Compound Interest Calculations: Without Formula

Question 1: Find the amount and the compound interest on Rs. 10,000 at 8 per cent per annum and in 2 years.
\displaystyle \text{Answer:}
\displaystyle \text{For 1st year: } P = \text{Rs. } 10000;\; R=8\%;\; T=1 \text{ year }
\displaystyle I = \frac{10000 \times 8 \times 1}{100} = \text{Rs. } 800
\displaystyle \text{Amount } = 10000 + 800 = \text{Rs. } 10800
\displaystyle \text{For 2nd year: } P = \text{Rs. } 10800;\; R=8\%;\; T=1 \text{ year }
\displaystyle I = \frac{10800 \times 8 \times 1}{100} = \text{Rs. } 864
\displaystyle A = 10800 + 864 = \text{Rs. } 11664
\displaystyle \text{Compound Interest } = 11664 - 10000 = \text{Rs. } 1664
\\
Question 2: Find the amount and the compound interest on Rs. 10,000 at 8 per cent per annum and in 1 year; interest being compounded half-yearly.
\displaystyle \text{Answer:}
\displaystyle \text{For 1st } \frac{1}{2} \text{ year: } P = \text{Rs. } 10000;\; R=8\%;\; T=\frac{1}{2} \text{ year }
\displaystyle I = \frac{10000 \times 8 \times 1}{100 \times 2} = \text{Rs. } 400
\displaystyle \text{Amount } = 10000 + 400 = \text{Rs. } 10400
\displaystyle \text{For 2nd } \frac{1}{2} \text{ year: } P = \text{Rs. } 10400;\; R=8\%;\; T=\frac{1}{2} \text{ year }
\displaystyle I = \frac{10400 \times 8 \times 1}{100 \times 2} = \text{Rs. } 416
\displaystyle A = 10400 + 416 = \text{Rs. } 10816
\displaystyle \text{Compound Interest } = 10816 - 10000 = \text{Rs. } 816
\\
Question 3: Calculate the compound interest accrued on Rs. 16,000 in 3 years, when the rates of interest for successive years are 10%, 12% and 15% respectively.
\displaystyle \text{Answer:}
\displaystyle \text{For 1st year: } P = \text{Rs. } 16000;\; R=10\%;\; T=1 \text{ year }
\displaystyle I = \frac{16000 \times 10 \times 1}{100} = \text{Rs. } 1600
\displaystyle \text{Amount } = 16000 + 1600 = \text{Rs. } 17600
\displaystyle \text{For 2nd year: } P = \text{Rs. } 17600;\; R=12\%;\; T=1 \text{ year }
\displaystyle I = \frac{17600 \times 12 \times 1}{100} = \text{Rs. } 2112
\displaystyle A = 17600 + 2112 = \text{Rs. } 19712
\displaystyle \text{For 3rd year: } P = \text{Rs. } 19712;\; R=15\%;\; T=1 \text{ year }
\displaystyle I = \frac{19712 \times 15 \times 1}{100} = \text{Rs. } 2956.80
\displaystyle A = 19712 + 2956.80 = \text{Rs. } 22668.80
\displaystyle \text{Compound Interest } = 22668.80 - 16000 = \text{Rs. } 6668.80
\\
Question 4: A man borrows Rs. 8,000 at 10% compound interest payable every six months. He repays Rs. 2500 at the end of every six months. Calculate the third payment he has to make at the end of 18 months in order to clear the entire loan.
\displaystyle \text{Answer:}
\displaystyle \text{For 1st six months: } P = \text{Rs. } 8000;\; R=10\%;\; T=\frac{1}{2} \text{ year }
\displaystyle I = \frac{8000 \times 10 \times 1}{100 \times 2} = \text{Rs. } 400
\displaystyle \text{Amount } = 8000 + 400 = \text{Rs. } 8400
\displaystyle \text{Balance } = 8400 - 2500 = \text{Rs. } 5900
\displaystyle \text{For 2nd six months: } P = \text{Rs. } 5900;\; R=10\%;\; T=\frac{1}{2} \text{ year }
\displaystyle I = \frac{5900 \times 10 \times 1}{100 \times 2} = \text{Rs. } 295
\displaystyle A = 5900 + 295 = \text{Rs. } 6195
\displaystyle \text{Balance } = 6195 - 2500 = \text{Rs. } 3695
\displaystyle \text{For 3rd six months: } P = \text{Rs. } 3695;\; R=10\%;\; T=\frac{1}{2} \text{ year }
\displaystyle I = \frac{3695 \times 10 \times 1}{100 \times 2} = \text{Rs. } 184.75
\displaystyle A = 3695 + 184.75 = \text{Rs. } 3879.75
\displaystyle \text{Final payment } = \text{Rs. } 3879.75
\\
Question 5: On a certain sum of money, invested at the rate of 5% per annum compounded annually, the difference between the interest of the first year and the interest of the third year is Rs. 61.50. Find the sum.
