Question

Q. Create a new paragraph style and perform the following operations/ function:
a) Left margin -> 0.5
b) Right margin -> 5
c) First line indent -> 1.5
d) Center aligns, Italics after entering the following text.
Select this paragraph style then write the following text & then copy the the second paragraph into the file ' polo'
The Times of India a group is a professionally managed growth oriented organization, employing a motivated team of result- oriented people.
Since our operations are extensively computerized the incumbent should have a good grasp of it

@Sway lol agreedddd (≧∇≦)/​

Answer 1

Answer:

refers to the attachment

Answer 2

$\large\underline{\sf{Solution-}}Solution−</p><p>Given that,</p><p></p><p>The population of a town increases by 5% every year and the population is 94,500 now.</p><p></p><p>We have to find</p><p></p><p>The population one year ago.</p><p></p><p>So, it means we have</p><p>\\ \\</p><p>Present population, A = 94500</p><p></p><p>Increase in population, r = 5 % per annum</p><p></p><p>Time, n = 1 year</p><p>\\ \\</p><p>We know,</p><p>If P is the population of the town increases every year at the rate of r % per annum for n years, then total population A is given by</p><p>\\ \\</p><p>\begin{gathered}\rm \: \boxed{ \rm{ \:A \: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} }} \\ \end{gathered}A=P[1+100r]n</p><p>\\ \\</p><p>So, on substituting the values, we get</p><p>\begin{gathered}\rm \: 94500 = P {\bigg[1 + \dfrac{5}{100} \bigg]}^{1} \\ \end{gathered}94500=P[1+1005]1</p><p>\\ \\</p><p>\begin{gathered}\rm \: 94500 = P {\bigg[1 + \dfrac{1}{20} \bigg]} \\ \end{gathered}94500=P[1+201]</p><p>\begin{gathered}\rm \: 94500 = P {\bigg[ \dfrac{20 + 1}{20} \bigg]} \\ \end{gathered}94500=P[2020+1]</p><p>\begin{gathered}\rm \: 94500 = P {\bigg[ \dfrac{21}{20} \bigg]} \\ \end{gathered}94500=P[2021]</p><p>\begin{gathered}\rm \: P = 94500 \times \frac{20}{21} \\ \end{gathered}P=94500×2120</p><p>\begin{gathered}\rm\implies \:\boxed{ \rm{ \:P \: = \: 90000 \: }} \\ \end{gathered}⟹P=90000</p><p>\rule{190pt}{2pt}</p><p>Additional information :-</p><p>1. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is given by</p><p>\begin{gathered}\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: \: }} \\ \end{gathered}Amount=P[1+100r]n</p><p>2. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded semi - annually for n years is given by</p><p>\begin{gathered}\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: \: }} \\ \end{gathered}Amount=P[1+200r]2n</p><p>3. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded quarterly for n years is given by</p><p>\begin{gathered}\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} \: \: }} \\ \end{gathered}Amount=P[1+400r]4n</p><p>4. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded monthly for n years is given by</p><p>\begin{gathered}\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{1200} \bigg]}^{12n} \: \: }} \\ \end{gathered}Amount=P[1+1200r]12n$