Question 1: Find the money, invested at \displaystyle 10\% compounded annually, on which the sum of interest for the first year and the third year is \displaystyle Rs. 1768 .

Answer:

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

\displaystyle \text{Therefore Interest } = \frac{x \times 10 \times 1}{100} = \text{ Rs. } 0.1x

\displaystyle \text{and, Amount } = x+0.1x = \text{ Rs. } 1.1x

\displaystyle \text{For 2nd year: } P = \text{ Rs. } 1.1x; R=10\% \text{ and } T=1 \text{ year }

\displaystyle \text{Therefore Interest } = \frac{1.1x \times 10 \times 1}{100} = \text{ Rs. } 0.11x

\displaystyle \text{and, Amount } = 1.10x + 0.11x = \text{ Rs. } 1.21x

\displaystyle \text{For 3rd year: } P = \text{ Rs. } 1.21x; R=10\% \text{ and } T=1 \text{ year }

\displaystyle \text{Therefore Interest } = \frac{1.21x \times 10 \times 1}{100} = \text{ Rs. } 0.121x

\displaystyle \text{and, Amount } = 1.21x + 0.121x = \text{ Rs. } 1.331x

\displaystyle \text{Given } 0.1x+0.121x= 1768 \Rightarrow x= \text{ Rs. } 8000

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Question 2: A sum of money is invested at compound interest payable annually. The interest in first two successive years is \displaystyle Rs. 1350 and \displaystyle Rs. 1440 respectively. Find;

  1. The rate of interest
  2. The original Sum
  3. The interest earned in the third year.

Answer:

Difference between the Compound interest of two successive years

\displaystyle = 1440-1350 = \text{ Rs. } 90

\displaystyle \Rightarrow \text{ Rs. } 90 \text{ is the interest on Rs. } 1350

\displaystyle \therefore \text{Rate of Interest } = \frac{100 \times I}{P \times T} \% = \frac{100 \times 90}{1350 \times 1} \% = 6 \frac{2}{3} \%

\displaystyle \text{For 1st year: } P = \text{ Rs. } P; R=6 \frac{2}{3} \% \text{ and } T=1 \text{ year }

\displaystyle \text{Interest } = \text{ Rs. } 1350 = P \times \frac{6\frac{2}{3}}{100} \times 1 \Rightarrow P= 20250

\displaystyle \text{and, Amount } = 20250 + 1350x = \text{ Rs. } 21600

\displaystyle \text{For 2nd year: } P = \text{ Rs. } 21600; R=6 \frac{2}{3} \% \text{ and } T=1 \text{ year }

\displaystyle \text{Therefore Interest } = \frac{21600 \times 6\frac{2}{3} \times 1}{100} = \text{ Rs. } 1440

\displaystyle \text{and, Amount } =21600 +1440 = \text{ Rs. } 23040

\displaystyle \text{For 3rd year: } P=Rs. 23040; R=6 \frac{2}{3} \% \text{ and } T=1 \text{ year }

\displaystyle \text{Therefore Interest } = \frac{23040 \times 6\frac{2}{3} \times 1}{100} = \text{ Rs. } 1536

\displaystyle \text{and, Amount } =23040 +1536 = \text{ Rs. } 24576

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Question 3: A sum of money amounts to \displaystyle Rs. 46305 in \displaystyle 1 year and to \displaystyle Rs. 48620.25 in \displaystyle 1\frac{1}{2} years at compound interest, compounded semi-annually. Find the sum and the rate of interest per annum.

Answer:

\displaystyle \text{Interest earned in 3rd half year } =48620.25-46305 = \text{ Rs. } 2315.25

\displaystyle P=Rs. 46305; I = \text{ Rs. } 2315.25 T = \frac{1}{2} ; R=x\%

\displaystyle \text{Interest }= \text{ Rs. } 2315.25=46305 \times \frac{x}{100} \times \frac{1}{2} \Rightarrow x= 10\%

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

\displaystyle \text{Interest } = P \times \frac{10}{100} \times \frac{1}{2} = 0.05P

\displaystyle \text{and, Amount } = P+0.05P = 1.05P

For 2nd half year: \displaystyle P = \text{ Rs. } 1.05P; R=10\% \text{ and } T= \frac{1}{2} \text{ year }

\displaystyle \text{Interest } = 1.05 \times \frac{10}{100} \times \frac{1}{2} = 0.0525P

\displaystyle \therefore \text{Amount }= 1.05P+0.0525P = 1.1025P = 46305 \Rightarrow P = \text{ Rs. } 42000

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Question 4: The cost of machine in \displaystyle Rs. 32,000 . Its value depreciates at the rate of \displaystyle 5\% every year. Find the total depreciation in its value by the end of \displaystyle 2 years.

