Class 9: Compound Interest 4.1

Compound Interest without Formula

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

Answer:

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

$\displaystyle \text{Principal } = \text{ Rs. } 3000$ $\displaystyle \text{ , Rate of Interest } = 5\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

$\displaystyle \text{Principal } = \text{ Rs. } 3150$ $\displaystyle \text{ , Rate of Interest } = 5\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

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

Answer:

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

$\displaystyle \text{Principal } = \text{ Rs. } 2000$ $\displaystyle \text{ , Rate of Interest } = 10\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

$\displaystyle \text{Principal } = \text{ Rs. } 2200$ $\displaystyle \text{ , Rate of Interest } = 10\%$ $\displaystyle \text{ per annum } T = 1$

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

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

For the the next six months: We have,

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

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

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

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

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

Answer:

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

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

$\displaystyle \text{Principal } = \text{ Rs. } 160000$ $\displaystyle \text{ , Rate of Interest } = 5\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

$\displaystyle \text{Principal } = \text{ Rs. } 168000$ $\displaystyle \text{ , Rate of Interest } = 5\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

$\displaystyle \text{Principal } = \text{ Rs. } 176400$ $\displaystyle \text{ , Rate of Interest } = 5\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

$\displaystyle \text{Principal } = \text{ Rs. } 185220$ $\displaystyle \text{ , Rate of Interest } = 5\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

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

Answer:

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

$\displaystyle \text{Principal } = \text{ Rs. } 6000$ $\displaystyle \text{ , Rate of Interest } = 5\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

$\displaystyle \text{Principal } = \text{ Rs. } 6300$ $\displaystyle \text{ , Rate of Interest } = 6\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

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Question 5: The simple interest on a certain sum of money for $\displaystyle 2$ years at $\displaystyle 6\%$ per annum is $\displaystyle Rs. 1680$ . 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 $\displaystyle Rs. P$ for $\displaystyle 2$ years at $\displaystyle 6\%$ per annum is $\displaystyle Rs. 1680$

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

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

Hence the Principal $\displaystyle = 14000 \text{ Rs. }$

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

$\displaystyle \text{Principal } = \text{ Rs. } 14000$ $\displaystyle \text{ , Rate of Interest } = 6\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

$\displaystyle \text{Principal } = \text{ Rs. } 14840$ $\displaystyle \text{ , Rate of Interest } = 6\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

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Question 6: A man invested $\displaystyle Rs. 8000$ for $\displaystyle 2$ years at $\displaystyle 10\%$ 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 $\displaystyle 2$ years.

Answer:

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

$\displaystyle \text{Principal } = \text{ Rs. } 8000$ $\displaystyle \text{ , Rate of Interest } = 10\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

$\displaystyle \text{Principal } = \text{ Rs. } 8800$ $\displaystyle \text{ , Rate of Interest } = 10\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

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

Answer:

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

$\displaystyle \text{Principal } = \text{ Rs. } 240000$ $\displaystyle \text{ , Rate of Interest } = 10\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

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

$\displaystyle \text{Principal } = \text{ Rs. } 259200$ $\displaystyle \text{ , Rate of Interest } = 10\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

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

Answer:

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

$\displaystyle \text{Principal } = \text{ Rs. } 15000$ $\displaystyle \text{ , Rate of Interest } = 14\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

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

$\displaystyle \text{Principal } = \text{ Rs. } 17100 - 4100 = 13000$ $\displaystyle \text{ , Rate of Interest } = 14\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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

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

Outstanding $\displaystyle \text{Principal } = \text{ Rs. } 14820 - 5220 = 9600 \text{ Rs. }$

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

Answer:

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

$\displaystyle \text{Principal } = \text{ Rs. } 5000$ $\displaystyle \text{ , Rate of Interest } = r\%$ $\displaystyle \text{ per annum } T = 1$

$\displaystyle \text{Therefore Interest } = \frac{P \times R \times T}{100}$

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

$\displaystyle \Rightarrow r = 6.5\%$

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

$\displaystyle \text{Principal } = \text{ Rs. } 5325$ $\displaystyle \text{ , Rate of Interest } = 6.5\%$ $\displaystyle \text{ per annum } T = 1$

