Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 14.1 8th Std Maths Answers Solutions Chapter 14 Compound Interest.
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.