Here, total number of days they worked = 72 days

 Let, the wife worked for x days

 The husband worked for (72 - x) days

 According to the question

$$\begin{gathered} 640x+(72-x)560 = 43,520 \\ \Rightarrow 640x+40,320-560x = 43,520 \\ \Rightarrow 80x = 43,520-40,320 \\ \therefore x= \frac{3200}{80}=40 \end{gathered}$$

So, the wife worked for 40 days and the husband worked for (72-40)= 32 days.

Here, the man's salary in 2015 was = 20,000 Tk.

 Here, the man's salary in 2016 was = $20,000+20,000 \times 10\% = 20,000 + 2,000 = 22,000 Tk.$

 Here, the man's salary in 2017 was = $22,000+22,000 \times 10\% = 22,000+ 2,200 = 24,200 Tk.$

 So, he earned in the years 2015 to 2017 inclusive = $20,000 + 22,000 + 24,200 = 66,200 Tk.$ (answer)

The sum of the odd numbers from 1 to 125
Inclusive = 1+3+5+...……+125
It's an arithmetic progression 

Where, a = 1; d=3-1=2

$$\begin{gathered} Let, n^{th} term=125 \\ a+ (n-1)d = 125 \\ \Rightarrow 1+ (n-1)2 = 125 \\ \Rightarrow 2n-2=125 - 1 \\ \Rightarrow 2n = 124 +2 \\ \therefore n=\frac{126}{2}=63 \\ Sn=\frac{n}{2}(2a+(n-1\left)d\right\} =\frac{63}{2} \{2x1+(63-1)\times 2)=\frac{63}{2}(2+124)=\frac{63}{2}\times 126= 3969 \end{gathered}$$

Again, sum of the add numbers form 169 to 209 inclusive = 169 + 171 + 173 +.......+ 209

Here, a = 169 

d = 171-169 = 2

$$\begin{gathered} Let, n^{th} term = 209 \\ \Rightarrow a+ (n-1)d = 209 \\ \Rightarrow 169+ (n-1)2=209 \\ \Rightarrow 2n-2=209-169 \\ \Rightarrow 2n=40+2 \\ \therefore n=\frac{42}{2}=21 \\ sn=\frac{n}{2}\left\{2a+\left(n-1\right)d\right\}=\frac{21}{2}\left\{2\times 169+\left(21-1\right)\times 2\right\}=\frac{21}{2}\left(338+40\right)=\frac{21}{2}\times 378=3969 \\ 1+3+5.........+125=169+171+173+......+209 \left[proved\right] \end{gathered}$$

$$\begin{gathered} L.H.S. \\ =a^{q-r}\times b^{r-p}\times c^{p-q} \\ ={\left(x{y}^{p-1}\right)}^{q-r}\times {\left(x{y}^{q-1}\right)}^{r-p}\times {\left(x{y}^{r-1}\right)}^{p-q}\left[putting the value of a,b,c\right] \\ =x^{\left(q-r\right)}\times y^{\left(p-1\right)\left(q-r\right)}\times x^{\left(r-p\right)}\times y^{\left(q-1\right)\left(r-p\right)}\times x^{\left(p-q\right)}\times y^{\left(r-1\right)\left(p-q\right)} \\ =x^{\left(q-r\right)}\times x^{\left(r-p\right)}\times x^{\left(p-q\right)}\times y^{\left(p-1\right)\left(q-r\right)}\times y^{\left(q-1\right)\left(r-p\right)}\times y^{\left(r-1\right)\left(p-q\right)} \\ =x^{q-r+r-p+p-q}\times y^{pq-pr-q+r}\times y^{qr-pq-r+p}\times y^{pr-qr-p+q} \\ =x^0\times y^{pq-pr-q+r+qr-pq-r+p+pr-p+q} \\ =x^0\times y^0 \\ =1\times 1 \\ =1 \\ = R.H.S \end{gathered}$$

$$\begin{gathered} Given, \frac{x}{2}+\frac{y}{3}=1........\left(i\right) \\ \&\frac{x}{3}+\frac{y}{2}=1.............\left(ii\right) \\ From equation \left(i\right)\frac{x}{2}+\frac{y}{3}=1 \\ \Rightarrow \frac{3x+2y}{6}=1 \\ \therefore 3x+2y=6 ............\left(iii\right) \\ From equation \left(ii\right) \frac{x}{3}+\frac{y}{2}=1 \\ \Rightarrow \frac{2x+3y}{6}=1 \\ \therefore 2x+3y=6 ...............\left(iv\right) \\ Now, multiplying equation \left(iii\right) by 3 and equation \left(iv\right) by 2 and subtracting equation \left(iv\right) from \left(iii\right), \\ we get \\ 9x+6y=18-4x+6y=125x=6 \end{gathered}$$

$\therefore x=\frac{5}{6}$

Putting value of x in equation (iii) we get

$$\begin{gathered} 3x+2y=6 \\ \Rightarrow 3\times \frac{6}{5}+2y=6 \\ \Rightarrow \frac{18}{5}+2y=6 \\ \Rightarrow 2y=6-\frac{18}{5} \\ \Rightarrow 2y=\frac{30-18}{5}=\frac{12}{5} \\ \therefore y=\frac{6}{5} \\ \therefore x=\frac{6}{5} \& y=\frac{6}{5} \end{gathered}$$

Here, r = Radius; h = Height
We know, the volume of a sphere = $\frac{4}{3}{\mathrm{\pi r}}^3$
The volume of hemisphere=$\frac{4}{2\times 3}{\mathrm{\pi r}}^3=\frac{2}{3}{\mathrm{\pi r}}^3$
We also know, the volume of cone =$\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$

Given, that hemisphere and cone have equal bases and the same height
Volume of hemisphere: volume of cone 3

$$\begin{gathered} =\frac{2}{3}{\mathrm{\pi r}}^3 : \frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}=\frac{2}{3}{\mathrm{\pi r}}^3 : \frac{1}{3}{\mathrm{\pi r}}^2\times \mathrm{h}=\frac{2}{3}{\mathrm{\pi r}}^3 : \frac{1}{3}{\mathrm{\pi r}}^3 \left[\mathrm{As} \mathrm{both} \mathrm{the} \mathrm{height} \mathrm{is} \mathrm{same}\right] \\ =\frac{2}{3} : \frac{1}{3} \end{gathered}$$ [dividing both by ${\mathrm{\pi r}}^3$

=2 : 1  [Multiplying both by 3]
Ratio of hemisphere and cone = 2 : 1

একজন সাহসীকে ধ্বংস করা যায়, কিন্তু পরাজিত করা যায় না। 

ভালবাসা এবং ভালোবাসা পাওয়াই জীবনের একমাত্র সুখ। 

গুণীজন কথা বলে কারণ তাদের কিছু বলার থাকে আর বোকারা কথা বলতে হয় বলে কথা বলে। 

শিক্ষা মনকে কখনো নিঃশেষ করে না।