Inverse Variation 2

Question 1 of 4

A supermarket determines that the demand for its chip varies inversely with the price of each packet. When the price per packet is

$ 2.60

$$2.60$ , the weekly demand is

420

$420$ packets.

(i)

$(i)$ Write the inverse variation equation that relates price “

p

$p$ ” and demand “

d

$d$ ”.

(i i)

$(ii)$ Find the weekly demand for chips when the price per packet drops to

$ 1.75

$$1.75$ .

Write fractions in the format “a/b”

$(i)$ $(i)$ Equation: $d =$ $d=$ (1092/p)

$(i i)$ $(ii)$ Weekly demand when price per packet is $$ 1.75$ $$1.75$ : (624) packets

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Inverse Variation Formula

y = \frac{k}{x}

$\color{#9a00c7}{y}=\frac{k}{\color{#004ec4}{x}}$

where

k \neq 0

$k≠0$ and is the constant of variation

Remember

An inverse variation is a relationship between two variables where if one decreases, the other increases. Similarly, if one variable increases, the other decreases.

First, solve for

k

$k$ , the constant of variation, by plugging in the known values to the Inverse Variation Formula.

$p$ $p$	$=$ $=$	$2.60$ $2.60$	$x$ $x$ variable
$d$ $d$	$=$ $=$	$420$ $420$	$y$ $y$ variable

$y$ $y$	$=$ $=$	$\frac{k}{\color{#004ec4}{x}}$	Inverse Variation Formula

$d$ $d$	$=$ $=$	$\frac{k}{\color{#004ec4}{p}}$	Variables replaced with $p$ $p$ and $d$ $d$

$420$ $420$	$=$ $=$	$\frac{k}{\color{#004ec4}{2.60}}$	Substitute values

$420$ $420$ $\times 2.60$ $times2.60$	$=$ $=$	$\frac{k}{2.60}$ $k/2.60$ $\times 2.60$ $times2.60$	Multiply $2.60$ $2.60$ to both sides

$1092$ $1092$	$=$ $=$	$k$ $k$
$k$ $k$	$=$ $=$	$1092$ $1092$

Next, rewrite the Inverse Variation Formula with

$k$ ,

$p$ and

$d$ substituted.

$y$	$=$	$k/x$

$d$	$=$	$1092/p$	Substitute $k$ , $d$ and $p$

Finally, use the new formula and substitute

$p=1.75$

$d$	$=$	$1092/p$	New formula

$d$	$=$	$1092/1.75$	Substitute $p=1.75$

$d$	$=$	$624$

$(i)$ Equation:

$d=1092/p$

$(ii)$ Weekly demand when price per packet is

$$1.75$ :

$624$ packets

Question 2 of 4

The cost per passenger of chartering a Learjet is inversely proportional to the number of passengers on the jet. If there are

$2$ passengers, the cost per passenger is

$$2100$ each. What is the cost per passenger when there are

$8$ passengers?

$$$ (525)

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Inverse Variation Formula

$\color{#9a00c7}{y}=\frac{k}{\color{#004ec4}{x}}$

where

$k≠0$ and is the constant of variation

Remember

An inverse variation is a relationship between two variables where if one decreases, the other increases. Similarly, if one variable increases, the other decreases.

First, solve for

$k$ , the constant of variation, by plugging in the known values to the Inverse Variation Formula.

$p$	$=$	$2$	No. of passengers
$c$	$=$	$$2100$	Cost per passenger

$c$	$=$	$\frac{k}{\color{#004ec4}{p}}$	Inverse Variation Formula

$2100$	$=$	$\frac{k}{\color{#004ec4}{2}}$	Substitute known values

$2100$ $times2$	$=$	$k/2$ $times2$	Multiply $2$ to both sides

$4200$	$=$	$k$
$k$	$=$	$4200$

Next, rewrite the Inverse Variation Formula with

$k$ substituted.

