Class 10: Compound Interest (Using Formula) – Miscellaneous Problems – Set 2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Calculate the amount and the compound interest on Rs. 7500 in 2 years and at 6% compounded annually.

Answer:

$\displaystyle P=7500 \text{ Rs.; } \hspace{1.0cm} r=6\%; \hspace{1.0cm} n=2 \text{ years }$

$\displaystyle \text{ Amount: } A=P \Big(1+ \frac{r}{100} \Big)^n = 7500 \Big(1+ \frac{6}{100} \Big)^2 = 8427 \text{ Rs. }$

$\displaystyle \text{Compound Interest: } A - P = 8427 - 7500 = 927 \text{ Rs. }$

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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.

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.

Answer:

$\displaystyle P=18000 \text{ Rs.; } \hspace{1.0cm} r=15\%; \hspace{1.0cm} n=2 \text{ years }$

$\displaystyle \text{ Amount: } A=P \Big(1+ \frac{r}{100} \Big)^n = 18000 \Big(1+ \frac{15}{100} \Big)^2 = 23805 \text{ Rs. }$

$\displaystyle \text{Compound Interest: } A - P = 23805-18000 = 5805 \text{ Rs. }$

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Question 4: What sum of money will amount to Rs. 9261 in 3 years at 5% per annum compound interest? [ ICSE Board 2009]

Answer:

$\displaystyle A=9261 \text{ Rs.; } \hspace{1.0cm} r=5\%; \hspace{1.0cm} n=3 \text{ years }$

$\displaystyle \text{ Amount: } 9261=P \Big(1+ \frac{5}{100} \Big)^3$

$\displaystyle P = 9261 \times \Big( \frac{20}{21} \Big)^2 = \text{Rs. } 8000$

$\displaystyle \text{ Required sum = Rs. } 8000$

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Question 5: On a certain sum, the compound interest in 3 years and at 10 per cent per annum amounts to Rs. 2317. Find the sum.

Answer:

$\displaystyle C.I.=2317 \text{ Rs.; } \hspace{1.0cm} r=10\%; \hspace{1.0cm} n=3 \text{ years }$

$\displaystyle \text{ Amount: } 2317=P \Bigg[ \Big(1+ \frac{10}{100} \Big)^3 - 1 \Bigg]$

$\displaystyle 2317 = P(1.331 - 1)$

$\displaystyle P = \frac{2317}{0.331} = \text{Rs. } 7000$

$\displaystyle \text{Required sum = Rs. } 7000$

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Question 6: On a certain sum, the compound interest in two years amounts to Rs. 2256. If the rates of interest for successive years are 8% and 10% respectively, find the sum.

Answer:

$\displaystyle \text{ Amount: } A=P \Big(1+ \frac{8}{100} \Big)\Big(1+ \frac{10}{100} \Big) = \frac{297}{250} P$

$\displaystyle \text{ Given C.I. = Rs. } 2256$

$\displaystyle \Rightarrow \frac{297}{250} P - P = 2256$

$\displaystyle \Rightarrow \frac{297P-250P}{250} = 2256$

$\displaystyle \Rightarrow P = \frac{2256 \times 250}{47} = \text{Rs. } 12000$

$\displaystyle \text{ Required sum = Rs. } 12000$

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Question 7: A person lends Rs. 15000 for 2 years at a certain rate of compound interest. If after 2 years, it amounts to Rs.16224; find the rate of interest.

