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Cencus population of a district in 1981 was 4.54 lacs, while in year 2001 it was 7.44 lacs. What was the estimated mid-year population of that district in year 2009.
Read Solution (Total 10)
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- 1981......>4.54
diff=20 diff=2.9
2001......>7.44
diff=8 this diff=(8*2.9)/20=1.16
2009......>x=?
hence x=7.44+1.16=8.6 lac - 10 years agoHelpfull: Yes(39) No(2)
- population in 1981 is 4.54 lac
population in 2001 is 7.44 lac
total increment in population in 20 yrs=7.44-4.54=2.9 lac
total increment in 1 yr=2.9/20 lac=0.145 lac
so total increment in 8 years(2009-2001)=8*0.145=1.16 lac
so the population in 2009 is=7.44+1.16= 8.6 lacs
..............answer
- 10 years agoHelpfull: Yes(37) No(1)
- 8.6 LAKH !!
P(estimate)=P1+n/N {P2-P1) WHERE n=NO. MONThs from P1 census to date of estiMate
N=NO. of months between census,P1= FIRST census,P2=second census..ANSWER IS 8.6 LAKH... - 10 years agoHelpfull: Yes(6) No(2)
- 8.57 lakh
- 10 years agoHelpfull: Yes(2) No(1)
- @Pratham - can you please answer with explanation???
As the method is needed.... - 10 years agoHelpfull: Yes(2) No(0)
- Ans 8.61 lak
1981 4.54 lack
2001. 7.44 lack
R= ? 7.44 =4.54((1+R/100)^20). R= 16/5
Then use same formula
We get 8.61 lak - 10 years agoHelpfull: Yes(2) No(1)
- every year 0.145 lacs population is increasing..
at the end of 8 and half years population is going to 8.67 lacs - 10 years agoHelpfull: Yes(2) No(0)
- in the year 1981 population is 4.54l
in 2001 population is 7.44L
so within 20 years(1981 to 2001)the population increased by 2.90L
so for 8 years(2001 to 2009)the population increases by,
(2.90*8)/20 =1.16
so in 2009,the population is 7.44L+1.16L=8.60L - 9 years agoHelpfull: Yes(1) No(0)
- 8,52,750 is the answer
- 10 years agoHelpfull: Yes(0) No(1)
- correct answer is 16 check it with 5 and 3
- 10 years agoHelpfull: Yes(0) No(1)
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