Question

The population of
a town increases by 5% during the first year and by 7% during the second year. At
the end of second year the population was 40446. What was the population at the beginning?
​

Answer 1

Answer:

5%+7%=12%

12/100×40446=4,853.52

increase population 4853 approx

40446-4853=35,593

Answer: 35593

Answer 2

GIVEN :-

• The population of town increses by 5% during first year and by 7% during the second year.
• At the end of second year , population was 40446.

TO FIND :-

• The population of town in the beginning.

SOLUTION :-

Let population in the beginning be 'x'.

Population is increased by 5% during 1st year.

Population after 1 year = x + 5% of x

$\\ \sf \: population(1 \: year) = x + \dfrac{5}{100} (x) \\ \\ \\ \sf \: population(1 \: year) = \cancel\dfrac{105x}{100} \\ \\ \\ \underline{\sf \: population(1 \: year) = \frac{21x}{20}}$

For the second year , population is increased by 7% .

Population after 2nd year

= (21x/20) + 7% of (21x/20)

Hence ,

$\\ \implies\sf \: 40446 = \dfrac{21x}{20} + \dfrac{7}{100} \left( \dfrac{21x}{20} \right) \\ \\ \\ \implies \sf \: 40446 = \dfrac{21x}{20} + \dfrac{147x}{2000} \\ \\ \\ \implies \sf \: 40446 = \dfrac{21x(100) + 147x}{2000} \\ \\ \\ \implies \sf \: 40446 = \dfrac{2100x + 147x}{2000} \\ \\ \\ \implies \sf \: 40446 \times 2000 = 2247x \\ \\ \\ \implies \sf \: 80892000 = 2247x \\ \\ \\\implies \sf \: x = \dfrac{80892000}{2247} \\ \\ \\ \implies \boxed{ \sf \: x = 36000}$

Hence , population at the beginning was 36000.