Question

in how many years will Rs 8000 amount to Rs 13824 at 20% per annum interest compounded annually?​

Answer 1

Answer:

Given :-

• A sum of Rs 8000 amount to Rs 13824 at 20% per annum.

To Find :-

• What is the time.

Formula Used :-

${\purple{\boxed{\large{\bold{P\bigg(1 + \dfrac{r}{100}\bigg)^{n} =\: A}}}}}$

where,

• P = Principal
• r = Rate of Interest
• n = Time
• A = Amount

Solution :-

Let, the time be n.

Given :

• Principal = Rs 8000
• Amount = Rs 13824
• Rate of Interest = 20%

According to the question by using the formula we get,

⇒ $\sf 8000\bigg(1 + \dfrac{20}{100}\bigg)^{n} =\: 13824$

⇒ $\sf 8000\bigg(\dfrac{100 + 20}{100}\bigg)^{n} =\: 13824$

⇒ $\sf 8000\bigg(\dfrac{120}{100}\bigg)^{n} =\: 13824$

⇒ $\sf \bigg(\dfrac{\cancel{120}}{\cancel{100}}\bigg)^{n} =\: \dfrac{13824}{8000}$

⇒ $\sf \bigg(\cancel{\dfrac{24}{20}}\bigg)^{n} =\: \bigg(\cancel{\dfrac{24}{20}}\bigg)^{3}$

➠ $\sf\bold{\green{n =\: 3\: years}}$

$\therefore$ The time is 3 years .

Answer 2

${\huge{\boxed{\underline{\mathrm{\orange{A}{\blue{n}{\red{s}{\green{w}{\pink{e}{\purple{r:-}}}}}}}}}}}$

In 3 years, ₹8,000 will amount to ₹13,824 at 20% p.a. when compounded annually.

Step-by-step explanation:

${\large{\boxed{\underline{\mathrm{\orange{Given:-}}}}}}$

• Principal = ₹8,000
• Rate of interest = 20%
• Amount = ₹13,824

${\large{\boxed{\underline{\mathrm{\pink{To \ Find:-}}}}}}$

• Time

${\large{\boxed{\underline{\mathrm{\purple{Formula \ Used:-}}}}}}$

${\bigstar} \ {\red{\boxed{\mathsf{A = P \bigg{(} 1+\dfrac{r}{100} \bigg{)}^n}}}}} \ {\bigstar}$

Where,

• P = Principal i.e. ₹8,000
• r = Rate of Interest i.e. 20%
• A = Amount

${\large{\boxed{\underline{\mathrm{\blue{Solution:-}}}}}}$

After inserting the formulas in the formula of amount, we get,

$\Longrightarrow {\mathsf{8000 \ {\bigg{(}} 1 + \dfrac{20}{100} {\bigg{)}}^n = 13824}}}$

$\Longrightarrow {\mathsf{8000 \ {\bigg{(}} \dfrac{100 + 20}{100} {\bigg{)}}^n = 13,824}}}$

$\Longrightarrow {\mathsf{8000 \ {\bigg{(}} \dfrac{120}{100} {\bigg{)}}^n = 13,824}}}$  

$\Longrightarrow {\mathsf{{\bigg{(}} \dfrac{120}{100} {\bigg{)}}^n = \dfrac{13,824}{8,000}}}}$

$\Longrightarrow {\mathsf{ {\bigg{(}} \dfrac{24}{20} {\bigg{)}}^n = \dfrac{13,824}{8,000}}}}$

$\Longrightarrow {\mathsf{{\bigg{(}} \dfrac{24}{20} {\bigg{)}}^3 = \dfrac{13,824}{8,000}}}}$

➠ n = 3 years

Hence,  ${\large{\boxed{\boxed{\underline{\mathsf{\orange{Time}{\blue{ \ is \ }{\green{3 \ years.}}}}}}}}}$