Question 1: The height of a plant is \displaystyle 80 cm and it is expected to grow at the rate of \displaystyle 20\% every month. What will be its height at the end of \displaystyle 3 \text{ months } ?


\displaystyle P=80 \text{ cm; }  r=20\%; n=3 \text{ months }

\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = 80 \Big(1+ \frac{20}{100} \Big)^3 = 138.24 \text{ cm }

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Question 2: The cost of a machine is supposed to depreciate each year by \displaystyle 12\% of its value at the beginning of the year. If the machine is valued at \displaystyle Rs.44,000 at the beginning of \displaystyle 2008 , find its value:

  1. At the end of \displaystyle 2009  
  2. At the beginning of \displaystyle 2007  


\displaystyle P=44000  \text{ Rs.; }  r=12\%; n=2 \text{ years }

At the end of \displaystyle 2009  

\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n = 44000 \Big(1- \frac{12}{100} \Big)^2 = 34073.6 \text{ Rs. per cm }

At the beginning of \displaystyle 2007  

\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = 44000 \Big(1+ \frac{12}{100} \Big)^1 = 49280 \text{ Rs. per cm }

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Question 3: The value of a machine is estimated to be \displaystyle Rs.27000 at the end of \displaystyle 2004 and \displaystyle Rs.21,870 at the beginning of \displaystyle 2007 . Supposing it depreciates at a constant rate per year of its value at the beginning of the year, calculate:

  1. The rate of depreciation;
  2. The value of the machine at the beginning of \displaystyle 2004 .


\displaystyle P=27000  \text{ Rs.; }  A= 21870  \text{ Rs.; }  r=r\%; n=2 \text{ years }

At the end of \displaystyle 2007  

\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow 21870= 27000 \Big(1- \frac{r}{100} \Big)^2 \Rightarrow r=10\%  

At the beginning of \displaystyle 2004  

\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow 27000= A \Big(1- \frac{10}{100} \Big)^1 \Rightarrow A=30000 \text{ Rs. }  

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Question 4: The value of in article decreased for two years at the rate of \displaystyle 10\% per year and then in the third year it increased by \displaystyle 10\% . Find the original value of the article, if its value at the end of \displaystyle 3 \text{ years } is \displaystyle Rs.40095 .


\displaystyle P=P  \text{ Rs.; }  A= A  \text{ Rs.; }  r=10\%; n=2 \text{ years }

At the end of \displaystyle 2^{nd} year  

\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow P= A \Big(1- \frac{10}{100} \Big)^2 \Rightarrow P=0.81A  

At the end of \displaystyle 3^{rd} Year  

\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow 40095= 0.81A \Big(1+ \frac{10}{100} \Big)^1 \Rightarrow A=45000 \text{ Rs. }  

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Question 5: According to a census taken towards the end of the year \displaystyle 2009 , the population of a rural town was found to be \displaystyle 64000 . The census authority also found that the population of this particular town had a growth of \displaystyle 5\% per annum. In how many years after \displaystyle 2009 did the population of this town reach \displaystyle 74088 ?


\displaystyle P=64000. A= 74088; r=5\%; n=n \text{ years }

At the end of \displaystyle n^{nd} year  

\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow 74088= 64000 \Big(1+ \frac{5}{100} \Big)^n \Rightarrow n=3  

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Question 6: The population of a town decreased by \displaystyle 12\% during \displaystyle 1998 and then increased by \displaystyle 8\% during \displaystyle 1999 . Find the population of the town, at the beginning of \displaystyle 1998 , if at the end of \displaystyle 1999 its population was \displaystyle 2,85,120 .


At the end of \displaystyle 1999 \text{ year }

\displaystyle P=P; A= 285120; r=8\%; n=1 \text{ years }

\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow 285120=P \Big(1+ \frac{8}{100} \Big)^1 \Rightarrow P=264000  

At the beginning of \displaystyle 1998 Year  

\displaystyle P=P; A= 26400; r=8\%; n=1 \text{ years }

\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow 264000= P \Big(1- \frac{12}{100} \Big)^1 \Rightarrow P=300000  

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Question 7: A sum of money, invested at compound interest, amounts to \displaystyle Rs.16,500 in \displaystyle 1 \text{ year } and to \displaystyle Rs.19965 in \displaystyle 3 \text{ years } . Find the rate per cent and the original sum of money.