Answer:

\displaystyle \text{For 1st year: } P = \text{ Rs. } 32000; R=5\% \text{ and } T=1 \text{ year }

\displaystyle \text{Depreciation } = 32000 \times \frac{5}{100} \times 1 = \text{ Rs. } 1600

and, \displaystyle \text{Value } = 32000-1600 = \text{ Rs. } 30400

\displaystyle \text{For 2nd year: } P = \text{ Rs. } 30400; R=5\% \text{ and } T=1 \text{ year }

\displaystyle \text{Depreciation } = 30400 \times \frac{5}{100} \times 1 = \text{ Rs. } 1520

and, \displaystyle \text{Value } = 30400-1520 = \text{ Rs. } 28880

\displaystyle \therefore \text{Total Depreciation } = 1600+1520 = \text{ Rs. } 3120

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Question 5: Find the sum, invested at \displaystyle 10\% compounded annually, on which the interest for the third year exceeds the interest of the first year by \displaystyle Rs. 252 .

Answer:

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

\displaystyle \text{Interest } = x \times \frac{10}{100} \times 1 = \text{ Rs. } 0.1x

\displaystyle \text{and, Amount } = x+0.1x = \text{ Rs. } 1.1x

\displaystyle \text{For 2nd year: } P = \text{ Rs. } 1.1x; R=10\% \text{ and } T=1 \text{ year }

\displaystyle \text{Interest } = 1.1x \times \frac{10}{100} \times 1 = \text{ Rs. } 0.11x

\displaystyle \text{and, Amount } = 1.1x+0.11x = \text{ Rs. } 1.21x

\displaystyle \text{For 3rd year: } P = \text{ Rs. } 1.21x; R=10\% \text{ and } T=1 \text{ year }

\displaystyle \text{Interest } = 1.21x \times \frac{10}{100} \times 1 = \text{ Rs. } 0.121x

\displaystyle \text{and, Amount } = 1.21x+0.121x = \text{ Rs. } 1.331x

\displaystyle \text{Given } 0.121x-0.1x=252 \Rightarrow x= \frac{252}{0.021} = \text{ Rs. } 12000

\displaystyle \\

Question 6: The compound interest, compounded annually, on a certain sum is \displaystyle Rs. 9680 in second year and is \displaystyle Rs. 10648 in third year. Calculate;

  1. The rate of interest, sum borrowed
  2. The interest of \displaystyle 1^{st} year.

Answer:

Difference between the Compound interest of two successive years

\displaystyle = 10648-9680 = \text{ Rs. } 968

\displaystyle \Rightarrow \text{ Rs. } 968 is the interest on \displaystyle Rs. 9680

\displaystyle \therefore \text{Rate of Interest } = \frac{100 \times I}{P \times T} \% = \frac{100 \times 968}{9680 \times 1} \% = 10\%

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

Therefore Interest on it for 1st year \displaystyle = 10\% \text{ of Rs. } 100 = \text{ Rs. } 10

\displaystyle \Rightarrow \text{Amount in one year }  = 100 + 10 = \text{ Rs. } 110

\displaystyle \therefore \text{Interest on it for } 2^{nd} \text{Year } = 10\% \text{ of } 110 = \text{ Rs. } 11

\displaystyle \Rightarrow \text{Amount in 2nd year } = 110 + 11 = \text{ Rs. } 121

When interest of 2nd year \displaystyle = \text{ Rs. } 11, \text{sum is Rs. } 100

\displaystyle \Rightarrow \text{When interest of 2nd year }= \text{ Rs. } 9680 , then

\displaystyle \text{sum }= \frac{100}{11} \times 9680 = \text{ Rs. } 88000

\displaystyle \text{Interest of 1st year } = 88000 \times \frac{10}{100} \times 1 = \text{ Rs. } 8800

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Question 7: A man borrows \displaystyle Rs. 10,000 at \displaystyle 10\% compound interest compounded yearly. At the end of each year, he pays back \displaystyle 30\% of the sum borrowed. How much money is left unpaid just after the second year?

