50 persons can do a work in 12 days by working 8 hours a day. Working how many hours per day can 60 persons finish the work in 16 days?

In 1 minute, first pipe can fill  $\frac{1}{12}$ part of tank.

Let, second pipe can empty the tank in x minutes.

∴ In 1 minute, it can empty  $\frac{1}{x}$ part of tank.

That means, the tank is filled in 1 minutes $\left(\frac{1}{12} - \frac{1}{x}\right)$ Part

According to question, $\frac{1}{12} - \frac{1}{x} = \frac{1}{96}$

$\Rightarrow \frac{1}{x} = \frac{1}{12} - \frac{1}{96} = \frac{8-1}{96} = \frac{7}{96}$

$\therefore x = \frac{96}{7} minutes$

∴ The tank contains $= 14 \times \frac{96}{7}$ = 192 liters.

$\frac{\log \sqrt{27}+log8-log\sqrt{1000}}{\log 1.2}$

$= \frac{\log \sqrt{3^3} + \log 2^3 - \log \sqrt{{\left(2 \times 5\right)}^3}}{\log {\displaystyle \frac{12}{10}}}$

$= \frac{\log 3^{{\displaystyle \frac{3}{2}}} + \log 2^3 - \log \sqrt{2^3 \times 5^3}}{\log {\displaystyle \frac{6}{5}}}$

$= \frac{{\displaystyle \frac{3}{2}\log 3 + 3 \log 2 - \log \left(2^{\frac{3}{2}}\times 5^{\frac{3}{2}}\right) }}{\log 6 - \log 5}$

$= \frac{{\displaystyle \frac{3}{2}\log 3 + 3 \log 2 - \log \left(2^{\frac{3}{2}}\times 5^{\frac{3}{2}}\right) }}{\log 6 \left(3 \times 2\right) - \log 5}$

$= \frac{{\displaystyle \frac{3}{2}\log 3 + 3 \log 2 - \frac{3}{2} \log 2 - \frac{3}{2} \log 5}}{\log 3+ \log 2 - \log 5}$

$= \frac{{\displaystyle \frac{3}{2}\log 3 + \frac{3}{2} \log 2 - \frac{3}{2} \log 5}}{\log 3+ \log 2 - \log 5}$

$= \frac{{\displaystyle \frac{3}{2}\left(\log 3 + \log 2 - \log 5\right)}}{\log 3+ \log 2 - \log 5} = \frac{3}{2} \left(\mathit{a}\mathit{n}\mathit{s}\mathit{w}\mathit{e}\mathit{r}\right)$

$\frac{1}{x-3}+\frac{1}{x-4}=\frac{1}{x-2}+\frac{1}{x-5}$

$\Rightarrow \frac{1}{x-3}- \frac{1}{x-5}=\frac{1}{x-2}-\frac{1}{x-4}$

$\Rightarrow \frac{x-5-x+3}{\left(x-3\right) \left(x-5\right)} = \frac{x-4 -x+2 }{\left(x-2\right) \left(x-4\right)}$

$\Rightarrow \frac{-2}{x^2 - 5x-3x + 15} = \frac{-2}{x^2 - 4x - 2x + 8}$

$\Rightarrow \frac{1}{x^2 - 8x + 15} = \frac{1}{x^2 - 6x + 8}$ [Dividing both sides by - 2]

$\Rightarrow x^2 - 8x + 15 = x^2 - 6x + 8$

$\Rightarrow -8x + 6x = 8-15$

$\Rightarrow -2x = -7$

 $\therefore x = \frac{-7}{-2} = \frac{7}{2} \mathbf{\left(}\mathit{a}\mathit{n}\mathit{s}\mathit{w}\mathit{e}\mathit{r}\mathbf{\right)}$