\displaystyle \text{Answer:}
\displaystyle \text{Let the sum (principal) } = \text{Rs. } 100
\displaystyle \text{C.I. of 1st year } = 5
\displaystyle \text{Amount after 1st year } = 105
\displaystyle \text{C.I. of 2nd year } = 5.25
\displaystyle \text{Amount after 2nd year } = 110.25
\displaystyle \text{C.I. of 3rd year } = 5.5125
\displaystyle \text{Difference } = 5.5125 - 5 = 0.5125
\displaystyle \text{If difference } = 0.5125,\; \text{sum } = 100
\displaystyle \text{If difference } = 61.50,\; \text{sum } = \frac{100}{0.5125} \times 61.50 = \text{Rs. } 12000
\\
Question 6: Mrs. Kapoor invested Rs. 6,000 every year at the beginning of the year, at 10% per annum compound interest. Calculate the amount of her total savings: (i) up to the end of the second year, (ii) at the beginning of the third year.
\displaystyle \text{Answer:}
\displaystyle \text{1st year: } I = \frac{6000 \times 10}{100} = 600,\; A = 6600
\displaystyle \text{2nd year principal } = 6600 + 6000 = 12600
\displaystyle I = \frac{12600 \times 10}{100} = 1260,\; A = 13860
\displaystyle \text{Savings at end of 2nd year } = \text{Rs. } 13860
\displaystyle \text{Beginning of 3rd year } = 13860 + 6000 = \text{Rs. } 19860
\\
Question 7: During every financial year, the value of a machine depreciates by 10%. Find the original value (cost) of a machine which depreciates by Rs. 2250 during the second year.
\displaystyle \text{Answer:}
\displaystyle \text{Let original cost } = \text{Rs. } 100
\displaystyle \text{Value after 1st year } = 90
\displaystyle \text{Depreciation in 2nd year } = 9
\displaystyle \text{If depreciation } = 9,\; \text{cost } = 100
\displaystyle \text{If depreciation } = 2250,\; \text{cost } = \frac{100}{9} \times 2250 = \text{Rs. } 25000
\\
Question 8: Calculate the difference between the compound interest and the simple interest on Rs. 4000 at 8 per cent per annum and in 2 years.
\displaystyle \text{Answer:}
\displaystyle \text{S.I. } = \frac{4000 \times 8 \times 2}{100} = 640
\displaystyle \text{1st year C.I.: } 4000 \rightarrow 4320
\displaystyle \text{2nd year C.I.: } 4320 \rightarrow 4665.60
\displaystyle \text{C.I. } = 4665.60 - 4000 = 665.60
\displaystyle \text{Difference } = 665.60 - 640 = \text{Rs. } 25.60
\\
Question 9: Ashok borrowed Rs. 16000 at 10% simple interest. He immediately invested this money at 10% compound interest compounded half-yearly. Calculate Ashok’s gain in 18 months.
\displaystyle \text{Answer:}
\displaystyle \text{S.I. } = \frac{16000 \times 10 \times 3}{100 \times 2} = 2400
\displaystyle \text{1st half-year: } 16000 \rightarrow 16800
\displaystyle \text{2nd half-year: } 16800 \rightarrow 17640
\displaystyle \text{3rd half-year: } 17640 \rightarrow 18522
\displaystyle \text{C.I. } = 18522 - 16000 = 2522
\displaystyle \text{Gain } = 2522 - 2400 = \text{Rs. } 122
\\
Question 10: A sum of money is invested at C.I. payable annually. The amounts of interest in two successive years are Rs. 2700 and Rs. 2880. Find the rate of interest.