Answer:

$\displaystyle P=15000 \text{ Rs.; } \hspace{1.0cm} A=16226 \text{ Rs. }; \hspace{1.0cm} n=2 \text{ years }$

$\displaystyle \Rightarrow 16224=15000 \Big(1+ \frac{r}{100} \Big)^2$

$\displaystyle \Big(1+ \frac{r}{100} \Big)^2 = \Big( \frac{26}{25} \Big)^2$

$\displaystyle 1 + \frac{r}{100} = \frac{26}{25}$

$\displaystyle r = 4\%$

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Question 8: In what period of time will Rs. 12,000 yield Rs. 3972 as compound interest at 10 per cent, if compounded on an yearly basis. [ ICSE Board 2011]

Answer:

$\displaystyle P=12000 \text{ Rs.; } \hspace{1.0cm} C.I.=3972 \text{ Rs. }; \hspace{1.0cm} r=10\% \text{ years }$

$\displaystyle \text{Amount: } A = P + I = 12000 + 3972 = 15972$

$\displaystyle \therefore 15972 = 12000 \Big( 1 + \frac{10}{100} \Big)^n$

$\displaystyle \Rightarrow \frac{15972}{12000} = \Big( \frac{11}{10} \Big)^3$

$\displaystyle \Rightarrow \Big( \frac{11}{10} \Big)^3= \Big( \frac{11}{10} \Big)^3$

$\displaystyle \Rightarrow n = 3$

$\displaystyle \text{Therefore Required Time = 3 years }$

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Question 9: 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.

Answer:

Let A’s share $\displaystyle = x$

Therefore B’s share $\displaystyle = ( 36465 - x)$

$\displaystyle \text{Given: Amount of A in 7 years = Amount of B in 5 years }$

$\displaystyle \Rightarrow x \Big( 1 + \frac{10}{100} \Big)^7 = ( 36465 - x) \Big( 1 + \frac{10}{100} \Big)^5$

$\displaystyle \Rightarrow x \Big( 1 + \frac{10}{100} \Big)^2 = ( 36465 - x)$

$\displaystyle \Rightarrow \frac{121}{100} x=36465 - x$

$\displaystyle \Rightarrow x = \text{Rs. } 16500$

$\displaystyle \Rightarrow 36465- x = 36465 - 16500 = \text{Rs. } 19965$

$\displaystyle \text{Therefore A's share= Rs. 16500 and B's share = Rs. 19965 }$

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Question 10: On what sum of money will the difference between the compound interest and the simple interest for 2 years be equal to Rs. 25, if the rate of interest charged for both is 5% p.a.? [ ICSE Board 2012]

Answer:

$\displaystyle C.I. = P \Bigg[ \Big( 1 + \frac{r}{100} \Big)^n - 1 \Bigg] = P \Bigg[ \Big( 1 + \frac{5}{100} \Big)^n - 1 \Bigg] = \frac{41P}{400}$

$\displaystyle S.I. = \frac{P \times 5 \times 2}{100} = \frac{P}{100}$

$\displaystyle \text{Given }C.I. - S. I. = 25$

$\displaystyle \Rightarrow \frac{41P}{400} - \frac{P}{10} = 25$

$\displaystyle \Rightarrow P = 10000$

$\displaystyle \text{Therefore Required sum = Rs. 10000 }$

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Question 11: Calculate the compound interest on Rs. 4000 in $\displaystyle 1\frac{1}{2}$ years at 10% per annum compounded half-yearly.

Answer:

$\displaystyle A = P \Big( 1 + \frac{r}{2 \times 100} \Big)^{n \times 2}$

$\displaystyle \Rightarrow 4000 \Big( 1 + \frac{10}{2 \times 100} \Big)^{\frac{3}{2} \times 2} = 4630.50$

$\displaystyle C.I. = A - P = 4630.50-4000 = \text{Rs. } 630.50$

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Question 12: Find the amount when Rs. 10000 is invested for $\displaystyle 2\frac{1}{2}$ years at 10% interest compounded yearly.