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

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

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Compound Interest Formula

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

Answer:

$\displaystyle \text{Principal } = \text{ Rs. } 16000$ $\displaystyle \text{ , Rate of Interest } = 15\%$ $\displaystyle \text{ per annum } n = 2$

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

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

$\displaystyle = 21160 \text{ Rs. }$

Therefore compound interest $\displaystyle = A - P = 21160 - 16000 = 5160 \text{ Rs. }$

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

Answer:

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

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

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

$\displaystyle = 25000 \times 1.08 \times 1.10 \times 1.12$

$\displaystyle = 33264 \text{ Rs. }$

Therefore compound interest $\displaystyle = A - P = 33264 - 25000 = 8264 \text{ Rs. }$

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

Answer:

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

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

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

$\displaystyle = 15000 \times (1.04)^3 = 16872.96 \text{ Rs. }$

Therefore compound interest $\displaystyle = A - P = 16872.96 - 15000 = 1872.96 \text{ Rs. }$

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Question 13: A man borrows $\displaystyle Rs. 1000$ at $\displaystyle 10\%$ per annum simple interest for $\displaystyle 3$ 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 $\displaystyle 3$ years?

Answer:

Simple Interest:

$\displaystyle \text{Principal } = \text{ Rs. } 1000$ $\displaystyle \text{ , Rate of Interest } : R=10\%,$ annually, $\displaystyle T= 3$ years

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

Compound Interest:

$\displaystyle \text{Principal } = \text{ Rs. } 1000$ $\displaystyle \text{ , Rate of Interest } = 10\%$ $\displaystyle \text{ per annum } n = 3$

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

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

$\displaystyle = 1331 \text{ Rs. }$

Therefore compound interest $\displaystyle = A - P = 1331 - 1000 = 331 \text{ Rs. }$

Therefore gain $\displaystyle = 331 - 300 = 31 \text{ Rs. }$

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

Answer:

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } = 10\%$ $\displaystyle \text{ per annum } n = 3 A = 4374 Rs$

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

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

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

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

Answer:

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

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

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

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

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

Answer:

Simple Interest:

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } : R=10\%,$ annually, $\displaystyle T= 3$ years

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

Compound Interest:

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } = 10\%$ $\displaystyle \text{ per annum } n = 3$

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

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

$\displaystyle A = 1.331P$

Compound Interest $\displaystyle = 1.331P-P = 0.331 P$

Given $\displaystyle 93 = 0.331P - 0.3P$

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

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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 $\displaystyle 10\%$ for a year is $\displaystyle Rs. 15$ . Find the sum of money lent out.

Answer:

Simple Interest:

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } : R=10\%,$ annually, $\displaystyle T= 1$ years

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

Compound Interest:

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

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

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

$\displaystyle A = 1.1025P$

Compound Interest $\displaystyle = 1.1025P-P = 0.1025 P$

Given $\displaystyle 15 = 0.1025P - 0.1P$

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

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

Answer:

Yearly:

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

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

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

$\displaystyle A = 1.20 \times 1.10 P = 1.32 P$

Therefore Compound Interest $\displaystyle = 1.32P - P = 0.32 P$

Half Yearly:

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

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

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

$\displaystyle A = 1.331P$

Therefore Compound Interest $\displaystyle = 1.331P - P = 0.331 P$

Given $\displaystyle 178.50 = 0.331P - 0.332P = 0.011$

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

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

Answer:

Compound Interest:

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } = 12\%$ $\displaystyle \text{ per annum } n = 2$

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

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

$\displaystyle A = 1.2544P$

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

Simple Interest:

Principal $\displaystyle = 3125$ $\displaystyle \text{ , Rate of Interest } : R=12\%,$ annually, $\displaystyle T= 2$ years

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

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

Answer:

Simple Interest:

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } : R=14\%,$ annually, $\displaystyle T= 2$ years

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

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

Compound Interest:

Principal $\displaystyle = 12500$ $\displaystyle \text{ , Rate of Interest } = 14\%$ $\displaystyle \text{ per annum } n = 2$

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

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

$\displaystyle A = 16245$

Compound Interest $\displaystyle = 16245 - 12500 = 3745 Rs.$

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

Answer:

Simple Interest:

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } : R=8\%,$ annually, $\displaystyle T= 2$ years

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

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

Compound Interest:

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

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

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

$\displaystyle A = 6084 \text{ Rs. }$

Compound Interest $\displaystyle = 6084 - 5625 = 459 Rs.$

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Question 22: A sum of money is lent out at compound interest for $\displaystyle 2 years$ years at $\displaystyle 20\%$ 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 $\displaystyle Rs. 482$ more by way of interest. Calculate the sum of money-lent out.

