Class 10: Compound Interest (Without Using Formula) – Miscellaneous Problems -Set 1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the amount and the compound interest on Rs. 10,000 at 8 per cent per annum and in 2 years.

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 10000; R=8\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } I = \frac{10000 \times 8 \times 1}{100} = \text{ Rs. } 800$

$\displaystyle \text{and, Amount } = P + I = 10000 + 800 = \text{ Rs. } 10800$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 10800; R=8\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest I } = \frac{10800 \times 8 \times 1}{100} = \text{ Rs. } 864$

$\displaystyle \text{and, Amount } A = 10800 + 864 = \text{ Rs. } 11664$

$\displaystyle \text{and Compound Interest } = A - P = 11664 - 10000 = \text{ Rs. } 1664$

$\\$

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.

Answer:

$\displaystyle \text{For 1st } \frac{1}{2} \text{ year: } P = \text{ Rs. } 10000; R=8\% \text{ and } T=\frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest } I = \frac{10000 \times 8 \times 1}{100 \times 2} = \text{ Rs. } 400$

$\displaystyle \text{and, Amount } = P + I = 10000 + 400 = \text{ Rs. } 10400$

$\displaystyle \text{For 2nd } \frac{1}{2} \text{ year: } P = \text{ Rs. } 10400; R=8\% \text{ and } T=\frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest I } = \frac{10400 \times 8 \times 1}{100 \times 2} = \text{ Rs. } 416$

$\displaystyle \text{and, Amount } A = 10400 + 416 = \text{ Rs. } 10816$

$\displaystyle \text{and Compound Interest } = A - P = 10816 - 10000 = \text{ Rs. } 816$

$\\$

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.

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 16000; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } I = \frac{16000 \times 10 \times 1}{100} = \text{ Rs. } 1600$

$\displaystyle \text{and, Amount } = P + I = 16000 + 1600 = \text{ Rs. } 17600$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 17600; R=12\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest I } = \frac{17600 \times 12 \times 1}{100} = \text{ Rs. } 2112$

$\displaystyle \text{and, Amount } A = 17600 + 2112 = \text{ Rs. } 19712$

$\displaystyle \text{For 3rd year: } P = \text{ Rs. } 19712; R=15\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest I } = \frac{19712 \times 15 \times 1}{100} = \text{ Rs. } 2956.80$

$\displaystyle \text{and, Amount } A = 19712 + 2956.80 = \text{ Rs. } 22668.80$

$\displaystyle \text{and Compound Interest } = A - P = 22668.80- 16000 = \text{ Rs. } 6668.80$

$\\$

Question 4: A person borrows Rs. 20,000 at 12 per cent C.I. If he repays Rs. 8,400 at the end of first year and Rs. 9,680 at the end of second year, find the amount of loan outstanding at the beginning of the third year. [ ICSE Board 2014]

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 20000; R=12\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } I = \frac{20000 \times 12 \times 1}{100} = \text{ Rs. } 2400$

$\displaystyle \text{and, Amount } = P + I = 20000 +24800 = \text{ Rs. } 22400$

Since, the man pays Rs. 8400 at the end of 1st year Principal for 2nd year $= 22400 - 8400 = \text{ Rs. }. 14000$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 14000; R=12\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest I } = \frac{14000 \times 12 \times 1}{100} = \text{ Rs. } 1680$

$\displaystyle \text{and, Amount } A = 14000 + 1680 = \text{ Rs. } 15680$

$\displaystyle \text{and Compound Interest } = A - P = 11664 - 10000 = \text{ Rs. } 1664$

Since, the man pays Rs. 9680 at the end of 2nd year Principal for 3rd year $= 15680- 9680 = \text{ Rs. } 6000$

$\\$

Question 5: 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.