Answer:

$\displaystyle A = P \Big( 1 + \frac{r}{100} \Big)^2 \Big( 1 + \frac{r}{2 \times 100} \Big)^{\frac{1}{2}\times 2}$

$\displaystyle \Rightarrow A = 10000 \Big( 1 + \frac{10}{100} \Big)^2 \Big( 1 + \frac{10}{2 \times 100} \Big) = \text{Rs. }12705$

$\displaystyle \text{Amount in } 2\frac{1}{2} \text{ years } = \text{Rs. }12705$

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Question 13: John borrowed Rs. 20000 for 4 years under the following conditions :
10% simple interest for the first $2\frac{1}{2}$ years.
10% C.I. for the remaining one and a half years on the amount due after $2\frac{1}{2}$ years.
The interest being compounded half yearly.
Find the total amount to be paid at the end of the four years,

Answer:

$\displaystyle \text{For first } 2\frac{1}{2} \text{ years: } \hspace{1.0cm} P = \text{Rs. }20000; \hspace{1.0cm} r = 10\% \text{ and } T = 2\frac{1}{2} \text{ years }$

$\displaystyle \text{Therefore Interest }= \frac{20000 \times 10 \times 5}{100 \times 2} = \text{Rs. }5000$

$\displaystyle \text{Therefore amount due after } 2 \frac{1}{2} years = 20000 + 5000 = \text{Rs. }25000$

$\displaystyle \text{For remaining } 1\frac{1}{2} \text{ years: } P = \text{Rs. }25000, \\ \\ n = 1\frac{1}{2} \text{ years and } r = 10\% \text{ per annum compounded half yearly. }$

$\displaystyle \text{Therefore } A = P \Big( 1 + \frac{r}{2 \times 100} \Big)^{n \times 2} = 25000 \Big( 1 + \frac{10}{2 \times 100} \Big)^{\frac{3}{2} \times 2} = \text{Rs. }28940.63$

$\displaystyle \text{Therefore the total amount to be paid at the end of 4 years } = \text{Rs. } 28940.63$

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Question 14: 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.

Answer:

$\displaystyle \text{Let the sum of money lent out be } \text{Rs. } x$

$\displaystyle \text{In 1st Case: } A_1 = x \Big( 1 + \frac{20}{100} \Big)^2 = \frac{36x}{25}$

$\displaystyle \text{In the 2nd case:} A_2 = x \Big( 1 + \frac{20}{2 \times 100} \Big)^{2 \times 2} = \frac{14641x}{10000}$

Given, C.I. in the 2nd case is Rs. 482 more than the C.I. in the 1st case.

For the same principal, amount in the 2nd case is Rs. 482 more than the amount in the 1st case.

$\displaystyle \Rightarrow \frac{14641x}{10000} - \frac{36x}{25} = 482$

$\displaystyle \Rightarrow x = \text{Rs. } 20000$

$\displaystyle \text{Therefore the sum of money lent out } = \text{Rs. } 20000$

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Question 15: 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

Answer:

$\displaystyle \text{Given :} P = \text{Rs. } 6400; C.I. = 1008.80 \text{ and } T = 18 \text{ months } = 1\frac{1}{2} \text{ years }$

$\displaystyle \text{Therefore Amount } A = P + C.I. = 6400 + 1008.80 = \text{Rs. } 7408.80$

$\displaystyle 7408.80 = 6400 \Big( 1 + \frac{r}{2 \times 100} \Big)^{\frac{3}{2} \times 2 }$

$\displaystyle \Rightarrow \Big( 1 + \frac{r}{200} \Big)^3 = \frac{7408.80}{6400} = \Big( \frac{21}{20} \Big)^3$

$\displaystyle \text{On solving we get } r = 10\%$

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Question 16: 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:
$(1.08)^4 = 1.3605 \hspace{0.5cm} (1.04)^8 = 1.3686 \hspace{0.5cm}(1.08)^8 = 1.8509 \hspace{0.5cm}(1.04)^4 = 1.1699$

Answer:

$\displaystyle \text{ Given: P = Rs. 4800, r = 8\% compounded half yearly and n = 4 years }$