\displaystyle \text{Answer:}
\displaystyle \text{Difference} = 2880 - 2700 = 180
\displaystyle \text{Thus } 180 \text{ is interest on } 2700 \text{ for 1 year}
\displaystyle \text{Rate} = \frac{100 \times 180}{2700} = 6\frac{2}{3}\%
\\
Question 11: A certain sum of money, placed out at compound interest, amounts to Rs. 6272 in 2 years and to Rs. 7024.64 in 3 years. Find the rate of interest and the sum of money.
\displaystyle \text{Answer:}
\displaystyle \text{Difference} = 7024.64 - 6272 = 752.64
\displaystyle \text{Interest on 6272 for 1 year} = 752.64
\displaystyle \text{Rate} = \frac{752.64}{6272} \times 100 = 12\%
\displaystyle \text{Let sum } = 100
\displaystyle \text{After 2 years } = 125.44
\displaystyle \text{When } 125.44 \rightarrow 6272,\; 100 \rightarrow P
\displaystyle P = \frac{100}{125.44} \times 6272 = \text{Rs. } 5000
\\
Question 12: A person invests Rs. 10000 for three years at a certain rate of interest compounded annually. At the end of one year this sum amounts to Rs. 11200. Calculate: (i) rate, (ii) amount after 2nd year, (iii) amount after 3rd year.
\displaystyle \text{Answer:}
\displaystyle \text{(i) } I = 11200 - 10000 = 1200
\displaystyle \text{Rate} = \frac{1200 \times 100}{10000} = 12\%
\displaystyle \text{(ii) } 11200 \rightarrow 12544
\displaystyle \text{(iii) } 12544 \rightarrow 14049.28
\\
Question 13: The simple interest on a certain sum computes to Rs. 600 in 3 years and the compound interest on the same sum, at the same rate and for 2 years computes to Rs. 410. Find the rate per cent.
\displaystyle \text{Answer:}
\displaystyle \text{S.I. for 1 year} = 200
\displaystyle \text{C.I. for 1st year} = 200
\displaystyle \text{C.I. for 2nd year} = 410 - 200 = 210
\displaystyle \text{Difference} = 210 - 200 = 10
\displaystyle \text{Rate} = \frac{100 \times 10}{200} = 5\%
\\
Question 14: The compound interest calculated yearly at 10% on a certain sum amounts to Rs. 665.50 in the fifth year. Calculate: (i) C.I. for 6th year, (ii) C.I. for 4th year.
\displaystyle \text{Answer:}
\displaystyle \text{(i) } 665.50 + 10\% \text{ of } 665.50 = 732.05
\displaystyle \text{(ii) Let } x = \text{C.I. for 4th year}
\displaystyle 665.50 = x + 0.1x \Rightarrow x = 605
\displaystyle \text{C.I. for 4th year} = \text{Rs. } 605
\\

Question 15: A sum of money, at compound interest, amounts to Rs. 8100 in 5 years and to Rs. 8748 in 6 years. Find: (i) the rate per cent (ii) amount in 7 years and (iii) amount in 4 years
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since amount in 5 years } = \text{Rs. } 8100
\displaystyle \text{Amount in 6 years } = \text{Rs. } 8748
\displaystyle \text{Therefore } 8748 - 8100 = \text{Rs. } 648 \text{ is the interest of 1 year on Rs. } 8100
\displaystyle \text{Therefore Rate } = \frac{648 \times 100}{8100 \times 1} = 8\%
\displaystyle \text{(ii) Amount in 7 years } = 8748 + 8\% \text{ of } 8748 = \text{Rs. } 9447.84
\displaystyle \text{(iii) Let amount in 4 years } = \text{Rs. } x
\displaystyle \text{Amount in 5 years} = \text{Amount in 4 years} + \text{interest on it for 1 year}
\displaystyle \Rightarrow 8100 = x + 8\% \text{ of } x
\displaystyle \Rightarrow 8100 = 1.08x
\displaystyle \Rightarrow x = 7500
\displaystyle \text{Therefore Amount in 4 years } = \text{Rs. } 7500
\\
Question 16: The cost of a machine depreciated by Rs. 4752 during the second year and by Rs. 4181.76 during the third year. Calculate: (i) the rate of depreciation; (ii) the original cost; (iii) the cost at the end of the third year.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Difference between depreciations of 2nd year and 3rd year } = 4752 - 4181.76 = \text{Rs. } 570.24
\displaystyle \text{Therefore depreciation of one year on Rs. } 4752 = \text{Rs. } 570.24
\displaystyle \text{Therefore rate of depreciation } = \frac{570.24}{4752} \times 100 = 12\%
\displaystyle \text{(ii) Let original cost of the machine be } \text{Rs. } 100
\displaystyle \text{Its value after one year } = 100 - 12\% \text{ of } 100 = \text{Rs. } 88
\displaystyle \text{Depreciation during 2nd year } = 12\% \text{ of } 88 = \text{Rs. } 10.56
\displaystyle \text{When depreciation during 2nd year } = \text{Rs. } 10.56,\; \text{original cost } = \text{Rs. } 100
\displaystyle \text{When depreciation during 2nd year } = \text{Rs. } 4752,\; \text{original cost } = \frac{100}{10.56} \times 4752 = \text{Rs. } 45000
\displaystyle \text{(iii) Total depreciation in 3 years } = 12\% \text{ of } 45000 + 4752 + 4181.76 = \text{Rs. } 14333.76
\displaystyle \text{Cost of machine at the end of the 3rd year } = 45000 - 14333.76 = \text{Rs. } 30666.24

Compound Interest Calculations: With Formula

Question 1: Calculate the amount and the compound interest on Rs. 7500 in 2 years and at 6% compounded annually.