\displaystyle P=16500; A= 19965; r=r\%; n=2 \text{ years }

\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow 19965=16500 \Big(1+ \frac{r}{100} \Big)^2 \Rightarrow r=10\%  

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Question 8: The difference between C.I. and S.I. on \displaystyle Rs.7500 for two years is \displaystyle Rs.12 at the same rate of interest per annum. Find the rate of interest.


Simple Interest

\displaystyle P=7500  \text{ Rs.; }  T=2 \text{ years;  }  r=x\%  

\displaystyle S.I=7500 \times \frac{x}{100} \times 2 = 150x  

Compound Interest

\displaystyle P=7500; A= A; r=x\%; n=2 \text{ years }

\displaystyle A= 7500 \times \Big(1+ \frac{x}{100} \Big)^2  

\displaystyle C.I. = 7500 \times \Big(1+ \frac{x}{100} \Big)^2 - 7500  

\displaystyle \text{Given  }  C.I. - S.I. = 12 \Rightarrow 7500 \times \Big(1+ \frac{x}{100} \Big)^2 - 7500 - 150x = 12  

\displaystyle \Rightarrow 7500+7500 \times \frac{x^2}{10000} + 7500 \times 2 \times \frac{x}{100} - 7500 -150x = 12  

\displaystyle \Rightarrow 0.75x^2=12  

\displaystyle \Rightarrow x= 4%  

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Question 9: A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in \displaystyle 10 \text{ years } . Find in how many years will the money become twenty-seven times of itself of the same rate of interest p.a.


After 10 Years

\displaystyle P=3A; A= A; r=x\%; n=10 \text{ years }

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

\displaystyle \Big(1+ \frac{x}{10} \Big)=3^{\frac{1}{10}} ... ... ... ... ... ... ... ... i)  

After n Years

\displaystyle P=27A; A= A; r=x\%; n=n \text{ years }

\displaystyle 27A=A \times \Big(1+ \frac{x}{100} \Big)^n  

\displaystyle 27=1 \times \Big(1+ \frac{x}{100} \Big)^n  

Substituting from i)

\displaystyle 27=1 \times \Big(3^{\frac{1}{10}} \Big)^n \Rightarrow n=30 \text{ years }

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Question 10: Sharma borrowed a certain sum of money at \displaystyle 10\% per annum compounded annually. If by paying \displaystyle Rs.19360 at the end of the second year and \displaystyle Rs.31,944 at the end of the third he clears the debt; find the sum borrowed by him.


\displaystyle P=x; A= A; r=10\%; n=2 \text{ years }

\displaystyle A= x \times \Big(1+ \frac{10}{100} \Big)^2 \Rightarrow A= 1.21x  

Third year

\displaystyle P=(1.21x-19360); A=31944 ; r=10\%; n=1 \text{ years }

\displaystyle 31944= (1.21x-19360) \times \Big(1+ \frac{10}{100} \Big)^1 \Rightarrow x= 40000 \text{ Rs. }  

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Question 11: The difference between compound interest for a year payable half-yearly and 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. [1998]


Simple Interest

\displaystyle P=x  \text{ Rs.; }  T=1 \text{ years;  }  r=10\%  

\displaystyle S.I=x \times \frac{10}{100} \times 1 = 0.1x  

Compound Interest

\displaystyle P=x; A= A; r=10\%; n=2 \text{ half years }  

\displaystyle A= x \times \Big(1+ \frac{10}{200} \Big)^2 = x \times \Big( \frac{21}{20} \Big)^2  

\displaystyle C.I. = x \times \Big( \frac{21}{20} \Big)^2 - x  

\displaystyle \text{Given  }  C.I. - S.I. = x \times \Big( \frac{21}{20} \Big)^2 - x - 0.1x = 15  

\displaystyle \Rightarrow x= 6000 \text{ Rs. }  

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Question 12: The ages of Person 1 and Person 2 are \displaystyle 16 \text{ years } and \displaystyle 18 \text{ years } respectively. In what ratio must they invest money at \displaystyle 5\% p.a. compounded yearly so that both get the same sum on attaining the age of \displaystyle 25 \text{ years } ?


Person 1:

\displaystyle P=x; A= P1; r=5\%; n=9 \text{ years }

\displaystyle P1= x \times \Big(1+ \frac{5}{100} \Big)^9  

Person 2:

\displaystyle P=x; A= P2; r=5\%; n=7 \text{ years }

\displaystyle P2= x \times \Big(1+ \frac{5}{100} \Big)^7  

\displaystyle \text{Given  }  P1=P2  

\displaystyle P1:P2 = x \times \Big(1+ \frac{5}{100} \Big)^9 : x \times \Big(1+ \frac{5}{100} \Big)^7 = 441:400