Question

23. The population of a town is 10000 now
and was 8000 two years ago. If it grows at
the same rate, what will it be 2 yr hence ?
(a) 1.2000 () 12500 (e) 12750 (d) 12755​

Answer 1

Answer:

12500

Step-by-step explanation:

Finding Rate of Interest 2 years ago:

P×R/100×T + P = 10000

8000×R/100×2 + 8000 = 10000

160R= 2000

R= 12.5

10000×12.5%×2 + 10000

= 2500 + 10000

=12500

So, after 2 years from now, the population of the city will be 12500.

Answer 2

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$</p><p>\blue{\underline {\underline {Question \: : }}}$

The population of a town is 10000 now, and was 8000 two years ago.

If it grows at the same rate, what will be the population after 2 years?

[a] 12000

[b] 12500

[c] 12750

[d] 12755

The Correct Answer is Option B. The Population of the town after 2 years is 12500.

This can be solved by using the S.I. Formula.

$\frac{prt}{100} + p \: = 10000$

$p \: = \: 8000$

$\frac{r}{50} = \frac{10}{8}$

$r = 12.5$

New Population = 12.5% of 10000

=125000.

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