When you're dealing with ratios, scale factors can be a very useful tool. Here's an explanation of how scale factors work!
Ratio problems often give us ratios and one or more absolute amounts, and we are asked to find all (or some) of the absolute amounts that are not given. A ratio doesn't tell us about the absolute amounts, but we can use a scale factor to "convert" a ratio to absolute amounts.
Don't worry if that sounds a bit confusing! We'll use an example to explain. This will make things much clearer!
Let's take this problem.
The ratio of boys to girls is 3/7. Great, so I know that for every 3 boys, there are 7 girls. But that doesn't tell me the actual amounts of boys and girls.
But I make a scale factor "n" and say the number of boys is 3n and the number of girls is 7n. Notice I have preserved the ratio. I just tagged an "n" onto the end of the numbers in the ratio. But now, these are actual amounts. While 3 and 7 were just part of the ratio of boys to girls, 3n is the actual number of boys and 7n is the actual number of girls.
You may say: So what? We don't know what n is!
Right! But these actual amounts 3n and 7n may be useful in setting up an equation.
And if we can set up an equation and solve for n, then we know the exact number of boys and girls!
Let's say that the problem tells us that there are 32 more girls than boys. This piece of information will allow us to find the exact value of n - and thus find the actual number of boys and the actual number of girls.
Girls = 7n
Boys = 3n
32 = Girls - Boys
32 = 7n - 3n
32 = 4n
n = 8
That's the scale factor (n = 8). And now we know that the actual number of girls is 7n = 7(8) = 56, and the actual number of boys is 3n = 3(8) = 24.
We call it a "scale factor" because it relates the ratio to the actual numbers. A scale factor isn't always the easiest approach, but it's often a good way to work with ratio problems.
Let's do another example problem. We'll do the last practice problem in the "Combining Ratios" lesson video (here's the GRE version, and here's the GMAT version).
The problem gives us a few pieces of information:
- 1 cup of butter goes into 12 cookies
- 1 cup of sugar goes into 8 cookies
- Kathy uses five more cups of sugar than butter. We are assuming that Kathy made as many cookies as possible with the sugar and butter she used.
The problem shows us:
- The butter/cookies ratio is 1:12
- The sugar/cookies ratio is 1:8
Let's express 1:12 as 2:24 and 1:8 as 3:24. I'm not changing the ratios, just turning them into equivalent ratio so I have 24 on the right of both. Now, we can say we have a ratio of:
2 cups butter : 3 cups sugar : 24 cookies
This is 3-way ratio, so I can't really convert it into a fraction. It's telling me the ratio that three parts are to each other.
How many cookies do we have? How much butter? How much sugar? We don't know yet. But we know they must be in that ratio.
Let's use a "scale factor" of n and solve for n:
2n : 3n : 24n
I'm not changing anything here, just multiplying everything in the ratio by n.
2n, 3n, and 24n represent the actual amounts of butter and sugar and cookies. I just need to figure out what n is, and then I can figure out how much I have of each.
The point of n is that I couldn't treat 2 : 3 : 24 as absolute amounts...2:3:24 is just the ratio. But by multiplying by n, now I can say that 2n, 3n, and 24n are the actual amounts.
So now let's set up an equation to solve for n and thereby figure out the exact amounts of everything:
I know the sugar is 5 more than the butter.
As we said above, sugar = 3n, butter is 2n
So 3n - 2n = 5
n = 5
Now I found n, I can multiply everything by the scale factor (n = 5).
butter : sugar : cookies
2n : 3n : 24n
2(5) : 3(5) : 24(5)
10 : 15 : 120
So we have 10 cups of butter, 15 cups of sugar, and 120 cookies.
That's scale factors for you. Now it's time for dessert! :D
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