Compound Interest without Formula

Question 1: Find the amount and the compound interest on Rs. \ 3000 , at 5\% per annum for 2 years, compounded annually.

Answer:

For the 1^{st} \ Year : We have,

Principal = Rs. 3000 , Rate of Interest = 5\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{3000 \times 5 \times 1}{100} = 150  \ Rs.

Therefore, the Amount at the end of 1^{st} year = 3000 + 150 = 3150 \ Rs.

 For the 2^{nd} \ Year : We have,

Principal = Rs. 3150 , Rate of Interest = 5\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{3150 \times 5 \times 1}{100} =157.50  \ Rs.

Therefore, the Amount at the end of 2^{nd} year = 3150 + 157.50 = 3307.50 \ Rs.

Compound Interest for 2 years = Amount at the end of 2^{nd} Year – Principal = 3307.50 - 3000 = 307.50 \ Rs.

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Question 2: Find the amount and the compound interest on Rs. \ 2000 at 10\% per annum for 2\frac{1}{2} years.

Answer:

For the 1^{st} \ Year : We have,

Principal = Rs. 2000 , Rate of Interest = 10\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{2000 \times 10 \times 1}{100} = 200  \ Rs.

Therefore, the Amount at the end of 1^{st} year = 2000 + 200 = 2200 \ Rs.

 For the 2^{nd} \ Year : We have,

Principal = Rs. 2200 , Rate of Interest = 10\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{2200 \times 10 \times 1}{100} =220  \ Rs.

Therefore, the Amount at the end of 2^{nd} year = 2200 + 220 = 2420 \ Rs.

For the the next six months: We have,

Principal = Rs. 2420 , Rate of Interest = 10\% per annum, T = \frac{1}{2}

Therefore Interest = \frac{P \times R \times T}{100} = \frac{2420 \times 10 \times \frac{1}{2}}{100} =121  \ Rs.

Therefore, the Amount at the end of 2\frac{1}{2} year = 2420 + 121 = 2541 \ Rs.

Compound Interest for 2\frac{1}{2} years = Amount at the end of 2\frac{1}{2} Year – Principal = 2541 - 2000 = 541 \ Rs.

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Question 3: Find the compound interest on Rs. \ 160000 for one year at the rate of 20\% , per annum, if the interest is compounded quarterly.

Answer:

Rate of Interest = \frac{20}{4} \% = 5\% per quarter

For the 1^{st} Quarter : We have,

Principal = Rs. 160000 , Rate of Interest = 5\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{160000 \times 5 \times 1}{100} = 8000  \ Rs.

Therefore, the Amount at the end of 1^{st} quarter = 160000 + 8000 = 168000 \ Rs.

For the 2^{nd} Quarter : We have,

Principal = Rs. 168000 , Rate of Interest = 5\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{168000 \times 5 \times 1}{100} = 8400  \ Rs.

Therefore, the Amount at the end of 2^{nd} quarter = 168000 + 8400 = 176400 \ Rs.

For the 3^{rd} Quarter : We have,

Principal = Rs. 176400 , Rate of Interest = 5\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{176400 \times 5 \times 1}{100} = 8820  \ Rs.

Therefore, the Amount at the end of 3^{rd} quarter = 176400 + 8820 = 185220 \ Rs.

For the 4^{th} Quarter : We have,

Principal = Rs. 185220 , Rate of Interest = 5\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{185220 \times 5 \times 1}{100} = 9261  \ Rs.

Therefore, the Amount at the end of 4^{th} quarter = 185220 + 9261 = 194481 \ Rs.

Compound Interest for the year = Amount at the end of 1^{st} Year – Principal = 194481 - 160000 = 34481 \ Rs.

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Question 4: Calculate the amount and the compound interest on Rs. \ 6000 for 2 years when the rates of interest for successive years are 5\% and 6\% respectively.

Answer:

For the 1^{st} \ Year : We have,

Principal = Rs. 6000 , Rate of Interest = 5\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{6000 \times 5 \times 1}{100} = 300  \ Rs.