Answer:

Compound Interest (yearly):

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } = 20\%$ $\displaystyle \text{ per annum } n = 2$

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

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

$\displaystyle A = 1.44P$

Compound Interest $\displaystyle = 1.44P - P = 0.44P$

Compound Interest (Half yearly):

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

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

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

$\displaystyle A = 1.4641$

Compound Interest $\displaystyle = 1.4641P - P = 0.4641P$

Given $\displaystyle 482 = 0.4641P - 0.44P \Rightarrow P = 20000 \text{ Rs. }$

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

Answer:

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

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

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

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

$\displaystyle \\$

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

Answer:

Compound Interest (yearly):

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } = r\%$ $\displaystyle \text{ per annum } n = 1 Amount = 9200 \text{ Rs. }$

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

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } = r\%$ $\displaystyle \text{ per annum } n = 3 Amount = 12167 \text{ Rs. }$

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

Dividing (ii) by (i) we get

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

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

$\displaystyle \Rightarrow r = 1.15 - 1 = 15\%$

Now substituting in (i)

$\displaystyle P = \frac{9200}{1.15} = 8000 \text{ Rs. }$

$\displaystyle \\$

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

Answer:

$\displaystyle 1^{st} Year$

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } = r\%$ $\displaystyle \text{ per annum } n = 1 Amount = A_1$

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

$\displaystyle 2^{nd} Year$

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } = r\%$ $\displaystyle \text{ per annum } n = 1 Amount = A_2$

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

$\displaystyle 3^{rd} Year$

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } = r\%$ $\displaystyle \text{ per annum } n = 1 Amount = A_3$

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

Given: $\displaystyle A_2 - A_1 = 880$ and $\displaystyle A_3 - A_2 = 968$

Therefore

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

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

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

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

Dividing (v) by (iv) we get

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

$\displaystyle \Rightarrow r = 1.1 - 1 = 10\%$

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

$\displaystyle \\$

Question 26: The compound interest on a sum of money for $\displaystyle 2$ years is $\displaystyle Rs. 1050$ and the simple interest on the same sum for the same period and at the same rate is $\displaystyle Rs. 1000$ . Find: (i) the rate of interest (ii) the sum.

Answer:

Compound Interest:

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } = r\%$ $\displaystyle \text{ per annum } T = 2 years$

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

Compound Interest $\displaystyle 1050 = P \Big( 1+ \frac{r}{100} \Big)^2 - P$

$\displaystyle \Rightarrow 1050 = P \Big[ (1+ \frac{r}{100})^2 - 1 \Big]$

$\displaystyle \Rightarrow 1050 = P (2+ \frac{r}{100} )( \frac{r}{100} )$ … … … … … (i)

Simple Interest:

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } : R=r\%,$ annually, $\displaystyle T= 2$ years

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

$\displaystyle 1000 = \frac{P \times r \times 2}{100} \Rightarrow 50000 = Pr$

$\displaystyle \Rightarrow P = \frac{50000}{r}$ … … … … … (ii)

Now solving (i) and (ii) we get

$latex \displaystyle 1050 = \frac{50000}{r} (2+ \frac{r}{100} )( \frac{r}{100} )

$\displaystyle \Rightarrow 1050 = 1000 + 5r$

$\displaystyle \Rightarrow r = 10\%$

$\displaystyle \Rightarrow P = \frac{50000}{10} = 5000 \text{ Rs. }$

$\displaystyle \\$

Question 27: Find the rate percent per annum if $\displaystyle Rs. 2000$ amounts to $\displaystyle Rs. 2662$ in $\displaystyle 1\frac{1}{2}$ years, interest being compounded half-yearly.

