Question

Calculate the compound interest accured on rs.16,000 in 3 years, when the rates of interest for successive years are 10%,12% and 15% respectively

Answer 1

Answer:

amount

=16000(1+10/100)(1+12/100)(1+15/100)

=16000(110/100)(112/100)(115/100)

=₹22668.8

CI=A-P

=22668.8-16000

=₹6668.8

Answer 2

Required answer:-

Question :

• Calculate the compound interest accured on rs.16,000 in 3 years, when the rates of interest for successive years are 10%,12% and 15% respectively.

Solution :

Given that,

• Principal = Rs. 16,000

• Time = 3 years

• Rates of successive years = 10% , 12% , 15%

Formulaes used,

$• \: \dfrac{Interest = Principal × Rate × Time }{100}$

$• \: Amount = \: Principal + Interest \: per \: year$

According to the question we have to calculate the compound interest.

In 1st year

P = Rs. 16,000 ; R = 10% ; T = 1 year

Therefore

$Interest = \: \frac{16000 \times 10 \times 1}{100} = Rs. \: 1600 \\ And, amount = Rs. 16000 + Rs. 1600 = Rs. 17600$

In 2nd year

The new Principal for second year will be Rs. 17,600

Thus,

P = Rs. 17,600 ; R = 12% ; T = 1 year

Therefore

$Interest = \frac{17600 × 12 × 1 }{100} = \: Rs. \: 2112 \\ And, \: amount \: = \: Rs. \: 17600 \: + \: Rs. \: 2112 \: = \: Rs. \: 19712$

In 3rd year

The new Principal for third year will be Rs. 19,712

Thus,

P = Rs. 19712 ; R = 15% ; T = 1 year

Therefore,

$Interest \: = \: \frac{19712 \times 15 \times 1}{100} = \: Rs.2956.80 \\ And, amount = Rs.19712 + Rs.2956.80 = Rs.22,668.80$

Finally,

$Compound \: Interest \: = Final \: amount \: - \: Initial \: Principal \\ Rs.22,668 - Rs.16,000 = Rs.6,668.80$

$Compound \: Interest \: is \: Rs. 6,668.80$