Therefore, the Amount at the end of 1^{st} year = 6000 + 300 = 6300 \ Rs.

 For the 2^{nd} \ Year : We have,

Principal = Rs. 6300 , Rate of Interest = 6\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{6300 \times 6 \times 1}{100} =378 \ Rs.

Therefore, the Amount at the end of 2^{nd} year = 6300 + 378 = 6678 \ Rs.

Compound Interest for 2 years = Amount at the end of 2^{nd} Year – Principal = 6678 - 6000 = 678 \ Rs.

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Question 5: The simple interest on a certain sum of money for 2 years at 6\% per annum is Rs. \ 1680 . Find the amount and the compound interest on the same sum, at the same rate and for the same time, compounded annually.

Answer:

Simple interest on Rs. P for 2 years at 6\% per annum is Rs. \ 1680

\Rightarrow \frac{P \times R \times T}{100} = 1680

\Rightarrow P = \frac{1680 \times 100}{6 \times 2} = 14000 \ Rs.

Hence the Principal = 14000 \ Rs.

For the 1^{st} \ Year : We have,

Principal = Rs. 14000 , Rate of Interest = 6\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{14000 \times 6 \times 1}{100} = 840  \ Rs.

Therefore, the Amount at the end of 1^{st} year = 14000 + 840 = 14840 \ Rs.

 For the 2^{nd} \ Year : We have,

Principal = Rs. 14840 , Rate of Interest = 6\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{14840 \times 6 \times 1}{100} =890.40  \ Rs.

Therefore, the Amount at the end of 2^{nd} year = 14840 + 890.40 = 15730.40 \ Rs.

Compound Interest for 2 years = Amount at the end of 2^{nd} Year – Principal = 15730.40 - 14000 = 1730.40 \ Rs.

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Question 6: A man invested Rs. \ 8000 for 2 years at 10\% per annum, compounded annually. Compute: (i) the amount at the end of first year. (ii) the compound interest for the second year. (iii) the compound interest for 2 years.

Answer:

(i)   For the 1^{st} \ Year : We have,

Principal = Rs. 8000 , Rate of Interest = 10\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{8000 \times 10 \times 1}{100} = 800  \ Rs.

Therefore, the Amount at the end of 1^{st} year = 8000 + 800 = 8800 \ Rs.

(ii)  For the 2^{nd} \ Year : We have,

Principal = Rs. 8800 , Rate of Interest = 10\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{8800 \times 10 \times 1}{100} =880  \ Rs.

Therefore, the Amount at the end of 2^{nd} year = 8800 + 880 = 9680 \ Rs.

(iii)  Compound Interest for 2 years = Amount at the end of 2^{nd} Year – Principal = 9680 - 8000 = 1680 \ Rs.

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Question 7: A person invests Rs. \ 240000 for 2 years at 10\% per annum compounded annually. If the income tax at 20\% is deducted at the end of each year on interest accrued, find the amount she received at the end of 2 years.

Answer:

For the 1^{st} \ Year : We have,

Principal = Rs. 240000 , Rate of Interest = 10\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{240000 \times 10 \times 1}{100} = 24000  \ Rs.

Income tax deducted = \frac{20}{100} \times 24000 = 4800 \ Rs.

Therefore, the Amount at the end of 1^{st} year = 240000 + 24000 - 4800 = 259200 \ Rs.

 For the 2^{nd} \ Year : We have,

Principal = Rs. 259200 , Rate of Interest = 10\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{259200 \times 10 \times 1}{100} =25920  \ Rs.

Income tax deducted = \frac{20}{100} \times 25920 = 5184 \ Rs.

Therefore, the Amount at the end of 2^{nd} year = 259200 + 25920 - 5184 = 279936 \ Rs.

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Question 8: A man borrows Rs. \ 15000 at 14\% per annum compounded annually. If he repays Rs. \ 4100 at the end of first year and Rs. \ 5220 at the end of second year, find the amount of the loan outstanding at the beginning of the third year.

