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 8th Std Maths Answers 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.