question_answer

                                                                  Principal = Rs. 1000, rate = 8% per annum. Fill in the following table and find which type of interest (simple or compound) changes in direct proportion with time period.

Time period	1 Year	2 Years	3 Years
Simple Interest (in Rs.)
Compound Interest (inRs.)

Answer:

Principal = Rs. 1000

Rate = 8%

Case I: t = 1 year

Simple Interest =  \[\frac{p\times r\times t}{100}=\frac{1000\times 1\times 8}{100}\]

= Rs. 80

Compound Interest = \[p{{\left( 1+\frac{r}{100} \right)}^{t}}-p\]

\[=1000\left( 1+\frac{8}{100} \right)-1000\]

\[=1000\left( \frac{108}{100} \right)-1000\]

\[=1080-1000\]

= Rs. 80

Case II:            t = 2 years

Simple Interest \[=\frac{p\times r\times t}{100}=\frac{1000\times 8\times 2}{100}\]

= Rs. 160

Compound Interest \[=p{{\left( 1+\frac{r}{100} \right)}^{t}}-p\]

\[=1000{{\left( 1+\frac{8}{100} \right)}^{2}}-1000\]

\[=1000{{\left( \frac{100+8}{100} \right)}^{2}}-1000\]

\[=1000\left( \frac{108\times 108}{100\times 100} \right)-1000\]

\[=\frac{108\times 108}{10}-1000\]

\[=1166.40-1000\]

= Rs. 166.40

Case III: t = 3 years,

Simple interest \[=\frac{p\times r\times t}{100}=\frac{1000\times 8\times 3}{100}\]

=Rs. 240

Compound interest \[=p{{\left( 1+\frac{r}{100} \right)}^{t}}-p\]

\[=1000{{\left( 1+\frac{8}{100} \right)}^{3}}-1000\]

\[=1000{{\left( \frac{108}{100} \right)}^{3}}-1000\]

\[=\frac{1000}{1000000}(1259712)-1000\]

\[=1259.71-1000\]

= Rs. 259.71

                                                                                                                                    The table is:

Time period	1 year	2 year	3 years
S.I.	80	160	240
C.I.	80	166.40	259.71