Answer:

For the 1^{st} \ Year : We have,

Principal = Rs. 15000 , Rate of Interest = 14\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{15000 \times 14 \times 1}{100} = 2100  \ Rs.

Therefore, the Amount at the end of 1^{st} year = 15000 + 2100 = 17100 \ Rs.

Amount repaid after 1^{st} \  Year = 4100 \  Rs.

For the 2^{nd} \ Year : We have,

Principal = Rs. 17100 - 4100 = 13000  , Rate of Interest = 14\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{13000 \times 14 \times 1}{100} = 1820  \ Rs.

Therefore, the Amount at the end of 2^{nd} year = 13000 + 1820 = 14820 \ Rs.

Amount repaid after 2^{nd} \  Year = 5220 \  Rs.

For the 3^{rd} \ Year : We have,

Outstanding Principal = Rs. 14820 - 5220 = 9600 \ Rs.

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Question 9: A person invested Rs. \ 5000 at a certain rate of interest compounded annually for two years. At the end of first year it amounts to Rs. \ 5325 . Calculate: (i) The rate of interest (ii) The amount at the end of second year

Answer:

For the 1^{st} \ Year : We have,

Principal = Rs. 5000 , Rate of Interest = r\% per annum, T = 1

Therefore Inerest = \frac{P \times R \times T}{100}

\Rightarrow 325 = \frac{5000 \times r \times 1}{100}

\Rightarrow r = 6.5\%

For the 2^{nd} \ Year : We have,

Principal = Rs. 5325  , Rate of Interest = 6.5\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{5325 \times 6.5 \times 1}{100} = 346.125  \ Rs.

Therefore, the Amount at the end of 2^{nd} year = 5325 + 346.125 = 5671.125 \ Rs.

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Compound Interest Formula

Question 10: Find the amount and compound interest on Rs. \ 16000 for 2 years at 15\% , interest being payable annually.

Answer:

Principal = Rs. 16000 , Rate of Interest = 15\% per annum, n = 2

A = P \Big( 1+ \frac{R}{100} \Big)^n

= 16000 \Big( 1+ \frac{15}{100} \Big)^2

= 21160 \ Rs.

Therefore compound interest = A - P = 21160 - 16000 = 5160 \ Rs.

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Question 11: Find the amount and compound interest on Rs. \ 25000 for 3 years compounded annually and the rate of interest being 8\%, 10\% and 12\% for three successive years respectively.

Answer:

Principal = Rs. 25000 , Rate of Interest :  R_1=8\%, R_2=10\% , R_3=12\% per annum,

A = P \Big(1+ \frac{R_1}{100} \Big).\Big(1+ \frac{R_2}{100} \Big).\Big(1+ \frac{R_3}{100} )

= 25000 \Big(1+ \frac{8}{100} \Big).\Big(1+ \frac{10}{100} \Big).\Big(1+ \frac{12}{100})

= 25000 \times 1.08 \times 1.10 \times 1.12

= 33264 \ Rs.

Therefore compound interest = A - P = 33264 - 25000 = 8264 \ Rs.

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Question 12: Compute the interest earned and amount due if a sum of Rs. \ 15000 is invested for 1\frac{1}{2} years at 8\% per annum compound interest, interest being compounded semi-annually.

Answer:

Principal = Rs. 15000 , Rate of Interest :  R=8\%, semi-annually, T=  1\frac{1}{2} years

A = P \Big( 1+ \frac{R}{100k} \Big)^{nk}

= 15000 \Big( 1+ \frac{8}{100 \times 2} \Big)^{1.5 \times 2}

= 15000 \times (1.04)^3 = 16872.96 \ Rs.

Therefore compound interest = A - P = 16872.96 - 15000 = 1872.96 \ Rs.