$c$	$=$	$k/p$

$c$	$=$	$4200/p$	Substitute $k$

Finally, use the new formula and substitute

$p=8$

$c$	$=$	$4200/p$	New formula

$c$	$=$	$4200/8$	Substitute $p=8$

$c$	$=$	$525$

Hence, the cost per passenger when there are

$8$ passengers on the Learjet is $$525$

$$525$

Question 3 of 4

When

$2$ blocks are balanced on a lever, their distances from the fulcrum are inversely proportional to their weights. If the

$3$ kg block is placed

$2.8$ metres from the pivot, how far should a

$5$ kg block be placed from the pivot in order to balance the two blocks?

Write answer in decimal form

(1.68) $m$

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Inverse Variation Formula

$\color{#9a00c7}{y}=\frac{k}{\color{#004ec4}{x}}$

where

$k≠0$ and is the constant of variation

Remember

An inverse variation is a relationship between two variables where if one decreases, the other increases. Similarly, if one variable increases, the other decreases.

First, solve for

$k$ , the constant of variation, by plugging in the known values to the Inverse Variation Formula.

$D$	$=$	$2.8$ m	Distance of the block to the pivot
$w$	$=$	$3$ kg	Weight of the block

$D$	$=$	$\frac{k}{\color{#004ec4}{w}}$	Inverse Variation Formula

$2.8$	$=$	$\frac{k}{\color{#004ec4}{3}}$	Substitute known values

$2.8$ $times3$	$=$	$k/3$ $times3$	Multiply $3$ to both sides

$8.4$	$=$	$k$
$k$	$=$	$8.4$

Next, rewrite the Inverse Variation Formula with

$k$ substituted.

$D$	$=$	$k/w$

$D$	$=$	$8.4/w$	Substitute $k$

Finally, use the new formula and substitute

$w=5$

$D$	$=$	$8.4/w$	New formula

$D$	$=$	$8.4/5$	Substitute $w=5$

$D$	$=$	$1.68$

Hence, the

$5$ kg block should be placed $1.68$

$m$ from the pivot in order to balance the two blocks.

$1.68$

$m$

Question 4 of 4

Two cats are in a room with a light bulb. The first cat, which is

$5m$ away from the light bulb, experiences a light intensity of

$90$ Lumens. Given that the light intensity varies inversely to the square of the distance, find the light intensity experienced by the second cat if it is

$10m$ away from the light bulb.

Use

$1$ decimal place

(22.5) Lumens

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Inverse Variation Formula

$\color{#9a00c7}{y}=\frac{k}{\color{#004ec4}{x}}$

where

$k≠0$ and is the constant of variation

Remember

An inverse variation is a relationship between two variables where if one decreases, the other increases. Similarly, if one variable increases, the other decreases.

First, solve for

$k$ , the constant of variation, by plugging in the known values to the Inverse Variation Formula.

$L$	$=$	$90$ Lumens	Light Intensity
$D$	$=$	$5$ m	Distance from the light bulb

$L$	$=$	$\frac{k}{\color{#004ec4}{D}^2}$	$L$ varies inversely to the square of $D$

$90$	$=$	$\frac{k}{\color{#004ec4}{5}^2}$	Substitute known values

$90$	$=$	$\frac{k}{25}$

$90$ $times25$	$=$	$k/25$ $times25$	Multiply $25$ to both sides

$2250$	$=$	$k$
$k$	$=$	$2250$

Next, rewrite the Inverse Variation Formula with

$k$ substituted.

$L$	$=$	$k/(D^2)$

$L$	$=$	$2250/(D^2)$	Substitute $k$

Finally, use the new formula and substitute

$D=10$

$L$	$=$	$2250/(D^2)$	New formula

$L$	$=$	$2250/(10^2)$	Substitute $D=10$

$L$	$=$	$2250/100$

$L$	$=$	$22.5$ Lumens

Hence, the light intensity experienced by the second cat is

$22.5$ Lumens

Inverse Variation 2

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