Answer:

Principal $\displaystyle = 2000 \text{ Rs. }$ $\displaystyle \text{ , Rate of Interest } : R=r\%,$ $\displaystyle \text{ semi-annually, } T= 1.5 years$ year $\displaystyle Amount = 2662 \text{ Rs. }$

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

$\displaystyle 26620 = 2000 \Big( 1+ \frac{r}{100 \times 2} \Big)^{1.5 \times 2}$

$\displaystyle 1.331 = \Big( 1+ \frac{r}{100 \times 2} \Big)^{1.5 \times 2}$

$\displaystyle 1.10 = 1 + \frac{r}{200} \Rightarrow r = 20\%$ per annum

$\displaystyle \\$

Question 28: Find the rate at which a sum of money will double itself in $\displaystyle 3$ years if the interest is compounded annually.

Answer:

Principal $\displaystyle = x \text{ Rs. }$ $\displaystyle \text{ , Rate of Interest } : R=r\%,$ annually, $\displaystyle n= 3$ year $\displaystyle Amount = 2x \text{ Rs. }$

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

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

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

$\displaystyle 1.2599 = 1 + \frac{r}{100} \Rightarrow r = 25.99\%$ per annum

$\displaystyle \\$

Question 29: Find the rate at which a sum of money will become four amount times the original in $\displaystyle 2$ years if the interest is compounded half yearly.

Answer:

Principal $\displaystyle = x \text{ Rs. }$ $\displaystyle \text{ , Rate of Interest } : R=r\%,$ annually, $\displaystyle n= 2$ year $\displaystyle Amount = 4x \text{ Rs. }$

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

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

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

$\displaystyle 1.4142 = 1 + \frac{r}{200} \Rightarrow r = 82.84\%$ per annum

$\displaystyle \\$

Question 30: A sum compounded annually becomes $\displaystyle \frac{25}{16}$ times of itself in $\displaystyle 2$ years. Determine the rate of interest.

Answer:

Principal $\displaystyle = x \text{ Rs. }$ $\displaystyle \text{ , Rate of Interest } : R=r\%,$ annually, $\displaystyle n= 2$ year $\displaystyle Amount = \frac{25}{16}x \text{ Rs. }$

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

$\displaystyle \frac{25}{16} x = x \Big( 1+ \frac{r}{100} \Big)^{2}$

$\displaystyle \frac{25}{16} = \Big( 1+ \frac{r}{100 \times 2} \Big)^2$

$\displaystyle \frac{5}{4} = 1 + \frac{r}{100} \Rightarrow r = 25\%$ per annum

$\displaystyle \\$

Question 31: Rishi invested $\displaystyle Rs. 30000$ in a finance company and received $\displaystyle Rs. 39930$ after $\displaystyle 1\frac{1}{2}$ years. Find the rate of interest per annum compounded semi-annually.

Answer:

Principal $\displaystyle = 30000 \text{ Rs. }$ $\displaystyle \text{ , Rate of Interest } : R=r\%,$ $\displaystyle \text{ semi-annually, } T= 1.5 years$ year $\displaystyle Amount = 39930 \text{ Rs. }$

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

$\displaystyle 39930 = 30000 \Big( 1+ \frac{r}{100 \times 2} \Big)^{1.5 \times 2}$

$\displaystyle 1.331 = \Big( 1+ \frac{r}{100 \times 2} \Big)^{1.5 \times 2}$

$\displaystyle 1.10 = 1 + \frac{r}{200} \Rightarrow r = 20\%$ per annum

$\displaystyle \\$

Question 32: In how much time would $\displaystyle Rs. 5000$ amounts to $\displaystyle Rs. 6655$ at $\displaystyle 10\%$ per annum compound interest?

Answer:

Principal $\displaystyle = 5000 \text{ Rs. }$ $\displaystyle \text{ , Rate of Interest } : R=10\%,$ annually, $\displaystyle T= n years$ year $\displaystyle Amount = 6655 \text{ Rs. }$

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

$\displaystyle 6655 = 5000 \Big( 1+ \frac{10}{100} \Big)^{n}$

$\displaystyle 1.331 = \Big( 1+ \frac{10}{100} \Big)^{n}$

$\displaystyle 1.1^3 = \Big( 1+ \frac{10}{100} \Big)^{n}$

$\displaystyle \Rightarrow n = 3 years$

$\displaystyle \\$

Question 33: In what time will $\displaystyle Rs. 4400$ become $\displaystyle Rs. 4576$ at $\displaystyle 8\%$ per annum interest compounded half-yearly?