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Question 13: A man borrows Rs. \ 1000 at 10\% per annum simple interest for 3 years. He immediately lends this money out at compound interest at the same rate and for the same time. What is his gain at the end of 3 years?

Answer:

Simple Interest:

Principal = Rs. 1000 , Rate of Interest :  R=10\%, annually, T=  3 years

Interest = \frac{P \times R \times T}{100} = \frac{1000 \times 10 \times 3}{100} = 300 \ Rs.

Compound Interest: 

Principal = Rs. 1000 , Rate of Interest = 10\% per annum, n = 3

A = P \Big( 1+ \frac{R}{100} \Big)^n

= 1000 \Big( 1+ \frac{10}{100} \Big)^3

= 1331 \ Rs.

Therefore compound interest = A - P = 1331 - 1000 = 331 \ Rs.

Therefore gain = 331 - 300 = 31 \ Rs.

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Question 14: What sum of money will amount to Rs. \ 4374 in 3 years at 12\frac{1}{2}\% per annum, compounded annually?

Answer:

Principal = P , Rate of Interest = 10\% per annum, n = 3 A = 4374 \ Rs

A = P \Big( 1+ \frac{R}{100} \Big)^n

4374 = P \Big( 1+ \frac{12.5}{100} \Big)^3

\Rightarrow P = \frac{4734}{1.125^3} = 3072 \ Rs.

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Question 15: What sum will become Rs. \ 9826 in 18 months if the rate of interest is 2\frac{1}{2}\% per is compounded half-yearly?

Answer:

Principal = P , Rate of Interest :  R=2\frac{1}{2}\%, semi-annually, T=  1\frac{1}{2} years  A = Rs. \ 9826

A= P \Big( 1+ \frac{R}{100k} \Big)^{nk}

9826 = P \Big( 1+ \frac{2.5}{100 \times 2} \Big)^{1.5 \times 2}

\Rightarrow P = \frac{9826}{1.0125^3} = 9466.54 \ Rs.

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Question 16: The difference between the compound interest and the simple interest on a certain sum at 10\% per annum for 3 years is Rs. \ 93 . Find the sum.

Answer:

Simple Interest:

Principal = P , Rate of Interest :  R=10\%, annually, T=  3 years

Interest = \frac{P \times R \times T}{100} = \frac{P \times 10 \times 3}{100} = \frac{3}{10} P \ Rs.  = 0.3P

Compound Interest:

Principal = P , Rate of Interest = 10\% per annum, n = 3

A = P \Big( 1+ \frac{R}{100} \Big)^n

A = P \Big( 1+ \frac{10}{100} \Big)^3

A = 1.331P

Compound Interest = 1.331P-P = 0.331 P

Given 93 = 0.331P - 0.3P

\Rightarrow P = \frac{93}{0.031} = 3000 \ Rs.

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Question 17: The difference between the compound interest for a year payable half-yearly and the simple interest on a certain sum of money lent out at 10\% for a year is Rs. \ 15 . Find the sum of money lent out.

Answer:

Simple Interest:

Principal = P , Rate of Interest :  R=10\%, annually, T=  1 years

Interest = \frac{P \times R \times T}{100} = \frac{P \times 10 \times 1}{100} = \frac{1}{10} P \ Rs.  = 0.1P

Compound Interest:

Principal = P , Rate of Interest :  R=10\%, semi-annually, T=  1 year  Amount = A

A= P \Big( 1+ \frac{R}{100k} \Big)^{nk}

A = P \Big( 1+ \frac{10}{100 \times 2} \Big)^{1 \times 2}

A = 1.1025P

Compound Interest = 1.1025P-P = 0.1025 P

Given 15 = 0.1025P - 0.1P

\Rightarrow P = \frac{15}{0.0025} = 6000 \ Rs.

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Question 18: On a certain sum, lent out at 20\% per annum for 1\frac{1}{2} , the difference between the compound interest reckoned yearly and the reckoned yearly half- is Rs. \ 178.75 . Find the sum.

