Question

Find the amount of ₹8000 for 2 years compounded annually and the rates being 9% per
annum during the first year and 10% per annum during the second year.
​

Answer 1

Answer:

Principal = Rs 8000 Time period = 2 years Rate of interest for 1st year = 9% Amount at the end of 1st year = Rs 8720 Rate of interest for 2nd year = 10% Principal, P = Rs 8720 Hence amount after 2 years is Rs 9592.

Answer 2

$\sf \blue{\underline{Given : }}$

• The amount of ₹8,000 is compounded annually for 2 years.
• For the first year it is compounded of about 9% p.a.(per annum).
• And for the second year it is compounded of about 10% p.a.(per annum).

$\sf \blue{\underline{To \: find :} }$

• The compound interest by maintaining the above statements.

$\sf \blue{\underline{Solution : }}$

NOTE :

This question can be solved by two method :

1. By using the formula.
2. Step by step explanation.

$\sf \blue{\underline{Procedure : }}$

We know that,

${ \underline{ \boxed{ \green{ \bf A(amount) = P {(1 + \frac{r}{100} )}^{n} }}}}$

$\sf \purple{\underline{\therefore \:Calculation \: for\: the\: {1}^{st} \: year :}}$

➠ P(principal) = ₹8,000

➠ R(rate) = 9%

➠ n(time) = 1 year.

Hence,

$\longmapsto \sf A \: = 8,000 {(1 + \frac{9}{100}) }^{1}$

$\sf \longmapsto A = 80\cancel{00} \times \frac{109}{1 \cancel{00}}$

$\sf \longmapsto A = (80 \times 109)$

$\bold \orange \dag{ \underline{ \boxed{ \blue{ \bf \therefore A = 8,720}}}} \bold \orange \dag$

Now,

We got an amount of ₹8,720 which will be the principal for the next year

$\sf \purple{\underline{\therefore \:Calculation \: for\: the\: {2}^{nd} \: year :}}$

Where,

➠ P(principal) = ₹8,720.

➠ R(rate) = 10%.

➠ n(time) = 1 year.

Hence,

$\sf \longmapsto A(amount) = 8720 {(1 + \frac{10}{100}) }^{1}$

$\sf \longmapsto A = 8720 {(1 + \frac{1}{10} )}^{1}$

$\sf \longmapsto A = 872 \cancel 0 \times \frac{11}{1 \cancel{0}}$

$\sf \longmapsto A = 872 \times 11$

$\bold \orange \dag{ \underline{ \boxed{ \blue{ \bf \therefore A = 9,592 \: ans.}}}} \bold \orange \dag$

$\gray\bigstar{ \underline{ \boxed{ \red { \bf {Alternative \: method : }}}}} \gray\bigstar$

By using the formula :

${ \blue{ \bf \longrightarrow A(amount) \: = p \times {(1 + \frac{r( {1}^{st} year)}{100}) }^{n} \times {(1 + \frac{r( {2}^{nd} year)}{100}) }^{n} }}$

Substituting the given cases :

$\sf \longmapsto A = 8000 {(1 + \frac{9}{100} )} \times (1 + \frac{1\cancel 0}{10 \cancel 0} )$

$\sf \longmapsto A = 8\cancel{000}\times \frac{109}{1\cancel{00}} \times \frac{11}{1\cancel 0}$

$\sf \longmapsto A =8 \times 109 \times 11$

$\bold \blue \dag{ \underline{ \boxed{ \orange{ \bf \therefore A = 9592 \: ans.}}}} \bold \blue \dag$