Answer:

\displaystyle \text{For 1st year: } P = \text{ Rs. } 10000; R=10\% \text{ and } T=1 \text{ year }

\displaystyle \text{Interest }= 10000 \times \frac{10}{100} \times 1 = \text{ Rs. } 1000

\displaystyle \text{and, Amount } = 10000+1000 = \text{ Rs. } 11000

\displaystyle \text{For 2nd year: } P = \text{ Rs. } (11000-3000)=8000; R=10\% \text{ and } T=1 \text{ year }

\displaystyle \text{Interest }= 8000 \times \frac{10}{100} \times 1 = \text{ Rs. } 800

\displaystyle \text{and, Amount } = 8000+800 = \text{ Rs. } 8800

\displaystyle \text{Amount left }= 8800-3000 = \text{ Rs. } 5800

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Question 8: A man borrows \displaystyle Rs. 10000 at \displaystyle 10\% compound interest compounded yearly. At the end of each year, he pays back \displaystyle 20\% of the amount for that year. How much money is left unpaid just after the second year?

Answer:

\displaystyle \text{For 1st year: } P = \text{ Rs. } 10000; R=10\% \text{ and } T=1 \text{ year }

\displaystyle \text{Interest }= 10000 \times \frac{10}{100} \times 1 = \text{ Rs. } 1000

\displaystyle \text{and, Amount } = 10000+1000 = \text{ Rs. } 11000

\displaystyle \text{For 2nd year: } P = \text{ Rs. } (11000-2000)=9000; R=10\% \text{ and } T=1 \text{ year }

\displaystyle \text{Interest } = 9000 \times \frac{10}{100} \times 1 = \text{ Rs. } 900

\displaystyle \text{and, Amount } = 9000+900 = \text{ Rs. } 9900

\displaystyle \text{Amount left } = 9900-2000 = \text{ Rs. } 7900

\displaystyle \\

Question 9: The population of a town increases \displaystyle 10\% every \displaystyle 3 years. If the present population of the town is \displaystyle 72600 , calculate:

  1. Its population after \displaystyle 2 years;
  2. Its population \displaystyle 2 years ago;

Answer:

\displaystyle \text{For 2 years ago: } P = \text{ Rs. } x; R=10\% \text{ and } T=1 \text{ year }

\displaystyle \text{Increase } = x \times \frac{10}{100} \times 1 = \text{ Rs. } 0.1x

\displaystyle \text{and, Population } = x+0.1x = \text{ Rs. } 1.1x

\displaystyle \text{For 1 year ago: } P = \text{ Rs. } 1.1x; R=10\% \text{ and } T=1 \text{ year }

\displaystyle \text{Increase } = 1.1x \times \frac{10}{100} \times 1 = \text{ Rs. } 0.11x

\displaystyle \text{and, Population } = 1.1x+0.11x = \text{ Rs. } 1.21x

\displaystyle \text{Given } 1.21x= 72600 \Rightarrow x= 60000

\displaystyle \text{For 1st year: } P = \text{ Rs. } 72600; R=10\% \text{ and } T=1 \text{ year }

\displaystyle \text{Increase } = 72600 \times \frac{10}{100} \times 1 = \text{ Rs. } 7260

\displaystyle \text{and, Population } = 72600+7260 = \text{ Rs. } 79860

\displaystyle \text{For 2nd year: } P = \text{ Rs. } 79860; R=10\% \text{ and } T=1 \text{ year }

\displaystyle \text{Increase } = 79860 \times \frac{10}{100} \times 1 = \text{ Rs. } 7986

\displaystyle \text{and, Population } = 79860+7986 = \text{ Rs. } 87846

\displaystyle \\

Question 10: 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. [2014]

Answer:

Difference between the Compound interest of two successive years

\displaystyle = 1452-1320 = \text{ Rs. } 132

\displaystyle \Rightarrow \text{ Rs. } 132 is the interest on \displaystyle Rs. 1320

\displaystyle \therefore \text{Rate of Interest } = \frac{100 \times I}{P \times T} \% = \frac{100 \times 132}{1320 \times 1} \% = 10\%

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

Therefore Interest on it for 1st year \displaystyle = 10\% \text{of } Rs. 100 = \text{ Rs. } 10

\displaystyle \Rightarrow \text{Amount in one year } = 100 + 10 = \text{ Rs. } 110

\displaystyle \therefore \text{Interest on it for } 2^{nd} \text{Year } = 10\% \text{of } 110 = \text{ Rs. } 11

\displaystyle \Rightarrow \text{Amount in 2nd year } = 110 + 11 = \text{ Rs. } 121

When interest of 2nd year \displaystyle = \text{ Rs. } 11, \text{sum is Rs. } 100

\displaystyle \Rightarrow \text{When interest of 2nd year }= \text{ Rs. } 1320 , then

\displaystyle \text{Sum }= \frac{100}{11} \times 1320 = \text{ Rs. } 12000

\displaystyle \\