Algebra Word Problem (Details below)

Question

Fergus served his infant a 20 ounce bowl of cream of wheat, and then left the kitchen to check on his daugther. The infant, who hated hot cereal, immediately began emptying his bowl by throwing 1/2 ounce spoonful of the cereal at the wall. If he had thrown one more spoonful per minute, and if his dad had stayed out out of the kitchen an extra two minutes, he would have just succeeded in disposing of his breakfast. As it turned out he had 8 ounces of Cream of Wheat left when Fergus returned. How long was Fergus out of the kitchen?

Answer 1

Finally, an Algebra problem with a very believable real-world context! To solve this, let:

r = Rate at which cereal is removed (ounces per minute)

t = Time that Fergus was out of the kitchen (minutes)

The amount emptied while Fergus is out of the kitchen is 20 - 8 = 12 ounces at a rate of r ounces per minute, for t minutes. Therefore:

rt = 12

If he had thrown one more spoonful (1/2 ounces) per minute, then it would have taken (t+2) minutes at a rate of (r + 1/2) ounces per minute to empty the total of 20 ounces, thus:

(r + 1/2)(t + 2) = 20

It's useful to multiply this by 2 to clear the fraction, so we need to solve the simultaneous system:

rt = 12

(2r + 1)(t + 2) = 40

There are several way to do so. From the first equation we have r = 12/t, and substituting gives:

(24/t + 1)(t + 2) = 40

FOIL the left-hand side, simplify, then multiply by t:

24 + 48/t + t + 2 = 40

48/t + t - 14 = 0

48 + t^2 - 14t = 0

This can be solved by factoring or the Quadratic Formula. There are two solutions:

t = 6, which gives r = 12/6 = 2

t = 8, which gives r = 12/8 = 3/2

Both of these are valid solutions; in either case the infant threw 24 spoonfuls (12 ounces total) during the time Fergus was gone. If his daughter needed more attention, go with t = 8 minutes. If the infant really hated hot cereal, go with the higher rate of throwing (thus t = 6 minutes). But we'll never really know for sure....