compound interest practice set 14.1 compound interest practice set 14.1 Standard 8
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Step-by-step explanation:
Given Compound interest practice set 14.1 compound interest practice set 14.1 Standard 8
- Find the amount and compound interest for P = 2000, R = 5% p.a and time = 2 years
- So A = P( 1 + R / 100)^n
- = 2000 (1 + 5/100)^2
- = 2000 (105 / 100)^2
- = 2000 (21 / 20)^2
- = 5 x 441
- A = Rs 2205
- I = A – P
- = 2205 – 2000
- I = Rs 205
- Similarly we can do for Rs 5000 and Rs 4000
- So A = P( 1 + R / 100)^n
- = 5000 (1 + 8/100)^3
- = 5000 (108 / 100)^3
- = 5000 (27 / 25)^3
- = 200 x 27 x 27 / 25 x 27 / 25
- A = Rs 6,298.56
- I = A – P
- = 6298.56 – 5000
- I = Rs 1298.56
- Sameer rao has taken a loan of Rs 12,500 at the rate of 12% p.a for 3 years, if the interest is compounded annually then how many rupees should he pay to clear his loan?
- A = p(1 + r/100)^n
- = 12,500 (1 + 12 / 100)^3
- = 12,500 (112 / 100)^3
- = 12,500 (28 / 25)^3
- = 12,500 x 28 / 25 x 28/25 x 28/25
- = Rs 17,561.60
Sameer rao should pay Rs 17,561.60 to clear his loan
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