Question

The difference between the compound
interest and simple interest on a certain sum
at 10% p.a. for 2 years is 1262. Find the
sum if interest is compounded annually.​

Answer 1

Answer:

.I.−S.I.=18

(P(1+

100

R

)

n

−P)−(

100

P×R×n

)=18

Given:-

P=?

R=6%

n=2 years

P(1+

100

6

)

2

−P−

100

P×6×2

=18

P(

100

106

)

2

−P−

25

3P

=18

1.1236P−P−0.12P=18

0.0036P=18

P=

0.0036

18

Answer 2

Step-by-step explanation:

Given:-

• The difference between the Compound Interest and Simple Interest is 1262.

• Rate = 10%

• Time = 2 years

To Find:-

• The sum of money (Principal)

Solution:-

Let the Principal be " P "

$\sf Simple \: Interest = \dfrac{P\times T\times R}{100}$

$\sf \dashrightarrow SI = \dfrac{P\times 2\times 10}{100}$

$\sf \dashrightarrow \red{ SI = \dfrac{P}{5} }$

$\sf Amount = </p><p>P \bigg(1 + \dfrac{r}{100} \bigg)^{n}$

$\sf Compound\: Interest = Amount - Principal$

$\sf \dashrightarrow CI= </p><p>P \bigg(1 + \dfrac{r}{100} \bigg) ^{n} - P$

$\sf \dashrightarrow CI= </p><p>P \bigg(1 + \dfrac{10}{100} \bigg) ^{2} - P$

$\sf \dashrightarrow CI= </p><p>P \bigg( \dfrac{110}{100} \bigg) ^{2} - P$

$\sf \dashrightarrow CI= </p><p>P \bigg( \dfrac{11}{10} \bigg) ^{2} - P$

$\sf\dashrightarrow CI= </p><p> \dfrac{121P}{100} - P$

$\sf\dashrightarrow \pink{ CI= </p><p> \dfrac{21P}{100}}$

$\sf Given, \: Compound \: Interest - Simple \: Interest = Rs.1262$

$\sf \dashrightarrow \: CI - SI= 1262$

$\sf \dashrightarrow \dfrac{21P}{100} - \dfrac{P}{5} = 1262$

$\sf \dashrightarrow \dfrac{21P - 20P}{100} = 1262$

$\sf \dashrightarrow \dfrac{P}{100} = 1262$

$\sf \dashrightarrow P = 126200$

$\dashrightarrow \underline{ \boxed{ \purple{ \textsf{ \textbf{Principal = Rs.126200}}}}}$