\displaystyle \text{Answer:}
\displaystyle P=7500 \text{ Rs.; } \hspace{1.0cm} r=6\%; \hspace{1.0cm} n=2 \text{ years }
\displaystyle \text{ Amount: } A=P \Big(1+ \frac{r}{100} \Big)^n = 7500 \Big(1+ \frac{6}{100} \Big)^2 = 8427 \text{ Rs. }
\displaystyle \text{Compound Interest: } A - P = 8427 - 7500 = 927 \text{ Rs. }
\\
Question 2: Calculate the amount and the compound interest on Rs. 12.000 in 3 years when the rates of interest for successive years are 8%, 10% and 15% respectively.
\displaystyle \text{Answer:}
\displaystyle \text{ Amount: } A=P \Big(1+ \frac{r_1}{100} \Big)\Big(1+ \frac{r_2}{100} \Big)\Big(1+ \frac{r_3}{100} \Big)
\displaystyle =12000 \Big(1+ \frac{8}{100} \Big) \Big(1+ \frac{10}{100} \Big) \Big(1+ \frac{15}{100} \Big)
\displaystyle = \text{Rs. } 16394.40
Question 3: Calculate the compound interest on Rs. 18000 in 2 years at 15% per annum.
\displaystyle \text{Answer:}
\displaystyle P=18000 \text{ Rs.; } \hspace{1.0cm} r=15\%; \hspace{1.0cm} n=2 \text{ years }
\displaystyle \text{ Amount: } A=P \Big(1+ \frac{r}{100} \Big)^n = 18000 \Big(1+ \frac{15}{100} \Big)^2 = 23805 \text{ Rs. }
\displaystyle \text{Compound Interest: } A - P = 23805-18000 = 5805 \text{ Rs. }
\\
Question 4: Divide Rs. 36465 between A and B so that when their shares are lent out at 10 per cent compound interest per year, the amount that A receives in 7 years is the same as what B receives in 5 years.
\displaystyle \text{Answer:}
\displaystyle \text{Let A's share } = x
\displaystyle \text{Therefore B's share } = (36465 - x)
\displaystyle \text{Given: Amount of A in 7 years = Amount of B in 5 years }
\displaystyle \Rightarrow x \Big( 1 + \frac{10}{100} \Big)^7 = (36465 - x) \Big( 1 + \frac{10}{100} \Big)^5
\displaystyle \Rightarrow x \Big( 1 + \frac{10}{100} \Big)^2 = (36465 - x)
\displaystyle \Rightarrow \frac{121}{100} x=36465 - x
\displaystyle \Rightarrow x = \text{Rs. } 16500
\displaystyle \Rightarrow 36465- x = 36465 - 16500 = \text{Rs. } 19965
\displaystyle \text{Therefore A's share= Rs. 16500 and B's share = Rs. 19965 }
\\
Question5: Calculate the compound interest on Rs. 4000 in \displaystyle 1\frac{1}{2} years at 10% per annum compounded half-yearly.
