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

$P=80\ cm; \ r=20\%; \ n=3 \ months$

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

$\\$

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

1. At the end of $2009$
2. At the beginning of $2007$

$P=44000\ Rs.; \ r=12\%; \ n=2 \ Years$

At the end of $2009$

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

At the beginning of $2007$

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

$\\$

Question 3: The value of a machine is estimated to be $Rs.27000$ at the end of $2004$ and $Rs.21,870$ at the beginning of $2007$ . 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 $2004$ .

$P=27000\ Rs.; A= 21870 \ Rs.; \ r=r\%; \ n=2 \ Years$

At the end of $2007$

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

At the beginning of $2004$

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

$\\$

$\\$

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

$P=P\ Rs.; A= A \ Rs.; \ r=10\%; \ n=2 \ Years$

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

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

At the end of $3^{rd} \ Year$

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

$\\$

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

$P=64000. A= 74088; \ r=5\%; \ n=n \ Years$

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

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

$\\$

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

At the end of $1999 \ year$

$P=P; A= 285120; \ r=8\%; \ n=1 \ Years$

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

At the beginning of $1998 \ Year$

$P=P; A= 26400; \ r=8\%; \ n=1 \ Years$

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

$\\$

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

$P=16500; A= 19965; \ r=r\%; \ n=2 \ Years$

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

$\\$

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

Simple Interest

$P=7500 \ Rs.; T=2 \ Years; r=x\%$

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

Compound Interest

$P=7500; A= A; \ r=x\%; \ n=2 \ Years$

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

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

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

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

$\Rightarrow 0.75x^2=12$

$\Rightarrow x= 4%$

$\\$

Question 9: A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in $10$ 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

$P=3A; A= A; \ r=x\%; \ n=10 \ Years$

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

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

After n Years

$P=27A; A= A; \ r=x\%; \ n=n \ Years$

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

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

Substituting from i)

$27=1 \times (3^{\frac{1}{10}})^n \Rightarrow n=30 \ Years$

$\\$

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

$P=x; A= A; \ r=10\%; \ n=2 \ Years$

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

Third year

$P=(1.21x-19360); A=31944 ; \ r=10\%; \ n=1 \ Years$

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

$\\$

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 $10\%$ for a year is $Rs.15$ . Find the sum of money lent out. [1998]

Simple Interest

$P=x \ Rs.; T=1 \ Years; r=10\%$

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

Compound Interest

$P=x; A= A; \ r=10\%; \ n=2 \ half \ years$

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

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

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

$\Rightarrow x= 6000 \ Rs.$

$\\$

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

Person 1:

$P=x; A= P1; \ r=5\%; \ n=9 \ years$

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

Person 2:

$P=x; A= P2; \ r=5\%; \ n=7 \ years$

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

Given $P1=P2$

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

$\\$