Inverse Variation 2
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Question 1 of 4
1. Question
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`, the weekly demand is `420` packets.`(i)` Write the inverse variation equation that relates price “`p`” and demand “`d`”.`(ii)` Find the weekly demand for chips when the price per packet drops to `$1.75`.Write fractions in the format “a/b”
`(i)` Equation: `d=` (1092/p)`(ii)` Weekly demand when price per packet is `$1.75`: (624) packets
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Inverse Variation Formula
$$\color{#9a00c7}{y}=\frac{k}{\color{#004ec4}{x}}$$where `k≠0` and is the constant of variationRemember
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` `=` `2.60` `x` variable `d` `=` `420` `y` variable `y` `=` $$\frac{k}{\color{#004ec4}{x}}$$ Inverse Variation Formula `d` `=` $$\frac{k}{\color{#004ec4}{p}}$$ Variables replaced with `p` and `d` `420` `=` $$\frac{k}{\color{#004ec4}{2.60}}$$ Substitute values `420``times2.60` `=` `k/2.60``times2.60` Multiply `2.60` to both sides `1092` `=` `k` `k` `=` `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
2. Question
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)
Hint
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Nice Job!
Incorrect
Inverse Variation Formula
$$\color{#9a00c7}{y}=\frac{k}{\color{#004ec4}{x}}$$where `k≠0` and is the constant of variationRemember
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
3. Question
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`
Hint
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Well Done!
Incorrect
Inverse Variation Formula
$$\color{#9a00c7}{y}=\frac{k}{\color{#004ec4}{x}}$$where `k≠0` and is the constant of variationRemember
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
4. Question
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
Hint
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Fantastic!
Incorrect
Inverse Variation Formula
$$\color{#9a00c7}{y}=\frac{k}{\color{#004ec4}{x}}$$where `k≠0` and is the constant of variationRemember
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`22.5` Lumens