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\text {Q1. Find the compound interest when principal = Rs 3000, rate = 5% per annum and time = 2 years.}
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\text {Sol. } \\
\text {Principal for the first year }=\text {Rs. } 3000 \\
\text {Interest for the first year Rs. } = 3000 \times 5 \times \frac{1}{100}=\text {Rs. } 150 \\
\text {Amount at the end of the first year }= 3000 +150=\text {Rs. } 3150 \\
\text {Principle Interest for the second year } =3150 \times 5 \times \frac{1}{100}=\text {Rs. } 157.50 \\
\text {Amount at the end of the second year }=3307.50 \\
\text {Compound interest} =3307.50 – 3000=\text {Rs. } 307.50
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\text {Q2. What will be the compound interest on Rs. 4000 in two years when rate of interest is 5% per annum? }
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\text {Sol. We know that amount A at the end of n years at the rate of R % per annum is given by A } =P(1+\frac{R}{100})^{n} \\
\text {Giver P =Rs. 4000, R=5 % p.a , n=2 years } \\
\Rightarrow \quad A= 4000(1+\frac{5}{100})^{2} \\
=4000(\frac{105}{100})^{2} \\
= 4410 \\
CI=A – P =4410 -4000 = \text {Rs.} 410 \\
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\text {Q3. Rohit deposited Rs. 8000 with a finance company for 3 years at an interest of 15% per annum. What is the compound interest that Rohit gets after 3 years?}
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\text {Sol. We know that amount A at the end of n years at the rate of R % per annum is given by A } =P(1+\frac{R}{100})^{n} \\
\text {Giver P =Rs. 8000, R=15 % p.a , n=3 years } \\
\Rightarrow \quad A= 8000(1+\frac{15}{100})^{3} \\
=8000(\frac{115}{100})^{3} \\
= 12167 \\
CI=A – P = 12167 – 8000 = \text {Rs.} 4167 \\
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\text {Q4. Find the compound interest on Rs. 1000 at the rate of 8% per annum for } 1 \frac {1}{2} \text { years when interest is compounded half yearly.}
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\text {Sol. Given P =Rs. 1000, R=8 % p.a , n=1.5 years } \\
\text {We know that } A =P(1+\frac{R}{2 \times 100})^{2n} \\
\Rightarrow \quad A= 1000(1+\frac{8}{200})^{3} \\
=1000(\frac{208}{100})^{3} \\
= 1124.86 \\
CI=A – P = 1124.86 – 1000 = \text {Rs.} 124.86 \\
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\text {Q5. Find the compound interest on Rs. 160000 for one year at the rate of 20% per annum, if the interest is compounded quarterly.}
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\text {Sol. Given P =Rs. 160000, R=20 % p.a , n=1 years } \\
\text {We know that } A =P(1+\frac{R}{4 \times 100})^{4n} \\
\Rightarrow \quad A= 160000(1+\frac{20}{400})^{4} \\
=160000(\frac{21}{20})^{4} \\
= 194481 \\
CI=A – P = 194481 – 160000 = \text {Rs.} 34481 \\
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\text {Q6. Swati took a loan of Rs. 16000 against her insurance policy at the rate of } 12 \frac {1}{2} \% \\
\text { per annum. Calculate the total compound interest payable by Swati after 3 years. }
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\text {Sol. We know that amount A at the end of n years at the rate of R % per annum is given by A } =P(1+\frac{R}{100})^{n} \\ \\\text {Giver P =Rs. 16000, R=12.5 % p.a , n=3 years } \\
\Rightarrow \quad A= 16000(1+\frac{12.5}{100})^{3} \\
=8000(\frac{112.5}{100})^{3} \\
= 22781.25 \\
CI=A – P = 22781.25 – 16000 = \text {Rs.} 6781.25 \\
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\text {Q7. Roma borrowed Rs. 64000 from a bank for } 1 \frac {1}{2} \text { years at the rate of 10% per annum. } \\
\text {Compare the total compound interest payable by Roma after } 1 \frac {1}{2} \text { years, if the interest is compounded half-yearly. }
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\text {Sol. Given P =Rs. 64000, R=10 % p.a , n=1.5 years } \\
\text {We know that } A =P(1+\frac{R}{2 \times 100})^{2n} \\
\Rightarrow \quad A= 64000(1+\frac{10}{200})^{3} \\
=64000(\frac{210}{100})^{3} \\
= 74088 \\
CI=A – P = 74088 – 64000 = \text {Rs.} 10088 \\
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\text {Q8. Mewa lal borrowed Rs. 20000 from his friend Rooplal at 18% per annum simple interest. } \\
\text {He lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years. }
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\text {Sol. Amount of SI, Mewa Lal need to pay } =\frac{P \times R \times T}{100}=\frac{20000 \times 18 \times 2}{100}=\text {Rs. } 7200 \\\text {CI he is going to receive } =A – P \\= 20000(1+\frac{18}{100})^{2} – 20000 \\
=27848-20,000 \\
=7848 \\