\displaystyle \text{Answer:}
\displaystyle A = P \Big( 1 + \frac{r}{2 \times 100} \Big)^{n \times 2}
\displaystyle \Rightarrow 4000 \Big( 1 + \frac{10}{2 \times 100} \Big)^{\frac{3}{2} \times 2} = 4630.50
\displaystyle C.I. = A - P = 4630.50 - 4000 = \text{Rs. } 630.50
\\
Question 6: Find the amount when Rs. 10000 is invested for \displaystyle 2\frac{1}{2} years at 10% interest compounded yearly.
\displaystyle \text{Answer:}
\displaystyle A = P \Big( 1 + \frac{r}{100} \Big)^2 \Big( 1 + \frac{r}{2 \times 100} \Big)^{\frac{1}{2}\times 2}
\displaystyle \Rightarrow A = 10000 \Big( 1 + \frac{10}{100} \Big)^2 \Big( 1 + \frac{10}{2 \times 100} \Big) = \text{Rs. } 12705
\displaystyle \text{Amount in } 2\frac{1}{2} \text{ years } = \text{Rs. } 12705
\\
Question 7: John borrowed Rs. 20000 for 4 years under the following conditions :
10% simple interest for the first \displaystyle 2\frac{1}{2} years.
10% C.I. for the remaining one and a half years on the amount due after \displaystyle 2\frac{1}{2} years.
The interest being compounded half yearly.
\displaystyle \text{Answer:}
\displaystyle \text{For first } 2\frac{1}{2} \text{ years: } P = \text{Rs. }20000; \hspace{1.0cm} r = 10\%; \hspace{1.0cm} T = 2\frac{1}{2} \text{ years }
\displaystyle \text{Therefore Interest } = \frac{20000 \times 10 \times 5}{100 \times 2} = \text{Rs. } 5000
\displaystyle \text{Therefore amount due after } 2\frac{1}{2} \text{ years } = 20000 + 5000 = \text{Rs. } 25000
\displaystyle \text{For remaining } 1\frac{1}{2} \text{ years: } P = \text{Rs. } 25000; \hspace{1.0cm} n = 1\frac{1}{2} \text{ years; } r = 10\% \text{ p.a. compounded half-yearly }
\displaystyle \text{Therefore } A = 25000 \Big( 1 + \frac{10}{2 \times 100} \Big)^{\frac{3}{2} \times 2} = \text{Rs. } 28940.63
\displaystyle \text{Therefore the total amount to be paid at the end of 4 years } = \text{Rs. } 28940.63
\\
Question 8: A sum of money is lent out at compound interest for two years at 20% per annum compound interest being reckoned yearly. If the same sum of money was lent out at compound interest at the same rate per cent per annum, compound interest being reckoned half-yearly, it would have fetched Rs. 482 more by way of interest in two years. Calculate the sum of money lent out.
\displaystyle \text{Answer:}
\displaystyle \text{Let the sum of money be Rs. } x
\displaystyle \text{In 1st Case: } A_1 = x \Big( 1 + \frac{20}{100} \Big)^2 = \frac{36x}{25}
\displaystyle \text{In the 2nd case: } A_2 = x \Big( 1 + \frac{20}{2 \times 100} \Big)^{2 \times 2} = \frac{14641x}{10000}
\displaystyle \text{Given difference in amounts } = 482
\displaystyle \Rightarrow \frac{14641x}{10000} - \frac{36x}{25} = 482
\displaystyle \Rightarrow x = \text{Rs. } 20000
\displaystyle \text{Therefore the sum of money } = \text{Rs. } 20000
\\
Question 9: A sum of Rs. 6400 earns a compound interest of Rs. 1008.80 in 18 months, when the interest is reckoned half-yearly. Find the rate of interest
\displaystyle \text{Answer:}
\displaystyle \text{Given: } P = \text{Rs. } 6400; \hspace{1.0cm} C.I. = 1008.80; \hspace{1.0cm} T = 18 \text{ months } = 1\frac{1}{2} \text{ years }
\displaystyle \text{Therefore Amount } A = 6400 + 1008.80 = \text{Rs. } 7408.80
\displaystyle 7408.80 = 6400 \Big( 1 + \frac{r}{2 \times 100} \Big)^{\frac{3}{2} \times 2}
