1.

10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum( find the population in 2001.

Answer»

less than 2003 in two years.

A(2003)= P(2001)(1+R/100)^n

54000 = P (2001)(1+5/100)²

54000 = P(2001)(1+1/20)²

54000= P(2001)(21/20)²

54000 = P(2001)((21×21)/(20×20))

P(2001)=( 54000×20×20)/21×21

Population in 2001 = 48979.59= 48980(approx)

ER. RAVI KUMAR ROY

(ii)ATQ, population is increasing. Therefore population in 2005,

A(2005)= P(1+R/100)^n

= 54000(1+5/100)²

= 54000(1+1/20)²

= 54000( 21/20)²

= 54000(21/20)× (21/20)

=( 54000× 441)/ 400

=( 540×441)/4

= 238140/4

= 59535

Population in 2005=59535

let population in 2001 =x54000=x[1+(5/100)]^254000=x[1+(1/20)]^254000=x(21/20)^254000×(20/21)^2=xx=54000×20/21×20/21x=48980 (approx)so, population in 2001 is 48980. answer

48980 the correct answer of the given question



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