\text {Hence, Mewa lal total gain in the whole transaction } =7848 – 7200 = \text {Rs. 648}
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\text {Q9. Find the compound interest on Rs. 8000 for 9 months at 20% per annum compounded quarterly. }
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\text {Sol. Given P =Rs. 8000, R=20 % p.a , n=9 months } = \frac {3}{4} yr \\
\text {We know that } A =P(1+\frac{R}{4 \times 100})^{4n} \\
\Rightarrow \quad A= 8000(1+\frac{20}{400})^{4 \times \frac {3}{4}} \\
=8000(\frac{21}{20})^{3} \\
= 9261 \\
CI=A – P = 9261 – 8000 = \text {Rs.} 1261 \\
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\text {Q10. Find the compound interest at the rate of 10% per annum for two years on that principal which in } \\
\text { two years at the rate of 10% per annum given Rs. 200 as simple interest. }
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\text {Sol. SI } =\frac{P \times R \times T}{100} \\
\therefore P=\frac{SI \times 100}{R \times T} \\
P=\frac{200 \times 100}{10 \times 2} \\
P=1000 \\
\text {We know that } A=P(1+\frac{R}{100})^{n} \\
\Rightarrow \quad A=1000(1+\frac{10}{100})^{2} \\
\Rightarrow \quad A = 1210 \\
CI=A – P = 1210 – 1000 = \text {Rs.} 210 \\
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\text {Q11. Find the compound interest on Rs. 64000 for 1 year at the rate of 10% per annum compounded quarterly. }
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\text {Sol. We konw to calculate the interest compounded quarterly }
A=P(1+\frac{R}{400})^{4 n} \\ \\\text {Here P =Rs. 64000, R=10 % p.a , n=1 yr } \\
\Rightarrow \quad A=64000(1+\frac{10}{400})^{4 \times 1} \\
\Rightarrow \quad A=64000(\frac{41}{40})^{4}
\Rightarrow \quad A= 70644.03 \\
CI=A – P = 70644.03 – 64000 = \text {Rs.} 6644.03 \\
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\text {Q12. Ramesh deposited Rs. 7500 in a bank which pays him 12% interest per annum compounded quarterly. }\\
\text {What is the amount which he receives after 9 months.}
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\text {Sol. We konw to calculate the interest compounded quarterly }
A=P(1+\frac{R}{400})^{4 n} \\ \\
\text {Here P =Rs. 7500, R=12 % p.a , n=9 month = \frac {3}{4} yr } \\
\Rightarrow \quad A=7500(1+\frac{12}{400})^{4 \times \frac {3}{4}} \\
\Rightarrow \quad A=7500(\frac{103}{100})^{3}
\Rightarrow \quad A= 8195.45\\
\therefore \text {Ramesh will receive Rs. 8195.45 after 9 months.}
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\text {Q13. Anil borrowed a sum of Rs. 9600 to install a hand pump in his dairy. If the rate of interest is } 5 \frac {1}{2} \% \text { per annum compounded annually, } \\
\text { determine the compound interest which Anil will have to pay after 3 years.}
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\text {Sol. We konw to calculate the interest compounded annually }
A=P(1+\frac{R}{100})^{n} \\ \\
\text {Here P =Rs. 9600, R=5.5 % p.a , n=3 yr } \\
\Rightarrow \quad A=9600(1+\frac{5.5}{100})^{3} \\
\Rightarrow \quad A=9600(\frac{211}{200})^{3} \\
\Rightarrow \quad A= 11272.72 \\
CI=A – P = 11272.72 – 9600 = \text {Rs.} 1672.72\\
\therefore \text {Anil has to pay an CI of Rs. 1672.72}
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\text {Q14. Surabhi borrowed a sum of Rs. 12000 from a finance company to purchase a refrigerator. } \\
\text {If the rate of interest is 5% per annum compounded annually, calculate the compound interest that Surabhi } \\
\text { has to pay to the company after 3 years.}
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\text {Sol. } \\\text {We konw to calculate the interest compounded annually }
A=P(1+\frac{R}{100})^{n} \\ \\\text {Here P =Rs. 12000, R=5 % p.a , n=3 yr } \\\Rightarrow \quad A=12000(1+\frac{5}{100})^{3} \\
\Rightarrow \quad A=12000(\frac{105}{100})^{3}
\Rightarrow \quad A= 13891.50\\CI=A – P = 13891.50 – 12000 = \text {Rs.} 1891.50\\
\therefore \text {Surabhi has to pay an CI of Rs. 1891.50}
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\text {Q15. Daljit received a sum of Rs. 40000 as a loan from a finance company. If the rate of interest } \\
\text { is 7% per annum compounded annually, calculate the compound interest that Daljit pays after 2 years.}
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\text {Sol. } \\\text {We konw to calculate the interest compounded annually }
A=P(1+\frac{R}{100})^{n} \\ \\\text {Here P =Rs. 40000, R=7 % p.a , n=7 yr } \\\Rightarrow \quad A=40000(1+\frac{7}{100})^{7} \\
\Rightarrow \quad A=40000(\frac{107}{100})^{7}
\Rightarrow \quad A= 45796\\CI=A – P = 45796 – 40000 = \text {Rs.} 5796\\
\therefore \text {Daljit has to pay an CI of Rs. 5796.}
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