Question

Three taps $A,B,C$ fill up a tank independently in $10$ hr, $20$ hr, $30$ hr, respectively. Initially, the tank is empty and exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour. What is the minimum number of hours required to fill the tank?

• A. $8$
• B. $9$
• C. $10$
• D. $11$

Answer 1

The correct option is A $8$

Let the volume of the tank be ${}^{\prime }{x}^{\prime }$ units

Given that

Taps $A,B,C$ fill the tank independently in $10hr,20hr,30hr$ respectively

Rate at which water flows through $A$ $=$ $v_A$ $=$ $\frac{Volumeofthetank}{Timetakentofill}$ $=$ $\frac{x}{10}$ $units/hr$

Similarly, $v_B$ $=$ $\frac{x}{20}$ $units/hr$

$v_C$ $=$ $\frac{x}{30}$ $units/hr$

Now, to fill the empty tank,

As every pair of taps have to be open at least for one hour,

For the first three hours, let us open different pair such that every pair is opened once,

Volume of the tank filled in the first three hours $=$ $(v_A$ $+$ $v_B)$ $+$ $(v_B$ $+$ $v_C)$ $+$ $(v_C$ $+$ $v_A)$

$=$ $\frac{11x}{30}$

Now let minimum number of hours be $n$ $=$ $m$ $+$ $3$,

As the pair $(A,B)$ fills that tank fastest, open the pair till the remaining tank is filled(i.e., for the next ${}^{\prime }{m}^{\prime }$ hours),

Volume of the tank filled in the next ${}^{\prime }{m}^{\prime }$ hours $+$ Volume of the tank filled for the first 3 hours $\ge$ Total volume of the tank

$\left(\right(v_A$ $+$ $v_B)$ $\times$ $m)$ $+$ $\frac{11x}{30}$ $\ge$ $x$

$\frac{3x}{20}$ $\times$ $m$ $\ge$ $\frac{19x}{30}$

$m$ $\ge$ $\frac{38}{9}$

$m$ $\ge$ $4.22$

As $n$ $=$ $m$ $+$ $3$, we get

$n$ $\ge$ $7.22$

$n$ $=$ $8$

Therefore, Minimum number of hours required to fill the tank is ${}^{\prime }{8}^{\prime }$.