Solve the following questions
Q1 amount 7850 ₹ , rate 5 pcpa , duration 2 years . Find principal.
Q2 principal 7000 , rate 10 pcpa, duration ? ,compound interest 700. Find period .
Q3 machine costs rupees 250000, depreciation 4 pcpa , duration 2 years . Find cost of machine after 2 years .
Answers
Answer:
Ans.1) Amount = P(1+i)ⁿ
where P= principal
i = rate of interest /100
n = number of years
∴ 7850 = P (1+0.05)²
∴ 7850 = P(1.05)²
∴ P = 7850 / 1.1025
= Rs. 7120.18
Ans. 2) Interest = P [ (1+i)ⁿ - 1]
∴ 700 = 7000 [ (1.1)ⁿ - 1]
∴ 0.1 = (1.1)ⁿ -1
∴ 0.1 + 1 = (1.1)ⁿ
∴ 1.1 = (1.1)ⁿ
∴ n = 1 year
Ans. 3) Cost of machine after 2 years = cost (1-i)ⁿ
= 250000 (1-0.04)²
= 250000 (0.96)²
= 250000 * 0.9216
= ₹ 230400