\displaystyle \Rightarrow \Big( 1 + \frac{r}{200} \Big)^3 = \frac{7408.80}{6400} = \Big( \frac{21}{20} \Big)^3
\displaystyle \Rightarrow r = 10\%
\\
Question 10: Calculate the amount when a sum of Rs. 4800 is invested at 8% per annum for 4 years, the C.I. being compounded half-yearly. Do not use mathematical tables, use the necessary information from the following:
\displaystyle (1.08)^4 = 1.3605 \hspace{0.5cm} (1.04)^8 = 1.3686 \hspace{0.5cm}(1.08)^8 = 1.8509 \hspace{0.5cm}(1.04)^4 = 1.1699
\displaystyle \text{Answer:}
\displaystyle \text{Given: } P = \text{Rs. } 4800,\; r = 8\% \text{ compounded half-yearly, } n = 4 \text{ years }
\displaystyle A = P \Big( 1 + \frac{r}{2 \times 100} \Big)^{n \times 2}
\displaystyle = 4800 \Big( 1 + \frac{8}{2 \times 100} \Big)^{4 \times 2}
\displaystyle = 4800 (1.04)^8 = 4800 \times 1.3686 = 6569.28
\\
Question 11: The simple interest on a sum of money for 2 years at 4% per annum is Rs. 340. Find : (i) the sum of money and (ii) the compound interest on this sum for one year payable half-yearly at the same rate
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given: } I = \text{Rs. } 340,\; T = 2 \text{ years, } R = 4\%
\displaystyle \therefore P = \frac{I \times 100}{R \times T} = \frac{340 \times 100}{4 \times 2} = 4250
\displaystyle \text{(ii) } C.I. = P \Big( 1 + \frac{r}{2 \times 100} \Big)^{n \times 2} - P
\displaystyle = 4250 \Big( 1 + \frac{4}{2 \times 100} \Big)^{1 \times 2} - 4250 = 4421.70 - 4250 = 171.70
\\
Question 12: The total number of industries in a particular portion of the country is approximately 1600. If the government has decided to increase the number of industries in the area by 20% every year; find the approximate number of industries after 2 years.
\displaystyle \text{Answer:}
\displaystyle \text{Number after 2 years} = 1600 \Big( 1 + \frac{20}{100} \Big)^2 = 2304
\\
Question 13: The cost of a machine depreciates by 10% every year. If its present worth is Rs. 18000; what will be its value after three years?
\displaystyle \text{Answer:}
\displaystyle \text{Value after 3 years} = 18000 \Big( 1 - \frac{10}{100} \Big)^3 = \text{Rs. } 13122
\\
Question 14: A machine depreciates every year at the rate of 20% of its value at the beginning of the year. The machine was purchased for Rs. 250000 when new, and the scrap value realized when sold was Rs. 128000. Find the number of years that the machine was used.
\displaystyle \text{Answer:}
\displaystyle \text{Let } n \text{ be the number of years}
\displaystyle 128000 = 250000 \Big( 1 - \frac{20}{100} \Big)^n
\displaystyle \Rightarrow \frac{128000}{250000} = \Big( \frac{4}{5} \Big)^n
\displaystyle \Rightarrow \Big( \frac{4}{5} \Big)^3 = \Big( \frac{4}{5} \Big)^n
\displaystyle \Rightarrow n = 3
\\
Question 15: The population of a town in China increases by 20% every year. If its present population is 2,16,000, find : (i) its population after 2 years, (ii) its population 2 years ago.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Population after 2 years} = 216000 \Big( 1 + \frac{20}{100} \Big)^2 = 311040
\displaystyle \text{(ii) } 216000 = P \Big( 1 + \frac{20}{100} \Big)^2
\displaystyle \Rightarrow P = 150000
\\
Question 16: A certain sum of money, lent out at compound interest, amounts to Rs. 14520 in 2 years and to Rs. 17569.20 in 4 years. Find the rate of interest per annum and the sum.
