Solve the following mathematical problems with sufficient calculations.

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.

$$\begin{gathered} 11.8 \\ 115-12-11-00 -1{\left(11\right)}_{10} = {\left(1011\right)}_{2}22222 \end{gathered}$$

$again, 0.8 \times 2 = 1.60$

$0.6 \times 2 = 1.20$

$0.2 \times 2 = 0.40$

$0.4 \times 2 = 0.80$

$0.8 \times 2 = 1.60$

$0.6 \times 2 = 1.20$

${(0.8)}_{10}=1100110011.....$

$\therefore {(11.8)}_{10} = {(1011.1100110011........)}_2$

$2^{nd} part {(10111.011)}_2$

$= (1 \times 2^4)+(0\times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (1 \times 2^0) + (0 \times 2^{-1}) + (1 \times 2^{-2}) + (1 \times 2^{-3})$

$= 16 +0 + 4 + 2 + 1 + 0 +\frac{1}{4} + \frac{1}{8} = 23+ \frac{3}{8} = 23.375$

প্রশ্নে বলা হচ্ছে যে, দৈনিক ৪ ঘণ্টা পরিশ্রম করে 50 জন লোক একটি কাজ 12 দিনে করতে পারে। 60 জন লোক দৈনিক কতঘণ্টা করে কাজ করে 16 দিনে কাজটি সম্পন্ন করতে পারে?

50 persons take 12 days working = 8 hours per day

1 person take 1 days working $= (8 \times 50 \times 12)$ hours per day

60 person take 16 days working $= \frac{8 \times 50 \times 12}{60 \times 16} = hours$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\Rightarrow -2x = -7$

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