Question

Find the compound interest on ₹ 24000 at 15% per annum compounded annually for 2 years 4 months.
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Answer 1

Step-by-step explanation:

Given:-

• Principal (P) = Rs.24000

• Rate % (r) = 15%

• Time (n) = 2 years four months = $2 \dfrac{1}{3}$ years.

To Find:-

• Compound Interest compounded annually.

Solution:-

Let us find Amount for 2 years

The formula for finding Amount is:-

$\boxed{\bf \leadsto Amount = P \bigg(1 + \dfrac{r}{100} \bigg) ^{n} }$

For 2 years:-

$\sf \dashrightarrow A = 24000\times \bigg(1 + \dfrac{15}{100} \bigg) ^{2}$

$\sf \dashrightarrow A = 24000\times \bigg( \dfrac{100 + 15}{100} \bigg) ^{2}$

$\sf \dashrightarrow A = 24000\times \bigg( \dfrac{115}{100} \bigg) ^{2}$

$\sf \dashrightarrow A = 24000\times \bigg( \dfrac{115}{100} \bigg) ^{2}$

$\sf \dashrightarrow A = 24000\times \dfrac{529}{400}$

$\sf \dashrightarrow A = 60\times 529$

$\sf \dashrightarrow A = 31740$

$\sf \therefore Amount \: after \: 2 \: years = Rs.31740$

Now, this Amount acts as the Principal for the next ⅓ years.

Now, Principal = Rs.31740 and Time = ⅓ years

SI for next ⅓ years :-

$\sf \dashrightarrow SI = \dfrac{P\times T \times R}{100}$

$\sf = \dfrac{31740\times 1 \times 15}{3 \times 100}$

$\sf = \dfrac{476100}{300}$

$\sf =1587$

Amount for 2⅓ years = 31740 + 1587 = Rs. 33327

Compound Interest = Amount - Principal

= 33327 - 24000

= 9327

$\underline{\boxed{\purple{\therefore \textsf{\textbf{Compound \: Interest = Rs.9327}}}}}$

Answer 2

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Given:-

Principal (P) = Rs.24000

Rate % (r) = 15%

Time (n) = 2 years four months = 2 \dfrac{1}{3}2

3

1

years.

To Find:-

Compound Interest compounded annually.

Solution:-

Let us find Amount for 2 years

The formula for finding Amount is:-

\boxed{\bf \leadsto Amount = P \bigg(1 + \dfrac{r}{100} \bigg) ^{n} }

⇝Amount=P(1+

100

r

)

n

For 2 years:-

\sf \dashrightarrow A = 24000\times \bigg(1 + \dfrac{15}{100} \bigg) ^{2}⇢A=24000×(1+

100

15

)

2

\sf \dashrightarrow A = 24000\times \bigg( \dfrac{100 + 15}{100} \bigg) ^{2}⇢A=24000×(

100

100+15

)

2

\sf \dashrightarrow A = 24000\times \bigg( \dfrac{115}{100} \bigg) ^{2}⇢A=24000×(

100

115

)

2

\sf \dashrightarrow A = 24000\times \bigg( \dfrac{115}{100} \bigg) ^{2}⇢A=24000×(

100

115

)

2

\sf \dashrightarrow A = 24000\times \dfrac{529}{400}⇢A=24000×

400

529

\sf \dashrightarrow A = 60\times 529⇢A=60×529

\sf \dashrightarrow A = 31740⇢A=31740

\sf \therefore Amount \: after \: 2 \: years = Rs.31740∴Amountafter2years=Rs.31740

Now, this Amount acts as the Principal for the next ⅓ years.

Now, Principal = Rs.31740 and Time = ⅓ years

SI for next ⅓ years :-

\sf \dashrightarrow SI = \dfrac{P\times T \times R}{100}⇢SI=

100

P×T×R

\sf = \dfrac{31740\times 1 \times 15}{3 \times 100}=

3×100

31740×1×15

\sf = \dfrac{476100}{300}=

300

476100

\sf =1587=1587

Amount for 2⅓ years = 31740 + 1587 = Rs. 33327

Compound Interest = Amount - Principal

= 33327 - 24000

= 9327

\underline{\boxed{\purple{\therefore \textsf{\textbf{Compound \: Interest = Rs.9327}}}}}

∴Compound Interest = Rs.9327