Lakshman Patel RJIT
asked
in Quantitative Aptitude
Apr 1, 2020
edited
Aug 29, 2020
by soujanyareddy13

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0 votes

Let say $A$ filled $10$ minutes and $B$ filled in $30$ minutes.

Capacity of tank $ = LCM(10,30) = 30$ litre

Efficiency of $A = 3$ litre/minute and efficiency of $B = 1$ litre/minute

Both the taps are kept open for $5$ minutes and then the first one $(A)$ is shut off.

Tank filled by $A\&B$ together for $5$ minute $ = 5\;\text{minute} \times 4 \;\text{litre/minute} = 20$ litre

Tank remaining to be filled $ = 30 – 20 = 10$ litre

Now, only second one $(B),$ filled the tank in $ = \dfrac{10\;\text{litre}}{1 \;\text{litre/minute}} = 10$ minute.

So, the correct answer is $(C).$

Capacity of tank $ = LCM(10,30) = 30$ litre

Efficiency of $A = 3$ litre/minute and efficiency of $B = 1$ litre/minute

Both the taps are kept open for $5$ minutes and then the first one $(A)$ is shut off.

Tank filled by $A\&B$ together for $5$ minute $ = 5\;\text{minute} \times 4 \;\text{litre/minute} = 20$ litre

Tank remaining to be filled $ = 30 – 20 = 10$ litre

Now, only second one $(B),$ filled the tank in $ = \dfrac{10\;\text{litre}}{1 \;\text{litre/minute}} = 10$ minute.

So, the correct answer is $(C).$

0 votes

Part filled in $1 $ minute: $\frac{1}{10}+\frac{1}{30}=\frac{2}{15}$

Part filled in $5 $ minute: $\frac{2}{15}\times 5=\frac{2}{3}$

Unfilled part= $1-\frac{2}{3}=\frac{1}{3}$

so this is filled by tape first by alone,hence will be filled in $30\times \frac{1}{3}=10$ minute.

So option (C) is correct.

Part filled in $5 $ minute: $\frac{2}{15}\times 5=\frac{2}{3}$

Unfilled part= $1-\frac{2}{3}=\frac{1}{3}$

so this is filled by tape first by alone,hence will be filled in $30\times \frac{1}{3}=10$ minute.

So option (C) is correct.