\displaystyle \text{Answer:}
\displaystyle P \Big( 1 + \frac{r}{100} \Big)^2 = 14520
\displaystyle P \Big( 1 + \frac{r}{100} \Big)^4 = 17569.20
\displaystyle \Rightarrow \Big( 1 + \frac{r}{100} \Big)^2 = \frac{17569.20}{14520} = \frac{121}{100}
\displaystyle \Rightarrow 1 + \frac{r}{100} = \frac{11}{10}
\displaystyle \Rightarrow r = 10\%
\displaystyle P \Big( \frac{121}{100} \Big) = 14520
\displaystyle \Rightarrow P = \text{Rs. } 12000
\displaystyle \text{Therefore rate = 10\% and sum = Rs. } 12000
\\
Question 17: The difference between the compound interest and the simple interest on Rs. 9500 for 2 years is Rs. 95 at the same rate of interest per year. Find the rate of interest
\displaystyle \text{Answer:}
\displaystyle \text{Let the rate of interest be } r\%
\displaystyle \text{S.I. in 2 years } = \frac{9500 \times r \times 2}{100} = 190r
\displaystyle \text{C.I. in 2 years } = 9500 \Big( 1 + \frac{r}{100} \Big)^2 - 9500
\displaystyle \text{Given } C.I. - S.I. = 95
\displaystyle \Rightarrow 9500 \Big( 1 + \frac{r}{100} \Big)^2 - 9500 - 190r = 95
\displaystyle \Rightarrow 100 \Big( 1 + \frac{r}{100} \Big)^2 - 100 - 2r = 1
\displaystyle \Rightarrow 100 + \frac{r^2}{100} + 2r - 100 - 2r = 1
\displaystyle \Rightarrow \frac{r^2}{100} = 1
\displaystyle \Rightarrow r = 10\%
\displaystyle \text{Therefore rate of interest } = 10\%
\\
Question 18: A sum of money lent out at C.I. at a certain rate per annum doubles itself in 5 years. Find in how many years will the money become eight times of itself at the same rate of interest p.a.
\displaystyle \text{Answer:}
\displaystyle \text{Let Principal = Rs. } x
\displaystyle \Rightarrow 2x = x \Big( 1 + \frac{r}{100} \Big)^5
\displaystyle \Rightarrow 2 = \Big( 1 + \frac{r}{100} \Big)^5
\displaystyle \text{For second part: } 8x = x \Big( 1 + \frac{r}{100} \Big)^n
\displaystyle \Rightarrow 8 = \Big( 1 + \frac{r}{100} \Big)^n
\displaystyle \Rightarrow (2)^3 = \Big( 1 + \frac{r}{100} \Big)^n
\displaystyle \Rightarrow \Big[ \Big( 1 + \frac{r}{100} \Big)^5 \Big]^3 = \Big( 1 + \frac{r}{100} \Big)^n
\displaystyle \Rightarrow \Big( 1 + \frac{r}{100} \Big)^{15} = \Big( 1 + \frac{r}{100} \Big)^n
\displaystyle \Rightarrow n = 15
\displaystyle \text{Hence required time = 15 years}
\\
Question 19: A man borrowed a sum of money and agrees to pay it off by paying Rs. 43200 at the end of the first year and Rs. 34992 at the end of the second year. If the rate of compound interest is 8% per annum, find the sum borrowed.
\displaystyle \text{Answer:}
\displaystyle \text{For Rs. 43200 after 1 year: } 43200 = P \Big( 1 + \frac{8}{100} \Big)
\displaystyle \Rightarrow P = 43200 \times \frac{100}{108} = \text{Rs. } 40000
\displaystyle \text{For Rs. 34992 after 2 years: } 34992 = P \Big( 1 + \frac{8}{100} \Big)^2
\displaystyle \Rightarrow P = 34992 \times \Big( \frac{100}{108} \Big)^2 = \text{Rs. } 30000
\displaystyle \text{Therefore sum borrowed } = 40000 + 30000 = \text{Rs. } 70000
\\
Question 20: A, B and C are three persons with ages 26 years, 27 years and 28 years respectively. In what ratio must they invest money at 10% p.a. compounded yearly so that each gets same sum at the age of his retirement (i.e. at the age of 58 years).
\displaystyle \text{Answer:}
\displaystyle x \Big( \frac{11}{10} \Big)^{32} = y \Big( \frac{11}{10} \Big)^{31} = z \Big( \frac{11}{10} \Big)^{30}
\displaystyle \Rightarrow x \Big( \frac{11}{10} \Big)^2 = y \Big( \frac{11}{10} \Big) = z
\displaystyle \Rightarrow x \frac{121}{100} = y \frac{11}{10} = z
\displaystyle \Rightarrow 121x = 110y = 100z = k
\displaystyle \Rightarrow x = \frac{k}{121},\; y = \frac{k}{110},\; z = \frac{k}{100}
\displaystyle \Rightarrow x : y : z = \frac{k}{121} : \frac{k}{110} : \frac{k}{100}
\displaystyle \Rightarrow x : y : z = 100 : 110 : 121
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