When three water pumps A, B and C are simultaneously turned

How long would it take to fill the water tank with only A, B or C respectively?
A+B+C = 4 hours
A+C = 6 hours
B+C = 6.7 hours
in 1 hour,
$$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{4}\ hours\ \ \ \ ----\ eqn\ \left(1\right)$$
$$\frac{1}{A}+\frac{1}{C}=\frac{1}{6}\ hours\ \ \ \ ----\ eqn\ \left(2\right)$$
$$\frac{1}{B}+\frac{1}{C}=\frac{1}{6.7}\ hours\ \ \ \ ----\ eqn\ \left(3\right)$$
From eqn (2)
$$\frac{1}{A}+\frac{1}{C}=\frac{1}{6}\ hours$$
$$\frac{1}{C}=\frac{1}{6}\ -\frac{1}{A}$$
Substituting this into eqn (1)
$$\frac{1}{A}+\frac{1}{B}+\frac{1}{6}-\frac{1}{A}=\frac{1}{4}$$
$$\frac{1}{B}+\frac{1}{6}=\frac{1}{4}$$
$$\frac{1}{B}=\frac{1}{4}-\frac{1}{6}=\frac{6-4}{24}=\frac{1}{12}$$
$$B=12\ hours$$

From Eqn (3)
$$\frac{1}{B}+\frac{1}{C}=\frac{1}{6.7}\ hours\ \ \ \ ----\ eqn\ \left(3\right)$$
Where 1/B = 1/12
$$\frac{1}{12}+\frac{1}{C}=\frac{1}{6.7}$$
$$\frac{1}{C}=\frac{1}{6.7}-\frac{1}{12}=\frac{12-6.7}{80.4}=\frac{5.3}{80.4}$$
$$C=15.2\ hours$$

From Eqn (2),
$$\frac{1}{A}+\frac{1}{C}=\frac{1}{6}\ hours\ \ \ \ ----\ eqn\ \left(2\right)$$
$$\frac{1}{A}+\frac{1}{15.2}=\frac{1}{6}$$
$$\frac{1}{A}=\frac{1}{6}-\frac{1}{15.2}=\frac{15.2-6}{91.2}=\frac{9.2}{91.2}$$
$$A=9.9\ hours\approx10hours$$
B=12 hours
$$C=15.2\ hours\approx15hours$$

Answer = option B

Thanks<i class="em em-pray">

=>

Assume A, B and C are the it takes machines A, B and C to fill the water tank, respectively.
Then we have 1/A + 1/B + 1/C = 1/4, 1/A + 1/B = 1/6 and 1/B + 1/C = 1/{6(2/3)} = 3/20.
When we add the last two equations, we have 1/A + 2/B + 1/C = 19/20.
When we subtract the first equation from the sum of last two equation, we have 1/C = (1/A + 1/B + 2/C) – (1/A + 1/B + 1/C) = 19/60 – 1/4=(19-15)/60=4/60=1/15.
We have 1/A = 1/10 and 1/B = 1/12.
Thus we have A = 10, B = 12 and C = 15.

Therefore, B is the answer.
Answer: B