Inverse Variation:
Practice Problems

$$ y = \frac k x $$

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

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

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

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

$$ s = \frac k t $$

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

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

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

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

$$ y = \frac k x $$

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

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

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

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

$$ I = \frac k R $$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The inverse variation equation will be

$$ t = \frac k T $$

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

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

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

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

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.

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

$$ t = \frac k w $$

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

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

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

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).

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 $$ F_g = \frac k {r^2} $$.

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

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

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

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

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.

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

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

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

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

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 $$ d = \frac k p $$.

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

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

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

The downloads should increase to approximately 16.13 million per month.

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

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

(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.

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).

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

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

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

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

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

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

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

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

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

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

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