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KG4Question 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?

Answer:

P=80\  cm; \ r=20\%; \ n=3 \ months

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

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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

Answer:

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

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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 .

Answer:

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.

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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 .

Answer:

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.

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EG10Question 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 ?

Answer:

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

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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 .

Answer:

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

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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.

Answer:

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\%

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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.

Answer:

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%

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EG3Question 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.

Answer:

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

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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.

Answer:

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. 

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

Answer:

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.  

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EG8Question 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?

Answer:

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

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