Answer:

Principal $\displaystyle = 4400 \text{ Rs. }$ $\displaystyle \text{ , Rate of Interest } : R=8\%,$ annually, $\displaystyle T= n years$ year $\displaystyle Amount = 4576 \text{ Rs. }$

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

$\displaystyle 4576 = 4400 \Big(1+ \frac{8}{100 \times 2} \Big)^{n \times 2}$

$\displaystyle 1.04 = \Big( 1+ \frac{4}{100} \Big)^{2n}$

$\displaystyle 1.04 = ( 1.04)^{2n}$

$\displaystyle \Rightarrow 2n = 1 year \Rightarrow n = \frac{1}{2} year$

$\displaystyle \\$

Question 34: The compound interest, calculated yearly, on a certain sum of money for the second year is $\displaystyle Rs. 1320$ and for the third year is $\displaystyle Rs. 1452$ . Calculate the rate of interest and the original sum of money.

Answer:

$\displaystyle 1^{st} Year$

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } = r\%$ $\displaystyle \text{ per annum } n = 1 Amount = A_1$

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

$\displaystyle 2^{nd} Year$

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } = r\%$ $\displaystyle \text{ per annum } n = 1 Amount = A_2$

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

$\displaystyle 3^{rd} Year$

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } = r\%$ $\displaystyle \text{ per annum } n = 1 Amount = A_3$

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

Given: $\displaystyle A_2 - A_1 = 1320$ and $\displaystyle A_3 - A_2 = 1452$

Therefore

$\displaystyle P \Big( 1+ \frac{r}{100} \Big)^2 - P \Big( 1+ \frac{r}{100} \Big)^1 = 1320$

$\displaystyle \Rightarrow P (1+ \frac{r}{100} )( \frac{r}{100} ) = 1320$ … … … … (iv)

$\displaystyle P \Big( 1+ \frac{r}{100} \Big)^3 - P \Big( 1+ \frac{r}{100} \Big)^1 = 1452$

$\displaystyle \Rightarrow P (1+ \frac{r}{100} )^2( \frac{r}{100} ) = 1452$ … … … … (v)

Dividing (v) by (iv) we get

$\displaystyle \frac{1452}{1320} = 1 + \frac{r}{100}$

$\displaystyle \Rightarrow r = 1.1 - 1 = 10\%$

Substituting in (iv) $\displaystyle P = \frac{1320 \times 100}{10 \times 1.1} = 12000 \text{ Rs. }$

$\displaystyle \\$

Question 35: On what sum of money will the difference between the compound interest and simple interest for $\displaystyle 2$ years be equal to $\displaystyle Rs. 25$ , if the rate of interest charged for both is $\displaystyle 5\%$ per annum?

Answer:

Compound Interest:

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } = 5\%$ $\displaystyle \text{ per annum } T = 2 years$

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

$\displaystyle A = P \Big( 1+ \frac{5}{100} \Big)^2 = 1.1025 P$

Compound Interest $\displaystyle = 1.1025 P - P = 0.1025P$

Simple Interest:

Principal $\displaystyle = P$ $\displaystyle \text{ , Rate of Interest } : R=5\%,$ annually, $\displaystyle T= 2$ years

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

$\displaystyle = \frac{P \times 5 \times 2}{100} \Rightarrow = 0.1P$

Given: $\displaystyle 25 = 0.1025 P - 0.1P \Rightarrow P = 10000 \text{ Rs. }$

$\displaystyle \\$

Question 36: Mr Kumar borrowed $\displaystyle Rs. 15000$ for two years. The rate of interest for the two successive years are $\displaystyle 8\%$ and $\displaystyle 10\%$ respectively. If he repays $\displaystyle Rs. 6200$ at the end of the first year, find the outstanding amount at the end of second year.

