SOLUTION: Reds varied inversely as yellows squared. When there were 10 reds, there were 100 yellows. How many reds were present when the yellows were reduced to 5?

Question 1202416: Reds varied inversely as yellows squared. When there were 10 reds, there were 100 yellows. How many reds were present when the yellows were reduced to 5?
Found 2 solutions by mananth, greenestamps:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!

Reds varied inversely as yellows squared. When there were 10 reds, there were 100 yellows. How many reds were present when the yellows were reduced to 5?
r 1/y^2

When there were 10 reds, there were 100 yellows
r = k * 1/y^2
10 = k * 1/(100)^2
k = 10*100^2
r = 10*10000 *1/5^2
r=4000

Answer by greenestamps(13209) (Show Source): You can put this solution on YOUR website!

Formally, inverse variation is defined as where k is a constant.

In practice, it is usually easier to work a problem involving inverse variation using the equivalent form

In this problem, reds (r) vary inversely as yellows squared (y^2). We need to find the number of reds when the number of yellows is 5, given that there are 100 yellows when there are 10 reds:

If your mental arithmetic is good, you can that answer quickly without using the formal formula. The number of yellows is reduced from 100 to 5, a reduction by a factor of 20, so the number of reds is increased by a factor of 20^2=400; the new number of reds is 10*400 = 4000.

ANSWER: 4000

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