Page 90 - ArithBook5thEd ~ BCC

Solution. The numerator of the first fraction (3) has a common factor with thedenominator of the
third fraction (9), so the product is equal to

✚ 3 ✚❃ 1 8 10 1 8 10
· · = · · .
4 5 ✚❃ 3 4 5 3
✚ 9
Continuing on, the numerator of the second fraction (8) has a common factor with the denominator of
the first fraction (4), so the product is equal to

1 ✚ 8 ✚❃ 2 10 2 10
· · =1 · · .
1
✚ 4 ✚❃ 5 3 5 3
Finally, the numerator of the third fraction (10) has a common factor with the denominator of the
second fraction (5), so (omitting the factor 1) the product is equal to

✚❃
2 ✚ 10 2 2 2
· = · .
1
✚ 5 ✚❃ 3 1 3
No further cancellation is possible. The final answer is now a simple product
2 2 4 1
· = =1 ,
1 3 3 3
which is already in lowest terms. These cancellations could have been done in a different order, or
(carefully) all at once.

Mixed numbers are multiplied by simply converting them into improper fractions.
3
2
1
Example 99. Find the product 2 · 1 · 2 . Express the result as a mixed number.
8 4 3
Solution. Rewriting each mixed number as an improper fraction, we have the product
19 5 8
· · .
8 4 3
Cancelling 8’s, we have
19 · 5 85 1
= =7 .
4 · 3 12 12

1
Example 100. Agas tank with a 13 -gallon capacity is only one third full. How much gas is in the
2
tank?
1
1
Solution. We need to find the product 1 · 13 . Converting 13 to the improper fraction 27 ,we have
2
2
3
2
1 1 1 27
· 13 = · .
3 2 3 2
Cancelling the common factor 3,
✚❃
1 ✚ 27 9 9
1 · 2 = 2 .
✁ 3 ✁✕
1
Converting 9 2 to a mixed number, we conclude that the tank contains 4 gallons of gas.
2
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