Question

standard: 6.2.1 solve addition and subtraction problems with fractions related to small engines
content objective: add and subtract fractions in oil measurement contexts
language objective: explain steps using fraction vocabulary (numerator, denominator, simplify, sum, difference)
essential question: what type of fraction problems are involved in small engine repair?

practice problems

1. a mechanic pours 1/4 quart of oil in the morning and 1/2 quart in the afternoon. how much oil was added in total?
2. a bottle holds 3/4 quart of oil. if the mechanic uses 1/2 quart, how much oil is left?
3. a mechanic uses 2/3 quart of fuel for one job and 1/6 quart for another. how much fuel was used in total?
4. a tank has 5/6 gallon of oil. after a repair, 2/3 gallon is left. how much oil was used?
5. a mechanic adds 2/5 gallon of oil to a mix in the morning and 1/10 gallon in the afternoon. how much oil was added altogether?

Explanation:

Step1: Find common denominator for problem 1

The common - denominator of 4 and 2 is 4. $\frac{1}{2}=\frac{2}{4}$. Then $\frac{1}{4}+\frac{2}{4}=\frac{1 + 2}{4}=\frac{3}{4}$ quart.

Step2: Find common denominator for problem 2

The common - denominator of 4 and 2 is 4. $\frac{1}{2}=\frac{2}{4}$. Then $\frac{3}{4}-\frac{2}{4}=\frac{3 - 2}{4}=\frac{1}{4}$ quart.

Step3: Find common denominator for problem 3

The common - denominator of 3 and 6 is 6. $\frac{2}{3}=\frac{4}{6}$. Then $\frac{4}{6}+\frac{1}{6}=\frac{4 + 1}{6}=\frac{5}{6}$ quart.

Step4: Find common denominator for problem 4

The common - denominator of 6 and 3 is 6. $\frac{2}{3}=\frac{4}{6}$. Then $\frac{5}{6}-\frac{4}{6}=\frac{5 - 4}{6}=\frac{1}{6}$ gallon.

Step5: Find common denominator for problem 5

The common - denominator of 5 and 10 is 10. $\frac{2}{5}=\frac{4}{10}$. Then $\frac{4}{10}+\frac{1}{10}=\frac{4+1}{10}=\frac{5}{10}=\frac{1}{2}$ gallon.

Answer:

1. $\frac{3}{4}$ quart
2. $\frac{1}{4}$ quart
3. $\frac{5}{6}$ quart
4. $\frac{1}{6}$ gallon
5. $\frac{1}{2}$ gallon