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?
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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):
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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
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.
