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AP ICET 2015 Paper

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Question : 90 of 200

Marks: +1, -0

Two pipes A and B can fill an empty tank independently in 3 hours and 4 hours, respectively. If the pipes A and B are opened in alternate hours starting with B, then the total time taken, in hours, to fill the tank is

$2\frac{1}{2}$
$3\frac{1}{2}$
$3\frac{2}{3}$
$2\frac{3}{4}$

Solution:

Part of tank filled in 2 hours

=\frac{1}{3}+\frac{1}{4}=\frac{17}{12}

Part of tank remaining empty after 2 hours

=\frac{5}{12}

Part of tank filled by B in the 3rd hour

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

Part of tank remaining empty after 3rd hour

=\frac{5}{12}-\frac{3}{12}=\frac{1}{6}

Time taken by A to fill

\frac{1}{6}

part of the tank

=\frac{\frac{1}{6}}{\frac{1}{3}}=\frac{1}{2}

hr

∴ Total time taken

=2+1+\frac{1}{2}=3\frac{1}{2}

hrs

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