Time, T = 0.35 + (69 / R) Years
When rate of interest is different for different years, say R1, R2, R3 and so on, the amount is calculated as A = P [1 + (R1 / 100)] [1 + (R2 / 100)] [1 + (R3 / 100)] …
Question 1 : Find the compound interest on Rs. 10,000 at 10% per annum for a time period of three and a half years.
Solution : Time period of 3 years and 6 months means for 3 years, the interest is compounded yearly and for the remaining 6 months, the interest is compounded compounded half yearly. This means that we have 3 cycles of interest compounded yearly and 1 cycle of interest compounded half yearly.
So, Amount = P [1 + (R / 100)]^{3} [1 + ( {R/2} / 100 )]
=> Amount = 10000 [1 + 0.1]^{3} [1 + 0.05]
=> Amount = 10000 (1.1)^{3} (1.05)
=> Amount = Rs. 13975.50
=> Compound Interest, CI = Amount – Principal = 13975.50 – 10000
Therefore, CI = Rs. 3975.50
Question 2 : If Rs. 5000 amounts to Rs. 5832 in two years compounded annually, find the rate of interest per annum.
Solution : Here, P = 5000, A = 5832, n = 2
A = P [1 + (R / 100)]^{n}
=> 5832 = 5000 [1 + (R / 100)]^{2}
=> [1 + (R / 100)]^{2} = 5832 / 5000
=> [1 + (R / 100)]^{2} = 11664 / 10000
=> [1 + (R / 100)] = 108 / 100
=> R / 100 = 8 / 100
=> R = 8 %
Thus, the required rate of interest per annum in 8 %
Question 3 : The difference between the SI and CI on a certain sum of money at 10 % rate of annual interest for 2 years is Rs. 549. Find the sum.
Solution : Let the sum be P.
R = 10 %
n = 2 years
SI = P x R x n / 100 = P x 10 x 2 / 100 = 0.20 P
CI = A – P = P [1 + (R / 100)]^{n} – P = 0.21 P
Now, it is given that CI – SI = 549
=> 0.21 P – 0.20 P = 549
=> 0.01 P = 549
=> P = 54900
Therefore, the required sum of money is Rs. 54,900
Question 4 : A sum of Rs. 1000 is to be divided among two brothers such that if the interest being compounded annually is 5 % per annum, then the money with the first brother after 4 years is equal to the money with the second brother after 6 years.
Solution : Let the first brother be given Rs. P
=> Money with second brother = Rs. 1000 – P
Now, according to the question,
P [1 + (5 / 100)]^{4} = (1000 – P) [1 + (5 / 100)]^{6}
=> P (1.05)^{4} = (1000 – P) (1.05)^{6}
=> 0.9070 P = 1000 – P
=> 1.9070 P = 1000
=> P = 524.38
Therefore, share of first brother = Rs. 524.38
Share of second brother = Rs. 475.62
Question 5 : A sum of money amounts to Rs. 669 after 3 years and to Rs. 1003.50 after 6 years on compound interest. Find the sum.
Solution : Let the sum of money be Rs. P
=> P [1 + (R/100)]^{3}= 669 and P [1 + (R/100)]^{6}= 1003.50
Dividing both equations, we get
[1 + (R/100)]^{3} = 1003.50 / 669 = 1.50
Now, we put this value in the equation P [1 + (R/100)]^{3}= 669
=> P x 1.50 = 669
=> P = 446
Thus, the required sum of money is Rs. 446
Question 6 : An investment doubles itself in 15 years if the interest is compounded annually. How many years will it take to become 8 times?
Solution : it is given that the investment doubles itself in 15 years.
Let the initial investment be Rs. P
=> At the end of 15 years, A = 2 P
Now, this 2 P will be invested.
=> Amount after 15 more years = 2 x 2 P = 4 P
Now, this 4 P will be invested.
=> Amount after 15 more years = 2 x 4 P = 8 P
Thus, the investment (P) will become 8 times (8 P) in 15 + 15 + 15 = 45 years
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