Ratios & Proportions: Combining Ratios with Different Multipliers - Equate the Common Member
A pilotless cargo carrier brings water, air and dry food to a space station. One-third of the cargo's weight is water and the ratio, by weight, of air to dry food is 2:3. If the cargo contains 1,800 pounds of water, how many pounds of air does it contain?
__Alternative explanation__:
If the water is one-third of the total at $$1{,}800$$ lbs, then the rest (air plus food) is two-thirds of the total, which is $$2\times1{,}800=3{,}600$$ lbs. Build a __Ratio Box__ with only the ratio of air to food to total air plus food:
| | A | : | F | : | A+F |
|------------|----------------|---|----------------|---|----------------|
| 1st ratio | $$\hspace{0.1in} 2$$ | : | $$\hspace{0.1in} 3$$ | : | $$\hspace{0.1in} 5$$ |
| Multiplier | $$\color{green}{\times 720}$$ | | $$\color{green}{\times 720}$$ | | $$\color{green}{\times 720}$$ |
| Real | $$1{,}440$$ | : | $$2{,}160$$ | : | $$3{,}600$$ |
In this table, the multiplier is
>$$\text{Multiplier} = \frac{3{,}600}{5} = \color{green}{720}$$.
Therefore, the weight of air is
>$$\text{Air} = 2 \times \color{green}{720} = 1{,}440 \text{ lbs}$$.
Incorrect.
[[snippet]]Incorrect.
[[snippet]]Incorrect.
[[snippet]]Incorrect.
[[snippet]]Correct.
[[snippet]]
First, insert the ratios
into the __Ratio Box__. Give each ratio a separate column, and add one
spare column for the combined ratios.
| | Water | : | Air | : | Food | : | Total |
|-----------|-------|---|-------|---|-------|---|-------------|
| 1st Ratio | $$\hspace{0.1in} 1$$ | | | | | : | $$\hspace{0.1in} 3$$ |
| 2nd Ratio | | | $$2$$ | : | $$\hspace{0.1in} 3$$ | | |
At first, it may seem that there is no common quantity to the two
ratios. But another glance at the ratio box shows you that the *partial
total of air plus food* is such a common quantity.
The cargo contains only three parts. Thus, in the 1st Ratio, if water takes one part of three, then the other parts, air and food, take the other two parts. In the 2nd Ratio, the total of air and food is just the sum of their ratio units $$2+3=5$$.
Now, equate the ratio units of air and food by expanding the 1st Ratio by {color:dark-green}5{/color} and the 2nd Ratio by {color:red}2{/color}.
| | W | : | A | : | F | : | A+F |
|-----------|----------------------|---|--------------------|---|--------------------|---|-----------------------|
| 1st ratio | $$1$$ | | | | | : | $$2$$ |
| 2nd ratio | | | $$2$$ | : | $$3$$ | : | $$5$$ |
| {color:dark-green}New 1st{/color} | $$\color{green}{5}$$ | | | | | {color:dark-green}:{/color} | $$\color{green}{10}$$ |
| {color:red}New 2nd{/color} | | | $$\color{red}{4}$$ | {color:red}:{/color} | $$\color{red}{6}$$ | {color:red}:{/color} | $$\color{red}{10}$$ |
| {color:dark-blue}Total{/color} | $$\color{blue}{5}$$ | {color:dark-blue}:{/color} | $$\color{blue}{4}$$ | {color:dark-blue}:{/color} | $$\color{blue}{6}$$ | {color:dark-blue}:{/color} | $$\color{blue}{10}$$ |
At this point, you may combine the two ratios and find the one multiplier for the whole __Ratio Box__: the new ratio is {color:dark-blue}5:4:6{/color} of water to air to food.
Find the multiplier by dividing the real value of the water by its number of rate units:
>$$\text{Multiplier} = \frac{1{,}800}{5}=360$$
Then calculate the weight of the air by multiplying its rate units by the multiplier:
>$$\text{Air} = 360\times \color{blue}{4} = 1{,}440 \text{ lbs}$$.
1,200
1,300
1,440
3,000
3,240
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