Question

Q . A person invested a sum of money
for 2 years at 10% per annum compound interest. After 2 years, he got back ₹ 14520. Find the sum of money he invested initially. [Hint : Let 100 be the sum invested.]



pls give me the correct answer with long explanation.​

Answer 1

Answer:

It is given that

Principal(P) = 10,000

Period (T)= 1 year

Sum amount (A)= 11200

Rate of interest =?

(i) We know that

Interest (I)= 11200- 10000= 1200

So the rate of interest

R= (1×100)/(P×T)

Substituting the values

R= (1200×100)/(1000×1)

So we get

R= 12% p.a

Therefore, the rate of interest per annum is 12% p.a

(ii) We know that

Period (T)= 2 years

Rate of interest (R) = 12% p.a.

Here

A =P(1+R/100)

1

Substituting the values

A = 10000(1+12/100)

2

By further calculation

A= 10000(28/25)

2

We can write it as

A= 10000×28/25×28/25

So we get

A=16×28×28

A=12544

Therefore, the amount at the end of the second year is 12544.

Step-by-step explanation:

This may help you

Answer 2

Answer:

$=\huge\mathrm{ 14520=P\left(1+\:\frac{10}{100}\:\right)\:^{2\:}\:=\:P=Rs.12000}$

Step-by-step explanation:

Given:

Amount in 2 years= RS. 14,520 ⇒ $P(1+\frac{R}{100} )^2 =$

14250..(I)

Amount in 4 years = Rs.  ⇒ $P(1+\frac{R}{100} )^4 =$

17569.20...(ii)

On dividing (ii) by (i), we get :

$\huge\frac{P(1+\frac{R}{100} )^4 }{P(1+\frac{R}{100} )^2} = \frac{17569.20}{14520}$

$(1+\frac{R}{100} )^2 = \frac{121}{100}$

⇒ $1+\frac{R}{100}$  $= \frac{11}{10}$

⇒ $r = 10$

$Now,\:P\left(1+\:\frac{r}{100}\:\right)\:^2=14520⇒P\left(1+\:\frac{10}{100}\:\right)\:^2\:=14520$

$= P\:\times\:\frac{121}{100}\:=14520\:and\:P=14520\times\:\frac{100}{121}=12000$

∴ Rate of interest = 10% per annum and sum = Rs. 12,000

Alternative method:

For the last 2 years :

P = Rs.14520, A = Rs. 17569.20 and n = 2 years

$Rs.17569.20=Rs.14520\left(1+\:\frac{r}{100}\:\right)\:^{2\:}\:A=P\left(1+\:\frac{r}{100}\right)\:^2\:$

$=\:\frac{14520}{17569.20}\:=\left(1+\:\frac{r}{100}\:\right)\:^2\:$

$\mathrm{On\:solving\:we\:get\::\:r\:=\:10\%For\:the\:first\:2\:years}$

$\mathrm{A\:=\:Rs.\:14520,\:P\:=\:?\:and\:r\:=\:10\%\:and\:n\:=\:2\:years}$

$= 14520=P\left(1+\:\frac{10}{100}\:\right)\:^{2\:}\:=\:P=Rs.12000$