﻿ Inverse Variation Formula. Practice Problems

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# Inverse Variation:Practice Problems

##### Problem 1
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*}

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

##### Problem 2
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*}

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

##### Problem 3
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*}

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

##### Problem 4
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*}

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

##### Problem 5
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*}

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

##### Problem 6
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*}

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

### Word Problems

##### Problem 7
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*}

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
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 $$\blue{V = 0.1}$$.

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

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

##### Problem 9
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*}

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

##### Problem 10
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

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

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
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*}

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

##### Problem 12
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.

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

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

##### Problem 13
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 $$\blue{p = \0.99}$$.

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

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.

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

##### Problem 14
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*}

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

##### Problem 15
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*}