Answer:

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

$\displaystyle \text{Principal } = \text{ Rs. } 15000$ $\displaystyle \text{ , Rate of Interest } = 8\%$ $\displaystyle \text{ per annum } T = 1$

$\displaystyle \text{Therefore Interest } = \frac{P \times R \times T}{100} = \frac{15000 \times 8 \times 1}{100} = 1200 \text{ Rs. }$

$\displaystyle \text{Therefore, the Amount at the end of } 1^{st} \text{ year } =15000 + 1200 = 16200 \text{ Rs. }$

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

$\displaystyle \text{Principal } = \text{ Rs. } 16200 - 6200 = 10000$ $\displaystyle \text{ , Rate of Interest } = 10\%$ $\displaystyle \text{ per annum } T = 1$

$\displaystyle \text{Therefore Interest } = \frac{P \times R \times T}{100} = \frac{10000 \times 10 \times 1}{100} =1000 \text{ Rs. }$

Therefore, the Amount at the end of $\displaystyle 2^{nd}$ year $\displaystyle = 10000 + 1000 = 11000 \text{ Rs. }$

$\displaystyle \\$

Depreciation Problems

Question 37: The value of a machine depreciates at the rate of $\displaystyle 10\%$ per annum. what will be its value $\displaystyle 2$ years hence if the present value is $\displaystyle Rs. 100000$ ? Also, find the total depreciation during this period.

Answer:

Present Value $\displaystyle V_0 = 100000 \text{ Rs. }$ Rate of Depreciation $\displaystyle = 10\%$ No of Years $\displaystyle = 2$

$\displaystyle V_2 = V_0 \Big( 1- \frac{R}{100} \Big)^n$

$\displaystyle V_2 = 100000 \Big( 1- \frac{10}{100} \Big)^2$

$\displaystyle V_2= 81000 \text{ Rs. }$

Total depreciation $\displaystyle = 100000 - 81000 = 19000 \text{ Rs. }$

$\displaystyle \\$

Question 38: Pritam bought a plot of land for $\displaystyle Rs. 640000$ . Its value is increasing by $\displaystyle 5\%$ of its previous value after every six months. What will be the value of the plot after $\displaystyle 2$ years?

Answer:

Present Value $\displaystyle = 640000 \text{ Rs. }$ $\displaystyle \text{ , Rate of Interest } = 5\%$ half yearly , $\displaystyle T = 2 years$

$\displaystyle V_2 = P \Big( 1+ \frac{r}{100k} \Big)^{nk}$

$\displaystyle V_2 = 640000 \Big( 1+ \frac{10}{100 \times 2} \Big)^{2 \times 2}$

$\displaystyle V_2 = 777924 \text{ Rs. }$

$\displaystyle \\$

Question 39: The value of a machine depreciates at the rate of $\displaystyle 10\%$ per annum. It was purchased $\displaystyle 3$ years ago. If its present value is $\displaystyle Rs. 43740$ , find its purchase price.

Answer:

Present Value $\displaystyle V_0 = 43740 \text{ Rs. }$ Rate of Depreciation $\displaystyle = 10\%$ No of Years $\displaystyle = 3$ Value 3 years back $\displaystyle V_{-3} = x$

$\displaystyle V_0 = V_{-3} \Big( 1- \frac{R}{100} \Big)^3$

$\displaystyle 43740 = V_{-3} \Big( 1- \frac{10}{100} \Big)^3$

$\displaystyle V_{-3} = 60000 \text{ Rs. }$

Total depreciation $\displaystyle = 60000 - 43740 = 16260 \text{ Rs. }$

$\displaystyle \\$

Question 40: The cost of a T.V. set was quoted $\displaystyle Rs. 17000$ at the beginning of 1999. In the beginning of 2000 the price was hiked by $\displaystyle 5\%$ . Because of decrease in demand the cost was reduced by $\displaystyle 4\%$ in the beginning of 2001. What was the cost of the T.V. set in 2001?

