Question

the present population of city is 27010. and this is increased by 2% every year. find the 2 years ago population.


solve step by step
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Answer 1

Answer:

Answer

Correct option is

D

2,50,000

Let the population before two years ago be x

Given the population increases by 10%

So increase in population after one year is

100

10

×x=0.1x

So population after one year is x+0.1x=1.1x

Increase in population after another one year is

100

10

×1.1x=0.11x

So population at present is 1.1x+0.11x=1.21x

But given population at present is 3,02,500

⟹1.21x=302500

⟹x=

1.21

302500

=2,50,000

Population of city two years ago is 2,50,000

Answer 2

${\large{\bf{\red{\underline{\underline{Given :}}}}}}$

The present population of city is 27010. and this is increased by 2% every year.

￣￣￣￣￣￣￣￣￣￣￣￣

${\large{\bf{\blue{\underline{\underline{To Find:}}}}}}$

Find the 2 years ago population.

￣￣￣￣￣￣￣￣￣￣￣￣

${\large{\bf{\orange{\underline{\underline{Formula \: used:}}}}}}$

${\sf{\pink{Population = \frac{present \: population \times rate \times time}{100} }}}$

￣￣￣￣￣￣￣￣￣￣￣￣

${\large{\bf{\green{\underline{\underline{Solution :}}}}}}$

Population after 1 year :-

Present population = 27,010

Rate = 2 %

Time = 1 year

Population after 1 year =

${\sf{\pink{ = \frac{present \: population \times rate \times time}{100} }}}$

${\cancel{\green{ = \frac{27010 \times 2 \times 1}{100} }}}$

${\red{\boxed{\mathfrak{= 540.2 }}}}$

Population after 2 years :-

Present population = 27,010 + 540.2

Rate = 2 %

Time = 1 year

${\sf{\pink{ = \frac{present \: population \times rate \times time}{100} }}}$

${\sf{\green{ = \frac{27550.2 \times 2\times 1}{100} }}}$

${\mathfrak{\red{\boxed{= 551.004}}}}$

Hence,

${\sf{\purple{Population \: increase \: in \: 1st \: year = 540.2}}}$

${\sf{\purple{Population \: increase \: in \: 2nd \: year = 551.004}}}$

${\sf{\purple{Total \: population \: increase = 540.2 + 551.004 = 1,091.204}}}</p><p>$

￣￣￣￣￣￣￣￣￣￣￣￣

${\mathfrak{\orange{ItzNobita5}}}$