Answer:

$\displaystyle \text{For 1st six months} P = \text{ Rs. } 8000; R=10\% \text{ and } T=\frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest } I = \frac{8000 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 400$

$\displaystyle \text{and, Amount } = P + I = 8000 + 400 = \text{ Rs. } 8400$

Money repaid $= \text{ Rs. } 2500$ Balance $= 8400 - 2500 = \text{ Rs. } 5900$

$\displaystyle \text{For 1nd six months} P = \text{ Rs. } 5900; R=10\% \text{ and } T=\frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest I } = \frac{5900 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 295$

$\displaystyle \text{and, Amount } A = 5900 + 295 = \text{ Rs. } 6195$

Money repaid $= \text{ Rs. } 2500$ Balance $= 6195 - 2500 = \text{ Rs. } 3695$

$\displaystyle \text{For 3rd six months} P = \text{ Rs. } 3695; R=10\% \text{ and } T=\frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest I } = \frac{3695 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 184.75$

$\displaystyle \text{and, Amount } A = 3695 + 184.75 = \text{ Rs. } 3879.75$

The 3rd instalment to be made to clear the entire loan $= \text{ Rs. } 3,879.75$

$\\$

Question 6: 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.

Answer:

$\displaystyle \text{Let the sum (principal) } = \text{ Rs. } 100$

$\displaystyle \text{C.I. of 1st year } = \frac{100 \times 5 \times 1}{100} = \text{ Rs. } 5$

$\displaystyle \text{And amount at the end of 1st year } = 100 + 5 = \text{ Rs. } 105$

$\displaystyle \text{Principal for 2nd year } = \text{ Rs. } 105$

$\displaystyle \text{C.I. of 2nd year } = \frac{105 \times 5 \times 1}{100} = \text{ Rs. } 5.25$

$\displaystyle \text{And amount at the end of 2ndyear } = 105 + 5.25 = \text{ Rs. } 110.25$

$\displaystyle \text{Principal for 3rd year } = \text{ Rs. } 110.25$

$\displaystyle \text{C.I. of 3rd year } = \frac{110.25 \times 5 \times 1}{100} = \text{ Rs. } 5.5125$

$\displaystyle \text{Now, when the difference of interest }= \text{ Rs. } 0.5125, \text{ sum } = \text{ Rs. } 100$

$\displaystyle \text{And, when the difference of interest }= \text{ Rs. } 61.50, \\ \\ \text{ sum }= \frac{100}{0.5125} \times 61.50 = \text{ Rs. }12000$

$\\$

Question 7: 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.

Answer:

For 1st Year :

$\displaystyle \text{Since, money invested at the beginning of the year } = \text{ Rs. } 6000$

$\displaystyle \text{Therefore Principal for 1st 1 year } = \text{ Rs. } 6000$

$\displaystyle \text{Therefore Interest I } = \frac{6000 \times 10 \times 1}{100 } = \text{ Rs. } 600$

$\displaystyle \text{and, Amount } A = 6000 + 600 = \text{ Rs. } 6600$

2nd Year :

Since, Rs. 6000 is invested again at the beginning of the second year,

Therefore, for the second year, principal $= 6600 + 6000 = \text{ Rs. }12600$

$\displaystyle \text{Therefore Interest I } = \frac{12600 \times 10 \times 1}{100 } = \text{ Rs. } 1260$

$\displaystyle \text{and, Amount } A = 12600 + 1260 = \text{ Rs. } 13860$

Amount of her total savings upto the end of the second year $= \text{ Rs. } 13860$

Since, Rs. 6000 is invested again at the beginning of the third year,

Amount of her total savings at the beginning of the third year $= 13860+6000 = \text{ Rs. }19860$

$\\$

Question 8: 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.

