EMAT 6600

Problem Solving

## Fractions

### Problems

A paint problem: To make a shade of orange paint that you
like, you must mix 2/3 of a bottle of red paint with each 4/5 of a bottle of
yellow paint that you use. You need 88 bottles of this orange paint. How many
bottles of red paint will you need and how many bottles of yellow paint will
you need? (All bottles are the same size.)

A problem about geese: A flock of geese on a pond were being
observed continuously. At 1:00 P.M., 1/5 of the geese flew away. At 2:00 P.M.,
1/8 of the geese that remained flew away. At 3:00 P.M., 3 times as many geese
as had flown away at 1:00 P.M. flew away, leaving 28 geese on the pond. At no
other time did any geese arrive or fly away. How many geese were in the
original flock?

### Solution 1 (ERROR)

When I first encounter this problem, I tried to do it
algebraically. It turned out pure
arithmetic is simpler.

First, since all the bottles are the same size, get the
fractions to have a common denominator.
2/3 and 4/5 are congruent to 10/15 and 12/15 respectively; the sum of
the two fractions makes just 1 bottle of orange (no, 22/15 bottle of orange). To make 88 bottle of orange, multiply both fractions by 88. We find that we would need 58 and 2/3
bottles of red paint and 70 and 2/5 bottles of yellow paint (We need to divide 88 by 22/15 to determine we need 60 combinations. Therefore 40 red and 48 yellow)

### Solution 2 (ERROR)

With this problem, I find it best to work backward in order
the find how many geese were in the original flock. Since there are 28 geese left and 28 represent 2/5 of the amount
that of geese that was there, we need to find the amount for the 3/5 that flew
away. We find that it is 42 geese is the 3/5.
Before 3:00 pm, there were 70 geese.
Similarly, 70 geese represent 7/8, and 1/8 is 10 of the geese. Thus, there were 80 geese before 2:00
pm. Carrying out the same procedure, 80
geese represent 4/5, so 1/5 is 20 geese.
Therefore, there were 100 geese in the original flock. (Answer: 280 geese)

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