}{\blue t}\\[6pt] s & = \frac{37.5}{\blue{2.25}} = \frac{3750}{225} = \frac{50} 3 = 16\,\frac 2 3 \approx 16.667 \end{align*} $$

Answer

$$ s = \frac{50} 3 $$ when $$ t = 2.25 $$ .

Problem 3

Suppose $$ y $$ varies inversely as $$ x $$, and $$ y = 0.025 $$ when $$ x = 2 $$. What is the value of $$ x $$ when $$ y = 0.4 $$?

Step 1

Write down the variation equation.

$$ y = \frac k x $$

Step 2

Use $$ \red{y = 0.025} $$ and $$ \blue{x = 2} $$ to find the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{0.025} & = \frac k {\blue 2}\\[6pt] \red{(0.025)}\blue 2 & = \frac k {\blue 2} \cdot \blue 2\\[6pt] 0.05 & = \frac k {\cancelred 2} \cdot \cancelred 2\\[6pt] k & = 0.05 = \frac 1 {20} \end{align*} $$

The updated variation equation is $$ y = \frac{0.05} x = \frac{1/20} x = \frac 1 {20x} $$

Step 3

Determine the value of $$ x $$ when $$ \red{y = 0.4} $$.

$$ \begin{align*} \red{0.4} & = \frac 1 {20x}\\[6pt] \red{0.4} \cdot (20x) & = 1\\[6pt] 8x & = 1\\[6pt] x & = \frac 1 8 = 0.125 \end{align*} $$

Answer

$$ x = 0.125 $$ when $$ y = 0.4 $$.

Problem 4

Suppose $$ I $$ is inversely proportional to $$ R $$ and when $$ R = 200 $$, $$ I = 35 $$. Determine the value of $$ R $$ when $$ I = 100 $$.

Step 1

Write down the variation equation.

$$ I = \frac k R $$

Step 2

Use $$ \red{I = 35} $$ and $$ \blue{R = 200} $$ to find the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{35} & = \frac k {\blue{200}}\\[6pt] \red{35}(\blue{200}) & = \frac k {\blue{200}}\cdot \blue{200}\\[6pt] 7000 & = \frac k {\cancelred{200}}\cdot \cancelred{200}\\[6pt] k & = 7000 \end{align*} $$

The updated variation equation is $$ I = \frac{7000} R $$.

Step 3

Determine the value of $$ R $$ when $$ \red{I = 100} $$.

$$ \begin{align*} \red{100} & = \frac{7000} R\\[6pt] 100R & = 7000\\[6pt] R & = \frac{7000}{100} = 70 \end{align*} $$

Answer

$$ R = 70 $$ when $$ I = 100 $$.

Problem 5

Suppose $$ y $$ varies inversely as the square-root of $$ x $$. If $$ y = 8 $$ when $$ x = 49 $$, determine the value of $$ y $$ when $$ x = 121 $$

Step 1

Write down the variation equation.

$$ y = \frac k {\sqrt x} $$

Step 2

Use $$ \red{y = 8} $$ and $$ \blue{x = 49} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red 8 & = \frac k {\sqrt{\blue{49}}}\\[6pt] \red 8 & = \frac k 7\\[6pt] \red 8 (7) & = \frac k 7\cdot 7\\[6pt] 56 & = \frac k {\cancelred 7}\cdot \cancelred 7\\[6pt] k & = 56 \end{align*} $$

The updated variation equation is $$ y = \frac{56}{\sqrt x} $$

Step 3

Determine the value of $$ y $$ when $$ \blue{x = 100} $$.

$$ \begin{align*} y & = \frac{56}{\sqrt{\blue{100}}}\\[6pt] & = \frac{56}{10}\\[6pt] & = 5.6 \end{align*} $$

Answer

$$ y = 5.6 $$ when $$ x = 100 $$.

Problem 6

Suppose $$ p $$ varies inversely as the cube of $$ q $$, and $$ p = 2 $$ when $$ q = 4 $$. Determine the value of $$ q $$ when $$ p = 8 $$.

Step 1

Write down the variation equation.

$$ p = \frac k {q^3} $$

Step 2

Use $$ \red{p = 2} $$ and $$ \blue{q = 4} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red 2 & = \frac k {\blue 4^3}\\[6pt] \red 2 & = \frac k {64}\\[6pt] \red 2 (64) & = \frac k {\cancelred{64}} \cdot \cancelred{64}\\[6pt] k & = 128 \end{align*} $$

The updated variation equation is $$ p = \frac{128}{q^3} $$.

