Math, asked by aaryanraj18, 8 months ago

The population of a country increased by an average of 2% per year from 2000 to 2003. If the population of this country was 2 000 000 on December 31, 2003, then the population of this country on January 1, 2000, to the nearest thousand would have been

A. 1 846 000
B. 1 852 000
C. 1 000 000
D. 1 500 000​

Answers

Answered by sahilnagarbowdi1605
1

Answer:

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Step-by-step explanation:

your answer

ANSWER

⇒ Let population of the country on one 1stJan,2000 be P

2000

.

⇒ Population on 31stDec,2003(P

2003

=2,000,000

⇒ Increase =2%=0.02

⇒ Final population after n years =(1+0.02)

n

=(1.002)

n

According to the question,

⇒ P

2003

=(1.02)

4

P

2000

[ As the 2000 to 2003 will be 4 years ]

⇒ P

2000

=

(1.02)

4

P

2003

=

(1.02)

4

2,000,000

=1847690.85

∴ Here, nearest thusand would be 18,47,000

Answered by abhijay1705
1

Answer:

Step-by-step explanation:

population of the country on Jan 1, 2000 =  P

r = 2%

total duration = 4 years,  :  2000, 2001, 2002, 2003

If we take the compounding of the rate of growth :

        2, 000, 000 = P (1 + 2/100)^4  = P (1.02)^4

              P =  1, 847,691

If we take the growth of the population similar to simple interest of money deposits, then :

          2, 000, 000 = P + P * 2/100 * 4

            P = 2, 000, 000 / 1.08 = 1, 851, 852

        In this case the answer is B

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