## Maharashtra State Board Class 8 Maths Solutions Chapter 14 Compound Interest Practice Set 14.1

Practice Set 14.1 Class 8 Question 1.

Find the amount and the compound interest.

No |
Principal (Rs) |
Rate (p.c.p.a.) |
Duration (years) |

i. | 2000 | 5 | 2 |

ii. | 5000 | 8 | 3 |

iii. | 4000 | 7.5 | 2 |

Solution:

i. Here P = Rs 2000, R = 5 p.c.p.a. and N = 2 years

= 5 × 441

∴ A = Rs 2205

I = Amount (A) – Principal (P)

= 2205 – 2000

= Rs 205

∴ The amount is Rs 2205 and the compound interest is Rs 205.

ii. Here, P = Rs 5000, R = 8 p.c.p.a. and N = 3 years

∴ A = Rs 6298.56

I = Amount (A) – Principal (P)

= 6298.56 – 5000

= Rs 1298.56

∴ The amount is Rs 6298.56 and the compound interest is Rs 1298.56.

iii. Here, P = Rs 4000, R = 7.5 p.c.p.a. and N = 2 years

∴A = Rs 4622.50

I = Amount (A) – Principal (P)

= 4622.50 – 4000

= Rs 622.50

∴The amount is Rs 4622.50 and the compound interest is Rs 622.50.

Compound Interest Practice Set 14.1 Question 2.

Sameerrao has taken a loan of Rs 12500 at the rate of 12 p.c.p.a. for 3 years. If the interest is compounded annually then how many rupees should he pay to clear his loan?

Solution:

Here, P = Rs 12,500, R = 12 p.c.p.a. and

N = 3 years

= 0.8 × 28 × 28 × 28

= Rs 17,561.60

Sameerrao should pay Rs 17,561.60 to clear his loan.

8th Standard Maths Practice Set 14.1 Question 3.

To start a business Shalaka has taken a loan of Rs 8000 at a rate of \(10\frac { 1 }{ 2 }\) p.c.p.a. After two years how much compound interest will she have to pay?

Solution:

Here, P = Rs 8000, N = 2 years and

I = Amount (A) – Principal (P)

= 9768.20 – 8000

= Rs 1768.20

∴ After two years Shalaka will have to pay Rs 1768.20 as compound interest.