Step 3

Determine the value of $$ q $$ when $$ \red{p = 8} $$.

$$ \begin{align*} \red 8 & = \frac{128}{q^3}\\[6pt] 8q^3 & = 128\\[6pt] q^3 & = \frac{128} 8\\[6pt] & = 16\\[6pt] q & = \sqrt[3]{16} = 2\sqrt[3] 2 \end{align*} $$

Answer

$$ q = 2\sqrt[3] 2 $$ when $$ p = 8 $$.

Problem 7

Speed and travel time are inversely related (the faster you go, the shorter the travel time). Suppose the daily trip to school normally takes 20 minutes when your speed is 30 miles per hour. How long should the trip take if you drive 45 miles per hour instead (and don't get pulled over for speeding)?

Step 1

Define variables to describe travel time and speed, and state the equation relating them.

Let's use $$ t $$ for time (in minutes).

Let's use $$ s $$ for speed (in miles per hour).

The equation we need then is $$ s = \frac k t $$.

Step 2

Use $$ \red{s = 30} $$ and $$ \blue{t = 20} $$to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{30} & = \frac k {\blue{20}}\\[6pt] \red{30}(\blue{20}) & = \frac k {\blue{20}}\cdot\blue{20}\\[6pt] 600 & = \frac k {\cancelred{20}}\cdot\cancelred{20}\\[6pt] k & = 600 \end{align*} $$

The updated variation equation is $$ s = \frac{600} t $$.

Step 3

Use the equation to determine the travel time for the trip when speed is 45 miles per hour.

$$ \begin{align*} \red{45} & = \frac{600} t\\[6pt] \red{45}t & = 600\\[6pt] t & = \frac{600}{\red{45}}\\[6pt] & = \frac{40} 3\\[6pt] & = 13\frac 1 3 \end{align*} $$

Answer

The trip will take $$ 13\frac 1 3 $$ minutes (or 13 minutes and 20 seconds) if you drive at 45 miles per hour.

Problem 8

The Ideal Gas Law from chemistry can be written as $$ P = \frac k V $$, where $$ P $$ is the pressure (in pascals) and $$ V $$ is the volume of the container (in cubic meters).

Suppose that when a particular gas is stored in a 0.25 cubic meter container, and exerts 40 pascals of pressure. How much pressure will the gas exert if it were transferred to a container that only holds a volume of $$ 0.1 $$ cubic meters.

Step 1

Use $$ \red{P = 40} $$ and $$ \blue{V = 0.25} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{P} & = \frac k {\blue{V}}\\[6pt] \red{40} & = \frac k {\blue{0.25}}\\[6pt] \red{40}(\blue{0.25}) & = \frac k {\blue{0.25}}\cdot\blue{0.25}\\[6pt] 10 & = \frac k {\cancelred{0.25}}\cdot\cancelred{0.25}\\[6pt] k & = 10 \end{align*} $$

So for this particular gas, $$ P = \frac{10} V $$

Step 2

Use the equation to determine the pressure when $$ \blue{V = 0.1} $$.

$$ P = \frac{10}{\blue{0.1}} = 100 $$

Answer

The gas exerts a pressure of 100 pascals in a 0.1 cubic meter container.

Problem 9

The time it takes a block of ice to melt varies indirectly with temperature. If a block of ice that is 1/8 cubic meters in volume requires 6 hours to melt when it is $$72 ^\circ F $$, at what temperature will the block melt in only 2 hours?

Step 1

Define variables to represent temperature and time and write down the inverse variation equation.

Let $$ t = $$ time (in hours)

Let $$ T = $$ temperature (in degrees Fahrenheit)

The inverse variation equation will be

$$ t = \frac k T $$

Step 2

Use $$ \red{t = 6} $$ and $$ \blue{T = 72} $$ to find the value of $$ k $$, then write down the updated version of the inverse variation equation.

$$ \begin{align*} \red t & = \frac k {\blue T}\\[6pt] \red 6 & = \frac k {\blue{72}}\\[6pt] \red 6(\blue{72}) & = \frac k {\blue{72}}\cdot\blue{72}\\[6pt] 432 & = \frac k {\cancelred{72}}\cdot\cancelred{72}\\[6pt] k & = 432 \end{align*} $$

The inverse variation equation is now $$ t = \frac{432} T $$.