Answer:

Let the original cost of the machine $\displaystyle = \text{ Rs. } 100$

Depreciation during 1st year $\displaystyle = 10% of \text{ Rs. } 100 = \text{ Rs. } 10$

Value of the machine at the beginning of 2nd year $\displaystyle = \text{ Rs. } 100- \text{ Rs. }10= \text{ Rs. } 90$

Depreciation during 2nd year $\displaystyle = 10% of \text{ Rs. } 90 = \text{ Rs. } 9$

Now, when depreciation during 2nd year $\displaystyle = \text{ Rs. } 9$

$\displaystyle \text{Original cost } = \text{ Rs. } 100$

Therefore when depreciation during 2nd year $\displaystyle = \text{ Rs. } 2250$

$\displaystyle \text{Original cost } = \frac{100}{9} \times 2250 = \text{ Rs. } 25000$

$\\$

Question 9: Calculate the difference between the compound interest and the simple interest on Rs. 4000 at 8 per cent per annum and in 2 years.

Answer:

$\displaystyle \text{For Simple Interest} P = \text{ Rs. } 4000; R=8\% \text{ and } T=2 \text{ year }$

$\displaystyle \text{Therefore Simple Interest I } = \frac{4000 \times 8 \times 2}{100 } = \text{ Rs. } 640$

$\displaystyle \text{Compound Interest for 1st Year } P = \text{ Rs. } 4000; R=8\% \text{ and } T=12 \text{ year }$

$\displaystyle \text{Therefore Interest I } = \frac{4000 \times 8 \times 1}{100 } = \text{ Rs. } 320$

$\displaystyle \text{and, Amount } A = 4000 + 320 = \text{ Rs. } 4320$

$\displaystyle \text{Compound Interest for 2nd Year } P = \text{ Rs. } 4320; R=8\% \text{ and } T=12 \text{ year }$

$\displaystyle \text{Therefore Interest I } = \frac{4320 \times 8 \times 1}{100 } = \text{ Rs. } 345.60$

$\displaystyle \text{and, Amount } A = 4320 + 345.60 = \text{ Rs. } 4665.60$

Required difference between C.I. and S.I. $= 665.60-640 = \text{ Rs. } 25.60$

$\\$

Question 10: 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.

Answer:

$\displaystyle \text{For Simple Interest} P = \text{ Rs. } 16000; R=10\% \text{ and } T=\frac{3}{2} \text{ year }$

$\displaystyle \text{Therefore Simple Interest I } = \frac{16000 \times 10 \times 3}{100 \times 2 } = \text{ Rs. } 2400$

Compound Interest

$\displaystyle \text{For 1st } \frac{1}{2} \text{ year: } P = \text{ Rs. } 16000; R=10\% \text{ and } T=\frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest } I = \frac{16000 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 800$

$\displaystyle \text{and, Amount } = P + I = 16000 + 800 = \text{ Rs. } 16800$

$\displaystyle \text{For 2nd } \frac{1}{2} \text{ year: } P = \text{ Rs. } 16800; R=10\% \text{ and } T=\frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest I } = \frac{16800 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 840$

$\displaystyle \text{and, Amount } A = 16800 + 840 = \text{ Rs. } 17640$

$\displaystyle \text{For 3rd } \frac{1}{2} \text{ year: } P = \text{ Rs. } 17640; R=10\% \text{ and } T=\frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest I } = \frac{17640 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 882840$

$\displaystyle \text{and, Amount } A = 17640 + 882 = \text{ Rs. } 18522$

Ashok’s gain in 18 months = C.I. earned – S.I. paid $= 2522 - 2400 = \text{ Rs. }122$

$\\$

Question 11: A sum of money is invested at C.I. payable annually. The amounts of interest in two successive years are Rs. 2700 and Rs. 2,880. Find the rate of interest.

Answer:

$\displaystyle \text{Difference between C.I. of two successive years} = 2880 - 2700 = \text{ Rs. } 180$

$\displaystyle \text{Therefore Rs. 180 is the interest of one year on Rs. 2700}$

$\displaystyle \text{Therefore Rate of Interest} = \frac{100 \times I}{P \times T} = \frac{100 \times 180}{2700 \times 1} = 6\frac{2}{3} \%$

Directly:

$\displaystyle \text{Rate of Interest} = \frac{\text{Difference in interest of two consecutive periods} \times 100 }{\text{C.I. of preceding year} \times \text{Time} } \\ \\ { \hspace{3.0cm} = \frac{(2880-2700) \times 100 }{2700 \times 1} = 6\frac{2}{3} \% }$

$\\$

Question 12: 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.

