$$\begin{gathered} Given that, \frac{2^{x+4}-4.2^{x+1}}{2^{x+2-1}}=\frac{2^x\times 2^4-4\times 2^x\times 2^1}{{\displaystyle \frac{2^{x+2}}{2^1}}} \\ =\frac{16\times 2^x-8\times 2^x}{2^{x+2-1}}=\frac{8\times 2^x}{2^{x+1}}=\frac{8\times 2^x}{2^x\times 2^1}=\frac{8\times 2^x}{2^{x\times 2}}=4 \end{gathered}$$

$$\begin{gathered} Let, the capacity of the \tan k be x liters. \\ By 1st pipe \\ In 15 minutes pipe fill=x part of the \tan k \\ \therefore In 1 minute pipe fill=\frac{x}{15} part of the \tan k \end{gathered}$$

$$\begin{gathered} According to question, \\ 135(\frac{x}{15}-20)=x \\ =135\left(\frac{x-300}{15}\right)=x \\ =9(x-300)=x \\ =9x-2700=x \\ =8x=2700 \\ \therefore x=\frac{2700}{8}=337.5 liters \\ \therefore The capacity of the \tan k is 337.5 liters. \end{gathered}$$

$$\begin{gathered} Let, she should invest x Tk. in BEXIMCO \\ \therefore She should invest = (50000 - x) Tk in DBBL \\ According to question \\ 6.5\% of x + 9\% of (50000 - x) = 4000 \\ =\frac{6.5c}{100}+\frac{9(50000-x)}{100}=4000 \\ =6.5x+450000-9x=4000000 \\ =6.5x-9x=450000-4000000 \\ =2.5x=50000 \\ \therefore x=\frac{50000}{2.5}=20000 \end{gathered}$$

$$\begin{gathered} \therefore The amount of BEXIMCO bond be 20,000 Tk. \\ \therefore DBBL amount would be = 50000 - 20000 = 30000Tk . \end{gathered}$$

$$\begin{gathered} Let, us depict the image according to question: \\ Now, Faiyaz s\tan ds at point A \\ And Ehan s\tan ds at point C \\ Here, direct dis\tan ce between them is AC \\ △ABC is a right angle triangle where AC = Hypotenuse \end{gathered}$$

$$\begin{gathered} \therefore Applying Pythagorean theorem, we get \\ {\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2 ={\left(AC\right)}^2={(50+70)}^2+{\left(90\right)}^2={\left(120\right)}^2+{\left(90\right)}^2 ={\left(AC\right)}^2=14400+8100=22500={\left(150\right)}^2 \therefore AC=150 \\ \therefore The straight dis\tan ce between Faiyaz and Ehan is 150 yards. \end{gathered}$$

$$\begin{gathered} Let, number of male be x \\ \therefore Number of female be = (1000000 - x) \\ According to question \\ 110\% of x + 94\% of (1000000 - x) = 1000000 \\ =\frac{110X}{100}+\frac{94(1000000-X)}{100}=1000000 \\ =\frac{11X}{10}+\frac{9.4(1000000-X)}{10}=1000000 \\ =11X+9400000-9.4X=10000000 \\ =1.6X=10000000-9400000=600000 \\ \therefore X=\frac{600000}{1.6}=375000 \\ \therefore Number of male = 3,75,000 \\ \therefore Number of female = 10,00,000-3,75,000 =6,25,000 \end{gathered}$$