Question

question 16 of 23 - multiple choice

a contractor had similar work to perform on each of 4 different jobsites. he uses gas-powered generators to provide power for his crews to work. when the jobs were done, he noted that the first jobsite used \\(142\frac{1}{4}\\) gallons of gas to power the generator; the second jobsite used 138-2/3 gallons; the third jobsite used \\(146\frac{1}{2}\\) gallons; and the 4th jobsite used \\(140\frac{3}{4}\\) gallons. what is the total amount of gas used?

a 566-3/16

b 568-3/16

c 566-1/6

d 568-1/6

Explanation:

Identify the given values and operation

We need to find the total amount of gas used across 4 jobsites. This requires adding the mixed numbers representing the gas used at each site.
Using the Fraction Word Problems knowledge point, we set up the sum:

$$ \text{Total} = 142\frac{1}{4} + 138\frac{2}{3} + 146\frac{1}{2} + 140\frac{3}{4} $$

Add the whole numbers

Using the Fraction Arithmetic knowledge point, we first sum the whole number parts:

$$ 142 + 138 + 146 + 140 = 566 $$

Add the fractional parts

Next, we find a common denominator for the fractions: \(\frac{1}{4}\), \(\frac{2}{3}\), \(\frac{1}{2}\), and \(\frac{3}{4}\). The least common multiple of 4, 3, and 2 is 12.
Convert each fraction:

$$ \frac{1}{4} = \frac{3}{12} $$

$$ \frac{2}{3} = \frac{8}{12} $$

$$ \frac{1}{2} = \frac{6}{12} $$

$$ \frac{3}{4} = \frac{9}{12} $$

Now, sum these converted fractions:

$$ \frac{3}{12} + \frac{8}{12} + \frac{6}{12} + \frac{9}{12} = \frac{26}{12} $$

Simplify the fractional sum

Convert the improper fraction \(\frac{26}{12}\) into a mixed number:

$$ \frac{26}{12} = 2\frac{2}{12} = 2\frac{1}{6} $$

Combine whole numbers and fractions

Add the sum of the whole numbers and the simplified fractional sum:

$$ 566 + 2\frac{1}{6} = 568\frac{1}{6} $$

This matches option D.

Note on Option B (\(568\frac{3}{16}\)): A common mistake leading to Option B is adding the numerators directly and adding the denominators directly without finding a common denominator:

$$ \frac{1+2+1+3}{4+3+2+4} = \frac{7}{13} \approx \frac{3}{16} $$

Alternatively, if a student incorrectly adds the numerators \(1 + 2 + 1 + 3 = 7\) and incorrectly adds the denominators \(4 + 3 + 2 + 4 = 13\), or makes a similar error in combining the fractional parts, they might arrive at a fraction near \(\frac{3}{16}\) while correctly carrying over the \(2\) to get \(568\).

Answer:

• (A) \(566\text{-}3/16\)
• (B) \(568\text{-}3/16\)
• (C) \(566\text{-}1/6\)
• (D) \(568\text{-}1/6\) (Correct answer)