Answer:

Yearly:

Principal = P , Rate of Interest :  R=20\%, annually, T=  1\frac{1}{2} year  Amount = A

A = P \Big(1+ \frac{R}{100} \Big) \Big(1 + \frac{\frac{R}{2}}{100} \Big)

A = P \Big(1+ \frac{20}{100} \Big) \Big(1 + \frac{10}{100} \Big)

A = 1.20 \times 1.10 P = 1.32 P

Therefore Compound Interest = 1.32P - P = 0.32 P

Half Yearly:

Principal = P , Rate of Interest :  R=20\%, semi-annually, T=  1\frac{1}{2} year  Amount = A

A= P \Big( 1+ \frac{R}{100k} \Big)^{nk}

A = P \Big( 1+ \frac{20}{100 \times 2} \Big)^{1.5 \times 2}

A = 1.331P

Therefore Compound Interest = 1.331P - P = 0.331 P

Given 178.50 = 0.331P - 0.332P = 0.011

\Rightarrow P = \frac{178.75}{0.011} = 16250 \ Rs.

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Question 19: The compound interest on a certain sum for 2 years at 12\% per annum is Rs. \ 795 . Find the simple interest on the same sum for the the same period and at the same rate.

Answer:

Compound Interest:

Principal = P , Rate of Interest = 12\% per annum, n = 2

A = P \Big( 1+ \frac{R}{100} \Big)^n

A = P \Big( 1+ \frac{12}{100} \Big)^2

A = 1.2544P

Compound Interest 795 = 1.2544P-P = 0.2544 P  \Rightarrow P = 3125

Simple Interest:

Principal = 3125 , Rate of Interest :  R=12\%, annually, T=  2 years

Interest = \frac{P \times R \times T}{100} = \frac{3125 \times 12 \times 2}{100} = 750 \ Rs.

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Question 20: The simple interest on a certain sum for 2 years at 14\% per annum is Rs. \ 3500 . Find the corresponding compound interest.

Answer:

Simple Interest:

Principal = P , Rate of Interest :  R=14\%, annually, T=  2 years

Interest = \frac{P \times R \times T}{100}

3500 = \frac{P \times 14 \times 2}{100} \Rightarrow P = 12500 \ Rs.

Compound Interest:

Principal = 12500 , Rate of Interest = 14\% per annum, n = 2

A = P \Big( 1+ \frac{R}{100} \Big)^n

A = 12500 \Big( 1+ \frac{14}{100} \Big)^2

A = 16245

Compound Interest = 16245 - 12500 = 3745 \ Rs.

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Question 21: The simple interest on a sum of money for 2 years at 8\% per annum is Rs. \ 900 . Find the compound interest on the sum at the same rate for one year, compounded half -yearly.

Answer:

Simple Interest:

Principal = P , Rate of Interest :  R=8\%, annually, T=  2 years

Interest = \frac{P \times R \times T}{100}

900 = \frac{P \times 8 \times 2}{100} \Rightarrow P = 5625 \ Rs.

Compound Interest:

Principal = 5625 , Rate of Interest :  R=8\%, semi-annually, T=  1 year  Amount = A

A= P \Big( 1+ \frac{R}{100k} \Big)^{nk}

A = 5625 \Big( 1+ \frac{8}{100 \times 2} \Big)^{1 \times 2}

A = 6084 \ Rs.

Compound Interest = 6084 - 5625 = 459 \ Rs.

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Question 22: A sum of money is lent out at compound interest for 2 \ years years at 20\% per annum interest being reckoned yearly. If the same sum of money is lent out at compound interest at the same rate per per annum, compound interest being reckoned half-yearly it will fetch Rs. \ 482 more by way of interest. Calculate the sum of money-lent out.

