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stepladder [879]
2 years ago
9

Marcellus has a pitcher that contained 5/8 gallon limeade. Marcellus drank 1/4 gallon of the limeade, and Marcy drank 1/8 gallon

. What fractional part of a gallon of limeade remained in the pitcher?
Mathematics
1 answer:
mash [69]2 years ago
8 0

Answer:

2/5

Step-by-step explanation:

Given Marcellus drank 1/4 gallon of the limeade and Marcy drank 1/8 gallon of the limeade

Given that only 5/8 portion of the pitcher contains limeade

Total amount drunk = amount drunk by Marcellus + amount drunk by Marcy

Total amount drunk = \frac{1}{4} +\frac{1}{8} = \frac{3}{8}

The amount of Limeade remained = The amount initially - the amount drunk

The amount remained =  \frac{5}{8} -\frac{3}{8} = \frac{1}{4}

Fractional part = amount remained / amount initially = \frac{\frac{1}{4} }{\frac{5}{8} } = \frac{2}{5}

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