Answer:

Cost of TV at the beginning of 1999 $\displaystyle = 17000 \text{ Rs. }$

Cost of TV at the beginning of 2000 $\displaystyle = 17000 \times 1.05 = 17850 \text{ Rs. }$

Cost of TV at the beginning of 2001 $\displaystyle = 17850 \times 0.96 = 17136 \text{ Rs. }$

$\displaystyle \\$

Question 41: Ashish staffed the business with an initial investment of $\displaystyle Rs. 500000$ . In the first he incurred a loss of $\displaystyle 4\%$ . However, during the second year he earned a profit of $\displaystyle 5\%$ which in third year rose to $\displaystyle 10\%$ . Calculate the net profit for the entire period of $\displaystyle 3$ years.

Answer:

Initial Investment at the beginning of 1st Year $\displaystyle = 500000 \text{ Rs. }$

Capital at the beginning of 2nd Year $\displaystyle = 500000 \times 0.96 = 480000 \text{ Rs. }$

Capital at the beginning of 3rd Year $\displaystyle = 480000 \times 1.05 = 504000 \text{ Rs. }$

Capital at the beginning of 4th Year $\displaystyle = 504000 \times 1.10 = 554400 \text{ Rs. }$

Net profit $\displaystyle = 554400 - 500000 = 54400 \text{ Rs. }$

$\displaystyle \\$

Population Questions

Question 42: The present population of a town is $\displaystyle 28000$ . If it increases at the rate of $\displaystyle 5\%$ per annum, what will be its $\displaystyle \text{Population after } 2$ years?

Answer:

Population $\displaystyle P = 28000$ , Rate $\displaystyle R = 5\%$ $\displaystyle \text{ per annum } n = 2$ years

$\displaystyle \text{Population after } n$ years $\displaystyle = P \Big(1+ \frac{R}{100} \Big)^{n}$

$\displaystyle \text{Population after } 2$ years $\displaystyle = 28000 \Big(1+ \frac{5}{100} \Big)^{2} = 30870$

$\displaystyle \\$

Question 43: The present population of a town is $\displaystyle 25000$ . It grows at first $\displaystyle 4\%, 5\%$ and $\displaystyle 8\%$ during year, second year and third year respectively. Find lts $\displaystyle \text{Population after } 3$ years.

Answer:

Current Population $\displaystyle P = 25000$ , Rate $\displaystyle R_1 = 4\%, R_2=5%, R_3 = 8\%$ $\displaystyle \text{ per annum } n = 3$ years

$\displaystyle \text{Population after } 3^{rd} Year = P \Big(1+ \frac{R_1}{100} \Big).\Big(1+ \frac{R_2}{100} \Big).\Big(1+ \frac{R_3}{100} \Big)$

$\displaystyle \text{Population after } 3^{rd} Year = 25000 \Big(1+ \frac{4}{100} \Big).\Big(1+ \frac{5}{100} \Big).\Big(1+ \frac{8}{100} \Big)$

$\displaystyle = 25000 \times 1.04 \times 1.05 \times 1.08 = 29484$

$\displaystyle \\$

Question 44: There is a continuous growth in population of a village at the rate of $\displaystyle 5\%$ per annum. If its present population is $\displaystyle 9261$ , what was it $\displaystyle 3$ years ago?

Answer:

Population 3 years ago $\displaystyle = P_{-3}$ Current Population $\displaystyle P_0 = 9261$ , Rate $\displaystyle R = 5\%$ $\displaystyle \text{ per annum } n = 3$ years

Therefore $\displaystyle P_{0} = P_{-3} \Big(1+ \frac{R}{100} \Big)^{n}$

$\displaystyle 9261 =P_{-3} \Big(1+ \frac{5}{100} \Big)^{3}$

$\displaystyle \Rightarrow P_{-3} = 8000$

$\displaystyle \\$

Question 45: In a factory the production of scooters rose to $\displaystyle 46305$ from $\displaystyle 40000$ in $\displaystyle 3$ years. Find the annual rate of growth of the production of scooters.

Answer:

$\displaystyle P = 40000 A = 46305 n = 3 years, Rate = R\%$

$\displaystyle 46305 = 40000 \Big(1+ \frac{R}{100} \Big)^{3}$

$\displaystyle 1.05 = 1 + \frac{R}{100}$

$\displaystyle \Rightarrow R = 5\%$

$\displaystyle \\$

Question 46: The population of a town increases at the rate of $\displaystyle 50$ per thousand. Its $\displaystyle \text{Population after } 2$ years will be $\displaystyle 22050$ . Find its present population.