Answer:

Compound Interest (yearly):

Principal = P , Rate of Interest = 20\% per annum, n = 2

A = P \Big( 1+ \frac{R}{100} \Big)^n

A = P \Big( 1+ \frac{20}{100} \Big)^2

A = 1.44P

Compound Interest = 1.44P - P = 0.44P

Compound Interest (Half yearly):

Principal = P , Rate of Interest :  R=20\%, semi-annually, T=  2 year  Amount = A

A= P \Big( 1+ \frac{R}{100k} \Big)^{nk}

A = P \Big( 1+ \frac{20}{100 \times 2} \Big)^{2 \times 2}

A = 1.4641

Compound Interest = 1.4641P - P = 0.4641P

Given 482 = 0.4641P - 0.44P \Rightarrow P = 20000 \ Rs.

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Question 23: What sum will amount to Rs. \ 2782.50 in 2 years at compound interest, if the rates are 5\% and 6\% for the successive years?

Answer:

Principal = P , Rate of Interest : R_1=5\%, R_2 = 6\% per annum, Amount = 2782.50 \ Rs.

A = P \Big(1+ \frac{R_1}{100} \Big).\Big(1+ \frac{R_2}{100} \Big)

2782.50 = P \Big(1+ \frac{5}{100} \Big).\Big(1+ \frac{6}{100} \Big)

\Rightarrow P = \frac{2782.50}{1.05 \times 1.06} = 2500 \ Rs.

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Question 24: A certain sum of money lent out at compound interest amounts to Rs. \ 9200 in one year and to Rs. \ 12167 in 3 years. Find ii) the rate of interest (ii) the original sum.

Answer:

Compound Interest (yearly):

Principal = P , Rate of Interest = r\% per annum, n = 1 Amount = 9200 \ Rs.

9200 = P \Big( 1+ \frac{r}{100} \Big)^1 … … … … (i)

Principal = P , Rate of Interest = r\% per annum, n = 3 Amount = 12167 \ Rs.

12167 = P \Big( 1+ \frac{r}{100} \Big)^3 … … … … (ii)

Dividing (ii) by (i) we get

\frac{12167}{9200} =  \Big( 1+ \frac{r}{100} \Big)^2

\Rightarrow 1.15 = \Big( 1+ \frac{r}{100} \Big)

\Rightarrow r = 1.15 - 1 = 15\%

Now substituting in (i)

P = \frac{9200}{1.15} = 8000 \ Rs.

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Question 25: The compound interest, calculated yearly on a certain sum of money for the second year is Rs. \ 880 and for the third year is Rs. \ 968 Calculate the rate of interest and the original money.

Answer:

1^{st} \ Year

Principal = P , Rate of Interest = r\% per annum, n = 1 Amount = A_1

A_1 = P \Big( 1+ \frac{r}{100} \Big)^1 … … … … (i)

2^{nd} \ Year

Principal = P , Rate of Interest = r\% per annum, n = 1 Amount = A_2

A_2 = P \Big( 1+ \frac{r}{100} \Big)^2 … … … … (ii)

3^{rd} \ Year

Principal = P , Rate of Interest = r\% per annum, n = 1 Amount = A_3

A_3 = P \Big( 1+ \frac{r}{100} \Big)^3 … … … … (iii)

Given: A_2 - A_1 = 880   and A_3 - A_2 = 968

Therefore

P \Big( 1+ \frac{r}{100} \Big)^2 - P \Big( 1+ \frac{r}{100} \Big)^1 = 880

\Rightarrow P (1+ \frac{r}{100} )( \frac{r}{100} ) = 880 … … … … (iv)

P \Big( 1+ \frac{r}{100} \Big)^3 - P \Big( 1+ \frac{r}{100} \Big)^1 = 968

\Rightarrow P (1+ \frac{r}{100} )^2( \frac{r}{100} ) = 968 … … … … (v)

Dividing (v) by (iv) we get

\frac{968}{880} = 1 + \frac{r}{100}

\Rightarrow r = 1.1 - 1 = 10\%

Substituting in (iv) P = \frac{880 \times 100}{10 \times 1.1} = 8000 \ Rs.

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Exercise 4(a) continued…

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