Answer:

$\displaystyle \text{Difference between the amounts of two consecutive years } \\ \\ = 7024.64 - 6272 = \text{ Rs. } 752.64$

$\displaystyle \text{Therefore Interest for one year on 6272 } = \text{ Rs. } 752.64$

$\displaystyle \text{Therefore Rate of Interest } = \frac{752.64}{6272} \times 100 = 12 \%$

Directly:

$\displaystyle \text{Rate of Interest} = \frac{\text{Difference between the amounts of two consecutive periods} \times 100 }{\text{Preceding amount} \times \text{Time} } \\ \\ { \hspace{3.0cm} = \frac{(7024.64 - 6272) \times 100 }{6272 \times 1} = 12 \% }$

$\displaystyle \text{Let the sum of money } = \text{ Rs. } 100$

$\displaystyle \text{Therefore Interest on it for 1st year } = 12\% \text{ of } \text{ Rs. } 100 = \text{ Rs. } 12$

$\displaystyle \text{Therefore Amount in one year } = 100 + 12 = \text{ Rs. } 112$

$\displaystyle \text{Similarly, amount in two years } = 112 + 12\% \text{ of } 112 = \text{ Rs. }125.44$

$\displaystyle \text{When amount in two years } = \text{ Rs. }125.44, \text{ sum } = \text{ Rs. } 100$

$\displaystyle \text{Therefore When amount in two years } = \text{ Rs. }6272, \\ \\ \text{ Sum } = \frac{100}{125.44} \times 6272 = \text{ Rs. } 5000$

$\\$

Question 13: 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) the rate of interest per annum.
(ii) the amount at the end of the second year.
(iii) the amount at the end of the third year [ ICSE Board 2006]

Answer:

(i) For the first year:

$\displaystyle \text{Principal } = \text{ Rs. }10000 \text{ and Amount }= \text{ Rs. }11200$

$\displaystyle \text{I = A - P } = 11200 - 10000 = \text{ Rs. }1200$

$\displaystyle \text{Rate } = \frac{I \times 100}{P \times T} \%$

$\displaystyle \text{Therefore Rate of Interest p.a. }= \frac{1200 \times 100}{10000 \times 1 } =12\%$

(ii) For the second year:

$\displaystyle \text{Principal } = \text{ Rs. }11200 , \text{ Rate } = 12\% \text{ and } T = 1 \text{ year }$

$\displaystyle \text{I } = \frac{P \times R \times T}{100} = \frac{11200 \times 12 \times 1}{100} = \text{ Rs. }1344$

$\displaystyle \text{Amount } = P + I = 11200 + 1344 = \text{ Rs. }12544$

(iii) For the third year:

$\displaystyle \text{Principal }= \text{ Rs. }12544 , \text{ Rate } = 12\% \text{ and } T = 1 \text{ year }$

$\displaystyle \text{I }= \frac{P \times R \times T}{100} = \frac{12544 \times 12 \times 1}{100} = 1505.28$

$\displaystyle \text{Amount = P + I } = 12544 + 1505.28 = \text{ Rs. }14049.28$

$\\$

Question 14: 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.

