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Re: The figure above shows the inside dimensions in feet of a certain room
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04 Aug 2017, 12:34
I've found these conversion problems particularly confusing, especially because there isn't a well-defined process for handling conversions (at least that I've seen).
Here's a process I've found useful for navigating conversions
Create a conversion ratio:
\(1m : 100cm\)
Raise both sides to the power of the operation (for volume operations, the power is three, for surface are operations, the power is 2, for regular ratios, the power is 1)
\((1m)^3:(100cm)^3 \rightarrow 1m^3:1,000,000cm^3\)
So something with a volume of \(15m^3\) is equivalently \(15,000,000cm^3\) in volume
This process can be reversed also. Say, for example, you know the ratio of area for two squares is \(1in^2:16in^2\). This represents a squared amount, so the ratio of these two square's sides is the square root of the ratio of area:
\(\sqrt{1in^2}:\sqrt{16in^2} \rightarrow 1in:4in\)
This process is fairly intuitive with base-10 metric conversions, but works the same with questions like this one.
Two \(8x10\) walls:
\(2*8*10 = 160ft^2\)
Two \(8x16\) walls:
\(2*8*16=256ft^2\)
One \(10x16\) wall:
\(10*16=160ft^2\)
So there are \(160+256+160=576ft^2\) of surface area for the actual room, and \(2304in^2\) for the model
\(576ft^2:2304in^2 \rightarrow 1ft^2:4in^2\)
This is a surface area (second dimension) problem with a ratio comparing already squared units, so to find the ratio of the un-squared units (first dimension), take the square root of the ratio:
\(\sqrt{1ft^2}:\sqrt{4in^2} \rightarrow 1ft:2in\)
So for every two inches in the model there is one foot in real life.
Answer D