Pipes and Cisterns Quiz Set 020

Let the time be x mins. Then sum of works done by X, Y and Z = 1. $x/5 + x/17 + x/4 = 1$. Solving, we get x = $1{167/173}$. Or use the shortcut ${abc}/{ab + bc + ca}$. Another thing, instead of solving the entire calculation, you can keep an eye on the options to find the nearest answer.

Work done by the leakage in 1 min is $1/5 + 1/8 - 1/15$ = ${31/120}$. This work is equivalent to a volume of 8 liters. So, the total volume is 8 × ${120/31}$ = ${960/31}$, which is same as: $30{30/31}$ liters.

Let the time taken by tap A be x mins. Then 19 minutes work of all the taps should add to 1. So we have, $19 × 1/x + 19 × 2/x + 19 × 4/x$ = 1, which is same as $19 × 7/x$ = 1. Solving, we get x = 133 mins.

Net part filling in one hour is $1/x - 1/y$ = $(y - x)/(xy)$. So complete filling occurs in $(xy)/(y - x)$ = ${6 × 15}/{15 - 6}$ = 10 mins.

Let the time taken by the slower pipe alone be x. Then 7 × $(1/x + 4/x)$ = 1. Solving for x, we get x = 7 × 5 = 35 mins.