The population of a city is 35000. On an increase of 6% in the number of men and an increase of 4% in the number of women, the population would become 36760. What was the number of women initially?

118000
219000
317000
420000

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let number of men in the population be x \\ Number of women = (35000 - x) \\ Increase in the number of men \\ = 6 % o f x = \frac{6 x}{100} \\ Increase in the number of women \\ = (3500 - x) \times \frac{4}{100} \\ Increase in whole population \\ = 36760 - 35000 = 1760 \\ Now, \\ \frac{6 x}{100} + [(35000 - x) \times \frac{4}{100}] = 1760 \\ [(6 x - 4 x) + 35000 \times \frac{4}{100}] = 1760 \\ 2 x + 35000 \times 4 = 1760 \times 100 \\ 2 x = 176000 - 35000 \times 4 \\ x = 18000 \\ Number of men = 18000 \\ Number of women \\ = 35000 - 18000 \\ = 17000 \end{aligned}

$\eqalign{ & {\text{Let number of men in the population be }}x \cr & {\text{Number of women}} = \left( {35000 - x} \right) \cr & {\text{Increase in the number of men}} \cr & = 6\% \,of\,x = \frac{{6x}}{{100}} \cr & {\text{Increase in the number of women}} \cr & = \left( {3500 - x} \right) \times \frac{4}{{100}} \cr & {\text{Increase in whole population}} \cr & = 36760 - 35000 = 1760 \cr & {\text{Now}}, \cr & \frac{{6x}}{{100}} + \left[ {\left( {35000 - x} \right) \times \frac{4}{{100}}} \right] = 1760 \cr & \left[ {\left( {6x - 4x} \right) + 35000 \times \frac{4}{{100}}} \right] = 1760 \cr & 2x + 35000 \times 4 = 1760 \times 100 \cr & 2x = 176000 - 35000 \times 4 \cr & x = 18000 \cr & {\text{Number}}\,{\text{of}}\,{\text{men}} = 18000 \cr & {\text{Number}}\,{\text{of}}\,{\text{women}} \cr & = 35000 - 18000 \cr & = 17000 \cr}$