Step 3

Determine the temperature required to have the block of ice melt in only $$ \blue{t = 2} $$ hours.

$$ \begin{align*} \blue 2 & = \frac{432} T\\[6pt] \blue 2T & = 432\\[6pt] T & = \frac{432}{\blue 2}\\[6pt] & = 216 \end{align*} $$

Answer

The 1/8 cubic meter block of ice will melt in only 2 hours if the temperature is $$ 216 ^\circ F $$.

Problem 10

Suppose 4 adults can build a shed in 16 hours. How much quicker would it be to have 5 adults building the shed?

Step 1

Confirm this is an inverse variation problem.

We note that (in an ideal world) having more adults working on a particular project will mean the task takes less time.

Since the time decreases as the number of workers increase, we can model this as an inverse variation.

Step 2

Define variables to represent the quantities involved, and write down the inverse variation equation.

Let $$ w = $$ the number of workers building the shed.

Let $$ t = $$ the number of hours it takes to complete the shed.

The inverse variation equation is

$$ t = \frac k w $$

Step 3

Use $$ \red{t = 16} $$ and $$ \blue{w = 4} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red t & = \frac k {\red w}\\[6pt] \red{16} & = \frac k {\blue 4}\\[6pt] \red{16}(\blue 4) & = \frac k {\blue 4}\cdot \blue 4\\[6pt] 64 & = \frac k {\cancelred 4}\cdot \cancelred 4\\[6pt] k & = 64 \end{align*} $$

The updated variation equation is $$ t = \frac{64} w $$.

Step 4

Determine the length of time required for $$\blue{w = 5}$$ people to build the shed.

$$ t = \frac{64}{\blue 5} = 12\frac 4 5 = 12.8 $$

Step 5

Determine the savings in time when 5 adults work on the shed versus only 4.

Note that the question didn't ask us, "How long does it take...", but rather "How much quicker..". This means we are interested in the difference between having 4 and 5 workers

$$ \mbox{Time saved} = (\mbox{time for 4 workers}) - (\mbox{time for 5 workers}) = 16 - 12.8 = 3.2 $$

Answer

Having 5 adults on the project instead of 4 will result in the shed be completed 3.2 hours more quickly (or 3 hours, 12 minutes more quickly).

Problem 11

The gravitational force (in newtons) between two objects is inversely proportional to square of the distance (in meters) between the centers of the objects.

Suppose the gravitational force between two particular objects is $$ 1.95\times 10^4 $$ newtons when they are separated by $$ 8.94\times 10^{12} $$ meters. If the gravitational force increases to $$ 4.1\times 10^6 $$ newtons, what has happened to the distance between the objects?

Step 1

Define variables to represent the two quantities under investigation. Then write down the variation equation.

Let $$ F_g $$ represent the force due to gravity, measured in newtons.

Let $$ r $$ represent the distance between the centers of the objects, measured in meters.

The variation equation is $$ F_g = \frac k {r^2} $$.

Step 2

Use $$ \red{F_g = 1.95\times 10^4} $$ and $$ \blue{r = 8.94\times 10^{12}} $$ to determine the value of $$ k $$, then write down the updated variation equation.

$$ \begin{align*} \red{F_g} & = \frac k {\blue r^2}\\[6pt] \red{1.95\times 10^4} & = \frac k {(\blue{8.94\times 10^{12}})^2}\\[6pt] \red{1.95\times 10^4} & = \frac k {\blue{79.9236\times 10^{24}}}\\[6pt] \red{1.95\times 10^4}(\blue{79.9236\times 10^{24}}) & = \frac k {\blue{79.9236\times 10^{24}}}\cdot\blue{79.9236\times 10^{24}}\\[6pt] 1.5585102 \times 10^{30} & = \frac k {\cancelred{79.9236\times 10^{24}}}\cdot\cancelred{79.9236\times 10^{24}}\\[6pt] k & = 1.5585102 \times 10^{30}\\[6pt] & \approx 1.56 \times 10^{30} \end{align*} $$

The updated variation equation is $$ \displaystyle F_g = \frac{1.56\times 10^{30}}{r^2} $$.

