A man's salary in 2015 was Tk. 20,000 per annul and it increased by 10% each year. Find how much he earned in the years 2015 to 2017 inclusive. (Series)

Here, Earned royalties = $1,00,000 Tk.$

Remaining royalties after tax = $1,00,000-1.00,000 \times 20\% = 80,000 Tk.$

Invested at higher rate=$50,000 Tk$

 Invested at lower rate = $80000 - 50000 = 30000Tk$

Let, Higher rate be = $x\%$

Lower rate be = $(x - 1)\%$

According to the question,

$$\begin{gathered} 50,000 \times x\%+30,000 \times (x - 1)\% = 6,100 \\ \Rightarrow 50.000 \times \frac{x}{100}+ 30000\times \frac{x - 1}{100}= 6,100 \\ \Rightarrow 500x+300(x - 1)= 6,100 \\ \Rightarrow 500x+300x-300 = 6,100 \\ \Rightarrow 800x=6,100+300 \\ \Rightarrow x=\frac{6400}{800}=8 \\ Higher rate=8-1=7\% \end{gathered}$$

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.

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