Cat and Mouse VariationReference > Mathematics > Algebra
Over on the Puzzler Forum (click Puzzler Forum link at the bottom of this lesson), I posed the following riddle:
If a cat and a half can catch a mouse and a half in a day and a half, how long will it take 6 cats to catch 12 mice?
This is a fun puzzle, and it is readily solved using the concept of direct variation and inverse variation.
The number of cats directly varies with the number of mice (in other words, if you double the number of mice, you double the number of cats and if you triple the number of mice, you triple the number of cats).
Also, the number of mice directly varies with the number of days (more mice requires more days).
Finally, the number of cats varies inversely with the number of days. If you have twice as many cats, it takes half as long to catch the mice.
So we solve this problem by holding one quantity constant while varying the other two.
We need to multiply the number of cats by 4, because there are six cats instead of 1.5. If we quadruple the number of cats, we quadruple the number of mice, so:
6 cats catch 6 mice in a day and a half.
Now we keep the number of cats constant and double the number of mice (because there are 12 instead of 6), which also doubles the number of days:
6 cats catch 12 mice in 3 days.
Now see if you can do it...