Step 3

Determine the distance between the two objects if the force of gravity is $$ \red{F_g = 4.1\times 10^6} $$ newtons.

$$ \begin{align*} \red{4.1\times 10^6} & = \frac{1.56\times 10^{30}}{r^2}\\[6pt] (\red{4.1\times 10^6})r^2 & = 1.56\times 10^{30}\\[6pt] r^2 & = \frac{1.56\times 10^{30}}{\red{4.1\times 10^6}}\\[6pt] r^2 & = \frac{1.56}{4.1}\times \frac{10^{30}}{10^6}\\[6pt] & = 0.3805\times 10^{24}\\[6pt] r & = \pm\sqrt{0.3805\times 10^{24}}\\[6pt] & = 0.6169\times 10^{12}\\[6pt] & = 6.169 \times 10^{12} \end{align*} $$

Answer

The distance decreases to $$ 6.169\times 10^{12} $$ meters.

Problem 12

In the study of electricity, Ohm's Law says the electrical current (measured in amps) across a conductor is inversely proportional to electrical resistance (measured in ohms).

Suppose a particular circuit has a variable resistor that is currently set to 360 ohm, and this results in 1/3 amps of current. If the resistance were decreased to 80 ohms, what would the resulting change in current be?

Step 1

Define variables to represent the quantities we are investigating, and then write down the appropriate variation equation.

Let $$ I = $$ electrical current, measured in amps.

Let $$ R = $$ electrical resistance, measured in volts.

The equation of variation is $$ I = \frac k R $$.

Note: This is Ohm's Law, and the constant of proportionality represents the voltage on the circuit.

Step 2

Use $$ \red{I = \frac 1 3} $$ and $$ \blue{R = 360} $$ to determine the value of $$ k $$. Then rewrite the variation equation.

$$ \begin{align*} \red I & = \frac k {\blue R}\\[6pt] \red{\frac 1 3} & = \frac k {\blue{360}}\\[6pt] \red{\frac 1 3}(\blue{360}) & = \frac k {\blue{360}}\cdot \blue{360}\\[6pt] 120 & = \frac k {\cancelred{360}}\cdot \cancelred{360}\\[6pt] k & = 120 \end{align*} $$

The variation equation is now $$ I = \frac{120} R $$.

Step 3

Use the equation to determine the current running through the circuit when the resistance drops to only $$ \blue{R = 80} $$ ohms.

$$ I = \frac{120}{\blue{80}} = \frac 3 2 $$

Answer

When the resistance decreases to 80 ohms, the current increases to 1.5 amps.

Problem 13

In economics, the basic Law of Demand tells us that as the price for a particular good (or service) increases, the demand for that good (or service) will decrease. This is an inverse variation relationship.

Suppose a new app is released for cell phones, and at a price of $$ \$4.99 $$ there are 3.2 million downloads each month. Later, the company that produced the app puts it on sale for $$ \$0.99 $$

(a) What will the accompanying change in the number of downloads per month be?

(b) If the wanted to phase out the app (for a newer, better one in development), and wanted to decrease the number of downloads per month to only $$ 500{,}000 $$, what price should they charge for the app?

Step 1

Define variables for the quantities we are investigating, then write down the appropriate variation equation.

Let $$ p = $$ the price, in dollars, for which the app is being sold.

Let $$ d = $$ the number of downloads per month, measured in millions of downloads per month.

So, $$ d = 1 $$ represents 1 million downloads per month.

The variation equation is $$ d = \frac k p $$.

Step 2

Use $$ \red{d = 3.2} $$ and $$ \blue{4.99} $$ to determine the value of $$ k $$, then write down the updated variation equation.

$$ \begin{align*} \red d & = \frac k {\blue p}\\[6pt] \red{3.2} & = \frac k {\blue{4.99}}\\[6pt] \red{3.2}(\blue{4.99}) & = \frac k {\blue{4.99}}\cdot \blue{4.99}\\[6pt] 15.968 & = \frac k {\cancelred{4.99}}\cdot \cancelred{4.99}\\[6pt] k & = 15.968 \end{align*} $$

The variation equation becomes $$ d = \frac{15.968} p $$.

Step 2 Part (a)

Use the variation equation to determine the number of downloads per month the company can expect from putting the app on sale for $$ \blue{p = \$0.99} $$.

$$ d = \frac{15.968}{\blue{0.99}} = \frac{1596.8}{99} \approx 16.13 $$

The downloads should increase to approximately 16.13 million per month.