$\displaystyle A = P \Big( 1 + \frac{r}{2 \times 100} \Big)^{n \times 2}$

$\displaystyle = 4800 \Big( 1 + \frac{8}{2 \times 100} \Big)^{4 \times 2}$

$\displaystyle = 4800 (1.04)^8 = 4800 \times 1.3686 = 6569.28$

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Question 17: 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

Answer:

$\displaystyle \text{ (i) Given : I = Rs. 340, T = 2 years and R = 4 \% }$

$\displaystyle \therefore P = \frac{I \times 100}{R \times T} = \frac{340 \times 100}{4 \times 2} = 4250$

$\displaystyle \text{ (ii) } C.I. = P \Big( 1 + \frac{r}{2 \times 100} \Big)^{n \times 2} - P \\ \\ {\hspace{1.7cm} = 4250 \Big( 1 + \frac{4}{2 \times 100} \Big)^{1 \times 2} - 4250 = 4421.70 - 4250 = 171.70 }$

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Question 18: 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.

Answer:

$\displaystyle \text{ Number of industries after 2 years} \\ \\ = \text{(Original number of industries ) } \times \Big( 1 + \frac{r}{100} \Big)^n \\ \\ = 1600 \Big( 1 + \frac{20}{100} \Big)^2 = 2304$

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Question 19: 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?

Answer:

$\displaystyle \text{ Value after 3 years } = 18000 \Big( 1 - \frac{10}{100} \Big)^3 = \text{Rs. } 13122$

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Question 20: 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.

Answer:

Let the required number of years $= n$ years

$\displaystyle \text{The value of machine after n years } = \text{ (Its value when new) } \times \Big( 1 - \frac{r}{100} \Big)^n$

$\displaystyle \Rightarrow 128000 = 250000 \Big( 1 - \frac{20}{100} \Big)^n$

$\displaystyle \Rightarrow \frac{128000}{250000} = \Big( \frac{80}{100} \Big)^n = \Big( \frac{4}{5} \Big)^n$

$\displaystyle \Rightarrow \Big( \frac{4}{5} \Big)^3 = \Big( \frac{4}{5} \Big)^n$

$\displaystyle \Rightarrow n = 3$ years

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Question 21: 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.

Answer:

$\displaystyle \text{Population after n years = (Present population) } \Big( 1 + \frac{r}{100} \Big)^n$

$\displaystyle \text{Therefore population after 2 years } = 216000 \Big( 1 + \frac{20}{100} \Big)^2 = 311040$

$\displaystyle \text{Present population = (Population n years ago ) } \Big( 1 + \frac{r}{100} \Big)^2$

$\displaystyle \text{Therefore 216000 = (Population n years ago ) } \Big( 1 + \frac{r}{100} \Big)^2$

$\displaystyle \text{Therefore Population 2 years ago } = 150000$

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Question 22: 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.

Answer:

$\displaystyle \text{Amount in 2 years } = 14520 \Rightarrow P \Big( 1 + \frac{r}{100} \Big)^2 = 14520 \text{ ... ... ... i)}$

$\displaystyle \text{Amount in 4 years } = 17569.20 \Rightarrow P \Big( 1 + \frac{r}{100} \Big)^4 = 17569.20 \text{ ... ... ... ii)}$

Dividing ii) by i) we get

$\displaystyle \frac{P \Big( 1 + \frac{r}{100} \Big)^4}{P \Big( 1 + \frac{r}{100} \Big)^2} = \frac{17569.20}{14520}$

$\displaystyle \Rightarrow \Big( 1 + \frac{r}{100} \Big)^2 = \frac{121}{100}$

$\displaystyle \Rightarrow 1 + \frac{r}{100} = \frac{11}{10}$

$\displaystyle \Rightarrow r = 10\%$

From i) we get

$\displaystyle P \Big( 1 + \frac{10}{100} \Big)^2 = 14520$

$\displaystyle \Rightarrow P \Big( \frac{121}{100} \Big) = 14520$

$\displaystyle \Rightarrow P = \text{Rs. }12000$

$\displaystyle \text{Therefore the rate of interest is 10\% and the sum is Rs. 12000 }$