Answer:

Rate of Increase $\displaystyle = 5% A$ (population in 2 years) $\displaystyle = 22050, P$ (Initial population) $\displaystyle = P$

$\displaystyle 22050 = P \Big(1+ \frac{5}{100} \Big)^{2}$

$\displaystyle \Rightarrow P = 20000$

$\displaystyle \\$

Question 47: The count of bacteria in a culture grows by $\displaystyle 10\%$ in the first hour, decreases by $\displaystyle 8\%$ in the second hour and again increase by $\displaystyle 12\%$ in third hour. If the count of bacteria in the sample $\displaystyle 13125000$ , what will be the count of bacteria after $\displaystyle 3$ hour.

Answer:

Current bacteria Population $\displaystyle P = 13125000$ , Rate $\displaystyle R_1 = 10\%, R_2=-8%, R_3 = 12\%$ $\displaystyle \text{ per annum } n = 3$ hours

$\displaystyle \text{ Bacteria Population after } 3^{rd} hours = P \Big(1+ \frac{R_1}{100} \Big).\Big(1+ \frac{R_2}{100} \Big).\Big(1+ \frac{R_3}{100} \Big)$

$\displaystyle \text{ Bacteria Population after } 3^{rd} hours = 13125000 \Big(1+ \frac{10}{100} \Big).\Big(1+ \frac{-8}{100} \Big).\Big(1+ \frac{12}{100} \Big)$

$\displaystyle = 13125000 \times 1.10 \times 0.92 \times 1.12 = 14876400$

$\displaystyle \\$

Question 48: $\displaystyle 6400$ workers were employed to construct a river bridge in four years. At the end First year, $\displaystyle 25\%$ workers were retrenched. At the second year $\displaystyle 25\%$ of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by $\displaystyle 25\%$ the end of the third year. How many workers were working during the fourth year?

Answer:

Current worker Population $\displaystyle P = 6400$ , Rate $\displaystyle R_1 = -25\%, R_2=-25%, R_3 = 25\%$ $\displaystyle \text{ per annum } n = 3$ years

$\displaystyle \text{Worker Population after } 3^{rd} years = P \Big(1+ \frac{R_1}{100} \Big).\Big(1+ \frac{R_2}{100} \Big).\Big(1+ \frac{R_3}{100} \Big)$

$\displaystyle \text{Worker Population after } 3^{rd} years = 6400 \Big(1+ \frac{-25}{100} \Big).\Big(1+ \frac{-25}{100} \Big).\Big(1+ \frac{25}{100} \Big)$

$\displaystyle = 64000 \times 0.75 \times 0.75 \times 1.25 = 4500$

$\displaystyle \\$

Question 49: A man started a factory with an initial investment of $\displaystyle Rs. 100000$ . In the first year, he incurred a loss of $\displaystyle 5\%$ . However, during the second year; he earned a profit of $\displaystyle 10\%$ which in the third year rose to $\displaystyle 12\%$ . Calculate the profit for the entire period of three years.

Answer:

Initial Investment $\displaystyle P = 100000$ , Rate $\displaystyle R_1 = -5\%, R_2=10%, R_3 = 12\%$ $\displaystyle \text{ per annum } n = 3$ years

$\displaystyle \text{ Worker Population after } 3^{rd} years = P \Big(1+ \frac{R_1}{100} \Big).\Big(1+ \frac{R_2}{100} \Big).\Big(1+ \frac{R_3}{100} \Big)$

$\displaystyle \text{ Worker Population after } 3^{rd} years = 100000 \Big(1+ \frac{-5}{100} \Big).\Big(1+ \frac{10}{100} \Big).\Big(1+ \frac{12}{100} \Big)$

$\displaystyle = 100000 \times 0.95 \times 1.10 \times 1.12 = 117040 \text{ Rs. }$

Therefore net profit $\displaystyle = 117040 - 100000 = 17040 \text{ Rs. }$

$\displaystyle \\$

Question 50: The population of a city increases each year by $\displaystyle 4\%$ of what it had been at the beginning of each year. If the population in 1999 had been $\displaystyle 6760000$ , find the population of the city in (i) 2001 (ii) 1997.