Question

directions: solve each problem. show your work and write your answer in a complete sentence. problems 1. a mechanic has 3/4 quart of oil. if each repair uses 1/8 quart, how many repairs can be completed? 2. a hose is 2/3 meter long. if each cut is 1/6 meter, how many pieces can be cut? 3. a gasket strip is 5/6 foot long. if each gasket requires 1/12 foot, how many gaskets can be made? 4. a mechanic has 7/8 gallon of fuel. if each engine uses 1/4 gallon, how many engines can be serviced? 5. a belt is 3/5 meter long. if each section is 1/10 meter, how many sections are there? 6. a container holds 9/10 gallon of solvent. if each job needs 3/20 gallon, how many jobs can be done?

Explanation:

Step1: Recall division of fractions rule

To find the number of parts, divide the total amount by the amount per - part. The rule for dividing fractions is $\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}$.

Step2: Solve problem 1

The total amount of oil is $\frac{3}{4}$ quart and the amount per repair is $\frac{1}{8}$ quart. So, $\frac{3}{4}\div\frac{1}{8}=\frac{3}{4}\times\frac{8}{1}=6$.

Step3: Solve problem 2

The length of the hose is $\frac{2}{3}$ meter and the length per cut is $\frac{1}{6}$ meter. So, $\frac{2}{3}\div\frac{1}{6}=\frac{2}{3}\times\frac{6}{1}=4$.

Step4: Solve problem 3

The length of the gasket - strip is $\frac{5}{6}$ foot and the length per gasket is $\frac{1}{12}$ foot. So, $\frac{5}{6}\div\frac{1}{12}=\frac{5}{6}\times\frac{12}{1}=10$.

Step5: Solve problem 4

The amount of fuel is $\frac{7}{8}$ gallon and the amount per engine is $\frac{1}{4}$ gallon. So, $\frac{7}{8}\div\frac{1}{4}=\frac{7}{8}\times\frac{4}{1}=\frac{7}{2}=3.5$. Since we can't service half an engine, the number of engines is 3.

Step6: Solve problem 5

The length of the belt is $\frac{3}{5}$ meter and the length per section is $\frac{1}{10}$ meter. So, $\frac{3}{5}\div\frac{1}{10}=\frac{3}{5}\times\frac{10}{1}=6$.

Step7: Solve problem 6

The amount of solvent is $\frac{9}{10}$ gallon and the amount per job is $\frac{3}{20}$ gallon. So, $\frac{9}{10}\div\frac{3}{20}=\frac{9}{10}\times\frac{20}{3}=6$.

Answer:

1. The mechanic can complete 6 repairs.
2. 4 pieces can be cut from the hose.
3. 10 gaskets can be made.
4. 3 engines can be serviced.
5. There are 6 sections in the belt.
6. 6 jobs can be done.