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?

Accepted Solution

Answer:2/5Step-by-step explanation:Given Marcellus drank 1/4 gallon of the limeade and Marcy drank 1/8 gallon of the limeadeGiven that only 5/8 portion of the pitcher contains limeadeTotal amount drunk = amount drunk by Marcellus + amount drunk by MarcyTotal amount drunk = [tex]\frac{1}{4} +\frac{1}{8}[/tex] = [tex]\frac{3}{8}[/tex]The amount of Limeade remained = The amount initially - the amount drunkThe amount remained = Β [tex]\frac{5}{8} -\frac{3}{8}[/tex] = [tex]\frac{1}{4}[/tex]Fractional part = amount remained / amount initially = [tex]\frac{\frac{1}{4} }{\frac{5}{8} }[/tex] = [tex]\frac{2}{5}[/tex]