Math word problem

Question

Yellows vary inversely as reds cubed. When there are 16 yellows, there are 4 reds. How many yellows are present when there are 8 reds?

Answer 1

In order to handle a variation problem, the first job is to translate the sentence into an equation, and the words tell you exactly how to do it.

"Varies inversely" means the variable goes on the BOTTOM of a fraction with a constant on top. "Reds cubed" tells you the power. So if y is yellows and r is reds:

y = k/r^3

That k is called the constant of variation, and finding it is always the first real step, because until we know k we cannot answer anything.

We are told that when there are 16 yellows there are 4 reds, so we plug those in:

16 = k/4^3

Since 4^3 = 64:

16 = k/64

Multiplying both sides by 64:

k = 1024

Now we have the full relationship:

y = 1024/r^3

Remember, our goal is the number of yellows when there are 8 reds, so we substitute r = 8:

y = 1024/8^3 = 1024/512 = 2

So, our final answer is 2 yellows.

It is worth pausing on how dramatic that drop is. The reds only doubled, from 4 to 8, but the yellows fell from 16 all the way down to 2. That is what cubing in the denominator does. Doubling r makes r^3 eight times larger, so y ends up eight times smaller, and 16 divided by 8 is 2. Once you see that pattern you can often predict the answer before doing the arithmetic.

Answer 2

Let y be the yellows and x be the reds.

y = k/x³

When y=16 and 4 reds: 16=k/4³ --> k = 16(4)³

y = 16(4/x)³

When x = 8, y = 16(4/8)³=16(1/2)³=16/8= 2

There are 2 yellows.