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Question 23: 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

Answer:

$\displaystyle \text{Let the rate of interest by r\% }$

$\displaystyle \text{Therefore S.I. in 2 years } = \frac{9500 \times r \times 2}{100} = 190 r$

$\displaystyle \text{And, C.I. in two years }= A - P = 9500 \Big( 1 + \frac{r}{100} \Big)^2 - 9500$

$\displaystyle \text{Given C.I. - S.I. }= 95$

$\displaystyle \Rightarrow 9500 \Big( 1 + \frac{r}{100} \Big)^2 - 9500 - 190r = 95$

$\displaystyle \Rightarrow 100 \Big( 1 + \frac{r}{100} \Big)^2 - 100 - 2r = 1$

$\displaystyle \Rightarrow 100 + \frac{r^2}{100} + 2r - 100 - 2r = 1$

$\displaystyle \Rightarrow \frac{r^2}{100} = 1$

$\displaystyle \Rightarrow r = 10\%$

Therefore rate of interest $\displaystyle = 10%$

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Question 24: 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.

Answer:

$\displaystyle \text{ Let Principal = Rs. } x \Rightarrow \text{ Amount in 5 years = Rs. } 2x$

$\displaystyle A = P \Big( 1 + \frac{r}{100} \Big)^n$ $\displaystyle \Rightarrow 2x = x \Big( 1 + \frac{r}{100} \Big)^5$ $\displaystyle \Rightarrow 2 = \Big( 1 + \frac{r}{100} \Big)^5$

$\displaystyle \text{ For second part: } P = Rs. x \text{ and } A = Rs. 8x$

$\displaystyle A = P \Big( 1 + \frac{r}{100} \Big)^n$ $\displaystyle \Rightarrow 8x = x \Big( 1 + \frac{r}{100} \Big)^n$ $\displaystyle \Rightarrow 8 = \Big( 1 + \frac{r}{100} \Big)^n$

$\displaystyle \text{ i.e. } (2)^3 = \Big( 1 + \frac{r}{100} \Big)^n$

$\displaystyle \Rightarrow \Bigg[ \Big( 1 + \frac{r}{100} \Big)^5 \Bigg]^3 = \Big( 1 + \frac{r}{100} \Big)^n$

$\displaystyle \Rightarrow \Big( 1 + \frac{r}{100} \Big)^{15} = \Big( 1 + \frac{r}{100} \Big)^n$

$\displaystyle \Rightarrow n = 15$

Hence, the required time is 15 years.

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Question 25: 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.

Answer:

For the payment of $\text{ Rs. } 43200$ at the end of the first year:

$\displaystyle A = \text{ Rs. } 43200; n = 1 \text{ year and } r = 8\%. \text{ To find P. }$

$\displaystyle A = P \Big( 1 + \frac{r}{100} \Big)^n \Rightarrow 43200 = P \Big( 1 + \frac{8}{100} \Big)^1 \Rightarrow P = 43200 \times \frac{100}{108} = \text{ Rs. } 40000$

For the payment of $\text{ Rs. } 34992$ at the end of the second year:

$\displaystyle 34992 = P \Big( 1 + \frac{8}{100} \Big)^2 \Rightarrow P = 34992 \times \Big( \frac{100}{108} \Big)^2 = \text{ Rs. } 30000$

$\displaystyle \text{ Therefore sum borrowed } = 40000 + 30000 = \text{ Rs. } 70000$

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Question 26: 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).

Answer:

Let the investments made be x, y, and z respectively. Then,

$\displaystyle x \Big( 1 + \frac{10}{100} \Big)^{58-26} = y \Big( 1 + \frac{10}{100} \Big)^{58-27} = z \Big( 1 + \frac{10}{100} \Big)^{58-28}$