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The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

1) Find the population in 2001 and

2) What would be its population in 2005

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48735 and 59535

53020 and 65423

25410 and 45213

35465 and 61231

answer is A.

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Detailed Solution

There was a rise in the population to 54,000 in 2003.
For first part, we have the year 2001. So,
Time (n) = 2 years

r = 5 %

Population in 2003, P= 54000
Total population =x

P = \frac{x}{{(1 - \frac{r}{100})}^{n}}

54000 = \frac{x}{{(1 - \frac{5}{100})}^{(2)}}

54000 = \frac{x}{0.9025}

x = 48735

The population was 48,735 in 2001.
For second part we have the year 2005.
So,
Time (n) = 2 years

r = 5 %

Population in 2005=A
Total population = 54000
Now, population in 2005,

A = 54000 {(1 + \frac{5}{100})}^{(2)}

= 54000 (1.1025)

= 59, 535

So, 2005 will see a population of 59,535.
Correct option is 1.

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