Question

In how many years will 36,000 amount to
46,621.044 at 9% p.a. compounded annually?​

Answer 1

Question :-

• In how many years will 36,000 amount to 46,621.044 at 9% p.a. compounded annually?

Given :-

• Amount = Rs, 46621.044
• Principal = Rs, 36000
• Rate = 9%

To Find:-

• The time ( years) in which the amount ₹ 36000 amount will be compounded to ₹ 46621.044 at 9% p.a compounded annually

Solution :-

$\qquad \quad {:} \longrightarrow \boxed{\sf{\bf{Amount \: = \: Principal \: \times \: \bigg( 1\: + \: \dfrac{Rate}{100}\bigg)^{Time}}}}$

${:} \longrightarrow {\sf{\sf{46621.044 = 36000 \times { \bigg( \: 1 + \frac{9}{100} \bigg)}^{t}}}}$

${:} \longrightarrow {\sf{\sf{ \dfrac{46621.044}{36000} = { \bigg( \: 1 + \frac{9}{100} \bigg)}^{t} }}}$

${:} \longrightarrow {\sf{\sf{ \dfrac{1295029}{1000000} = { \bigg( \: \frac{109}{100} \bigg)}^{t} }}}$

${:} \longrightarrow {\sf{\sf{ { \bigg( \: \frac{109}{100} \bigg)}^{3} = { \bigg( \: \frac{109}{100} \bigg)}^{t} }}}$

${:} \longrightarrow {\sf{\sf{\bigg( \: \frac{109}{100} \bigg)}^{t} = { \bigg( \: \frac{109}{100} \bigg)}^{3} }}$

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So, the time required is 3 years.

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Answer 2

Amount =principal (1+rate/100)time

46621.044=36000(1+9/100)time

46621.044/36000=109/100)time

1295029/1000000=(109/100)time

(109/100)3=(109/100)time

(109/100)time=(109/100)3

so the time required is 3years