Step 2 Part (b)

Use the variation equation to determine the price which will cause the number of downloads per month to decrease to only 5 hundred thousand ($$ \red{d = 1/2} $$).

$$ \begin{align*} \red{\frac 1 2} & = \frac{15.968} p\\[6pt] \frac p 2 & = 15.968\\[6pt] p & = 31.936 \end{align*} $$

The price would need to be about $$ \$31.94 $$ to drive down the demand to only half a million.

Answer

(a) Demand will increase to about 16.13 million downloads per month.

(b) Increasing the price to about $$ \$31.94 $$ will drive down demand to $$ 500{,}000 $$ downloads per month.

Problem 14

When working with electrical circuits, it turns out that the electrical resistance is inversely proportional to the square of the current.

Suppose a particular circuit has a current measured at $$ 0.05 $$ amps with a resistance of 90 ohms.

(a) What will happen to the resistance if the current drops to only 0.01 amps?

(b) Suppose someone swaps out the 90-ohm resistor for one that is rated at only 60 ohms. What happens to the current?

Step 1

Define variables for the quantities, then write down the variation equation.

Let $$ R = $$ the electrical resistance measured in ohms.

Let $$ I = $$ the electrical current measured in amps.

The variation equation is $$ R = \frac k {I^2} $$

Note: In a previous question about electrical circuits we mentioned the constant of proportionality was in fact the voltage on the circuit. This time, the constant of proportionality represents the electrical power (in watts).

Step 2

Use $$ \red{R = 90} $$ and $$ \blue{I = 0.05} $$ to determine the value of $$ k $$, then write down the updated variation equation.

$$ \begin{align*} \red R & = \frac k {\blue I^2}\\[6pt] \red{90} & = \frac k {(\blue{0.05})^2}\\[6pt] \red{90}\cdot(\blue{0.0025}) & = \frac k {\blue{0.0025}}\cdot (\blue{0.0025})\\[6pt] 0.225 & = \frac k {\cancelred{0.0025}}\cdot (\cancelred{0.0025})\\[6pt] k & = 0.225 \end{align*} $$

The variation equation is now $$ \displaystyle R = \frac{0.225}{I^2} = \frac 9 {40I^2} $$

Step 3

Determine what happens to the current if the resistance is changed to $$ \red{R = 60} $$ ohms.

$$ \begin{align*} \red{60} & = \frac 9 {40I^2}\\[6pt] \red{60}I^2 & = \frac 9 {40}\\[6pt] I^2 & = \frac 9 {2400}\\[6pt] I & = \sqrt{\frac 9 {2400}}\\[6pt] & = \frac 3 {20\sqrt 6}\\[6pt] & \approx 0.0612 \end{align*} $$

Answer

The current increases to $$ \frac 3 {20\sqrt 6} $$ or approximately 0.0612 amps.

Problem 15

Suppose an object is falling toward the earth, and it's speed is inversely proportional to the square root of the distance from the earth's surface.

Also, suppose the object's speed is 60 meters/second when it is 900 meters above the earth's surface. How close will the object be when it's velocity reaches 100 meters/second?

Step 1

Define variables for the quantities we are interested in, then write down the variation equation.

$$ v = $$ velocity, in meters/second

$$ d = $$ distance from the earth's surface, in meters.

The variation equation is $$ \displaystyle v = \frac k {\sqrt d} $$.

Step 2

Use $$ \red{v = 60} $$ and $$ \blue{d =900} $$ to determine the value of $$ k $$, then write down the updated variation equation.

$$ \begin{align*} \red v & = \frac k {\sqrt{\blue d}}\\[6pt] \red{60} & = \frac k {\sqrt{\blue{900}}}\\[6pt] \red{60}(\blue{30}) & = \frac k {\blue{30}}\cdot\blue{30}\\[6pt] 1800 & = \frac k {\cancelred{30}}\cdot\cancelred{30}\\[6pt] k & = 1800 \end{align*} $$

The variation equation is now $$ v = \frac{1800}{\sqrt d} $$

Step 3

Use the equation to determine the distance of the object from the earth's surface when it's velocity reaches $$ \red{v = 100} $$ meters per second.

$$ \begin{align*} \red{100} & = \frac{1800}{\sqrt d}\\[6pt] \sqrt d & = \frac{1800}{\red{100}} = 18\\[6pt] d & = 18^2 = 324 \end{align*} $$

Answer

The object is only 324 meters above the earth's surface when it's velocity reaches 100 meters per second.