Answer:

$\displaystyle \text{Since, S.I. of 3 years } = \text{ Rs. } 600$

$\displaystyle \text{S.I. of 1 year }= \frac{600}{3} = \text{ Rs. } 200$

$\displaystyle \text{S.I. is same every year. For 1st year : C.I. = S.I. }$

$\displaystyle \text{C.I. for first year } = \text{ Rs. } 200$

$\displaystyle \text{Given, C.I. for two years } = \text{ Rs. } 410$

$\displaystyle \text{Therefore C.I for 2nd year } = 410 - 200 = \text{ Rs. } 210$

$\displaystyle \text{Difference between the C.I. of two successive years } = 210 - 200 = \text{ Rs. } 10$

Therefore Rs. 10 is the interest for one year on the interest of 1st year i.e. on Rs. 200

$\displaystyle \text{Therefore Rate } \% = \frac{100 \times I }{P \times T} = \frac{100 \times 10}{200 \times 1} = 5\%$

$\\$

Question 15: The compound interest calculated yearly at 10% on a certain sum of money amounts to Rs. 665.50 in the fifth year. Calculate :
(i) C.I. for the sixth year at the same rate and on the same sum.
(ii) C.I. for the fourth year on the same sum and at the same rate.

Answer:

$\displaystyle \text{(i) C.I. for 6th year = C.I. of 5th year + Interest on it for 1 year } \\ \\ = 665.50 + 10\% \text{ of } \text{ Rs. } 665.50 = \text{ Rs. } 732.05$

(ii) $\displaystyle \text{Let C.I. for 4th year } = x$

Since, C.I. for 5th year = C.I. of 4th year + Interest on it for 1 year.

$\displaystyle \Rightarrow 665.50 = x + 10% \text{ of } x$

$\displaystyle \Rightarrow x = 605$

$\displaystyle \Rightarrow C.I. \text{ of 4th year } = \text{ Rs. } 605$

$\\$

Question 16: Asum 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

Answer:

(i) Since Amount in 5 years = Rs. 8100

Amount 6 years = Rs. 8748

Therefore 8748-8100 = Rs. 648 is the interest of 1 year on Rs. 8100

$\displaystyle \text{Therefore Rate } = \frac{648 \times 100}{8100 \times 1} = 8\%$

$\displaystyle \text{(ii) Amount in 7 years = Amount in 6 years + Interest on it for 1 year} \\ \\ = 8748 + 8\% \text{ of } 8748 = \text{ Rs. } 9447.84$

(iii) $\displaystyle \text{Let amount in 4 years } = \text{ Rs. } x$

Amount in 5 years = Amount in 4 years + interest on it for 1 year

$\displaystyle \Rightarrow 8100 = x + 8\% \text{ of } x$

$\displaystyle \Rightarrow x - 750$

$\displaystyle \text{Therefore Amount in 4 years }= \text{ Rs. } 7500$

$\\$

Question 17: A sum of Rs. 9600 is invested for 3 years at 10% annum at compound interest.

(i) What is the sum due at the end of the first year ?

(ii) What is the sum due at the end of the second year.

(iii) Find the difference between the answers in (ii) and (i) and find the interest on this sum (difference) for one year.

(iv) Hence, write down the compound interest for the third year [ ICSE Board 1996]

Answer:

$\displaystyle \text{(i) Interest for the 1st year } = \frac{9600 \times 10 \times 1}{100} = \text{Rs. } 960$

$\displaystyle \text{The sum due at the end of the 1st year } = 9600 + 960 = \text{Rs. } 10560$

$\displaystyle \text{(ii) Interest for the 2nd year } = \frac{10560 \times 10 \times 1}{100} = \text{Rs. } 1056$

$\displaystyle \text{The sum due at the end of the 2nd year } = 10560 + 1056 = \text{Rs. } 11616$

$\displaystyle \text{(iii) Required difference } = 11616 - 10560 = \text{Rs. }1056$

$\displaystyle \text{Therefore Interest for 1 year on this difference }= \frac{1056 \times 10 \times 1}{100} = \text{Rs. }105.60$

$\displaystyle \text{(iv) C.I. for the 3rd year = C.I. of the 2nd year + interest on it for 1 year } \\ \\ = 1056 + 105.60 = \text{Rs. }1161.60$

$\\$

Question 18: 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.

Answer:

$\displaystyle \text{(i) Difference between depreciations of 2nd year and 3rd year } \\ \\ = 4752 - 4181.76 = \text{Rs. } 570.24$