Question

A sum of money is lent at 8% per annum compound interest. If the interest for the second year exceeds that for the first year by Rs. 32,find the sum of money.
If you guys know please answer this question... ​

Answer 1

$\bigstar\mid$ Given :

• A sum of money lent at 8% p.a. (per annum) compound interest.
• The interest for the second year exceeds by 32.

$\bigstar\mid$ To find :

✴️ The sum of money.

$\bigstar\mid$ Solution :

Let's assume that the sum is ₹x

Now,

✴️ We need to find the sum year by year.

For the first year :

We know that,

${ \underline{ \boxed{ \sf{Simple \: interest = \frac{p \times r \times t}{100} }}}}$

Where,

p (principal) = ₹x.

t (time) = 1 year.

r (rate) = 8% .

Procedure :

$\rm \implies \: S. I.= \frac{x \times 1 \times 8}{100}$

$\rm \implies \: S. I. = \frac{8x}{100}$

Again,

For the second year :

✴️ Taking p (principal) as :

➠ $\sf x + {\frac{8x}{100}}$

➠ $\sf \frac{100x + 8x}{100}$

➠ $\sf \frac{108x}{100}$

Interest for the second year :

➠ $\sf \: S. I. \: = \frac{ \frac{108x }{100} \times 8 \times 1}{100}$

➠ $\sf \: S. I. \: = \frac{108x \times 8 \times 1}{10000}$

➠ $\sf S. I. = \frac{864x}{100}$

✴ As it is told that the interest for the second year exceeds the interest of first year by 32.

$☯ \begin{gathered}\underline{\boldsymbol{According\: to \:the\: question :}}\\\end{gathered}$

$\sf : \implies \: \frac{864x}{10000} = \frac{8x}{100} + 32$

$\sf : \implies \: \frac{864x}{10000} - \frac{8x}{100} = 32$

$\sf : \implies \: \frac{864x - 800x}{10000} = 32$

$\sf : \implies \: \frac{64x}{10000} = 32$

$\sf : \implies \: x = \frac{10000 \times \cancel{32}} { \cancel{64}}$

$\sf : \implies \: x = { \cancel{\frac{10000}{2}}}$

$\bf \therefore \: x = ₹5000$

$\gray{ \underbrace{ \textbf{ \textsf{ \green{Required \:answer : }}}}} \\ { \underline{ \boxed{ \textbf{\textsf{ \pink{The \: sum \: of \: money \: is \: ₹5,000}}}}}}$

Answer 2

Answer:

Mark the first one as brainliest

Step-by-step explanation:

have a great day ahead