Question

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?

Answer 1

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 = $\frac{\frac{1}{4} }{\frac{5}{8} }$ = $\frac{2}{5}$