Question

c) Find the amount and compound interest on 16,000
for 3 years at 5% per annum compounded annually.​

Answer 1

Step-by-step explanation:

i=

100

p×r×t

For the first year,

P = 16000

r = 5

T = 1 ... (because we are first finding for 1st year)

i = \frac{16000 \times 5 \times 1}{100}i=

100

16000×5×1

i = \frac{80000}{100}i=

100

80000

i = 800i=800

New Principal = 16000 + 800

= 16800

Now for 2nd year,

P = 16800

r = 5

T = 1

i = \frac{p \times r \times t}{100}i=

100

p×r×t

i = \frac{16800 \times 5 \times 1}{100}i=

100

16800×5×1

i = \frac{8400}{100}i=

100

8400

i = 840i=840

New Principal = 16800 + 840

= 17640

For third year,

P = 17640

r = 5

T = 1

i = \frac{p \times r \times t}{100}i=

100

p×r×t

i = \frac{17640 \times 5 \times 1}{100}i=

100

17640×5×1

i = \frac{88200}{100}i=

100

88200

i = 882i=882

Answer 2

Step-by-step explanation:

Given:-

• Principal (P) = Rs.16000

• Rate (r) = 5%

• Time (n) = 3 years

To Find:-

• The Amount

• The Compound Interest compounded annually.

Solution:-

The formula for finding the Amount is:-

$\boxed{\bull\: \sf Amount = P \bigg(1 + \frac{r}{100}\bigg)^{n}}$

Substituting the values:-

$\sf \longrightarrow A = 16000\bigg(1 + \dfrac{5}{100}\bigg)^{3}$

$\sf \longrightarrow A = 16000\bigg( \dfrac{100 + 5}{100}\bigg)^{3}$

$\sf \longrightarrow A = 16000\bigg(\dfrac{105}{100}\bigg)^{3}$

$\sf \longrightarrow A = 16000\bigg( \dfrac{21}{20}\bigg)^{3}$

$\sf \longrightarrow A = 16000\times \dfrac{21 \times 21 \times 21}{20 \times 20 \times 20}$

$\sf \longrightarrow A = 16000\times \dfrac{9261}{8000}$

$\sf \longrightarrow A = 2\times 9261$

$\sf \longrightarrow A = 18522$

$\underline{\boxed{\therefore \sf Amount = Rs.18522}}$

Compound Interest = Amount - Principal

⇢ CI = 18522 - 16000

⇢ CI = 2522

$\underline{\boxed{\therefore \sf Compound\: Interest= Rs.2522}}$