Question

In what time will ₹10000 amout to ₹12000 at 12% per annum, compounded anually..

Need deeper explanation please help. ​

Answer 1

GIVEN :-

• Principal ( P ) = Rs. 10000
• Amount ( A ) = Rs. 12000
• Rate ( R ) = 12% Per annum.

TO FIND :-

• The time ( n ).

SOLUTION :-

As we know that,

$\implies \displaystyle \sf \:Amount = P\bigg\lgroup 1 + \dfrac{R}{100}\bigg\rgroup ^{n}$

$\implies \displaystyle \sf \:12000 = 10000 \bigg \lgroup1 + \frac{12}{100} \bigg \rgroup ^{n}$

$\implies \displaystyle \sf \: \frac{12000}{10000} = \bigg \lgroup \frac{100 + 12}{100} \bigg \rgroup ^{n}$

$\implies \displaystyle \sf \: \frac{12}{10} = \bigg \lgroup \frac{112}{100} \bigg \rgroup ^{n}$

$\implies \displaystyle \sf \: \frac{6}{5} = \bigg \lgroup \frac{28}{25} \bigg \rgroup ^{n}$

$\implies \displaystyle \sf \: (1.12) ^{1.6} = {(1.12)} ^{ n}$

$\implies \underline{ \boxed{ \displaystyle \sf \:n = 1.6 \: years}}$

Answer 2

Answer:

Given :-

• A sum of Rs 10000 amount to Rs 12000 at 12% per annum, compounded annually.

To Find :-

• What is the time.

Formula Used :-

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

where,

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

Solution :-

Let, the time be n

Given :

• Principal = Rs 10000
• Rate of Interest = 12%
• Amount = Rs 12000

According to the question by using the formula we get,

⇒ $\sf 10000\bigg(1 + \dfrac{12}{100}\bigg)^{n} =\: 12000$

⇒ $\sf \bigg(1 + \dfrac{12}{100}\bigg)^{n} =\: \dfrac{12000}{10000}$

⇒ $\sf \bigg(1 + \dfrac{12}{100}\bigg)^{n} =\: \dfrac{\cancel{12000}}{\cancel{10000}}$

⇒ $\sf \bigg(\dfrac{\cancel{112}}{\cancel{100}}\bigg)^{n} =\: \dfrac{\cancel{12}}{\cancel{{10}}}$

⇒ $\sf \bigg(\dfrac{28}{25}\bigg)^{n} =\: \dfrac{6}{5}$

⇒ $\sf (1.12)^{n} =\: (1.12)^{1.6}$

➠ $\sf\bold{\red{n =\: 1.6\: years}}$

$\therefore$ The time is 1.6 years .