Question

The population of a certain town was 50,000. In a year, male population was increased
by 5% and female population was increased by 3%. Now the population became
52020. Then what was the number of males and females in the previous year?​

Answer 1

Given -

=> Initial population of town was = 50,000 people

=> Final population of town became = 52,020 people when -

• Male population increases by 5 % and
• Female population increased by 3 %

To find -

=> Original number of males and females in the town.

Here -

=> We will basically equate the given information using a variable and find the required answer.

=> Let x be the number of males in the town originally.

=> Since original population numbered 50,000 originally.

=> Original number of females = (50,000-x)

For new male and female populations -

∵ Male population increases by 5 % of original male population -

∵ Original male population = x males

∴ New number of males = $x + \dfrac{5}{100} *x$ $=>$ $x+\dfrac{1}{20} *x => \dfrac{20x+x}{20} =>\dfrac{21x}{20}$

∵ Female population increases by 3 % of original female population -

∵ Original female population = (50,000-x) females

∴ New number of females = $(50000-x)+\dfrac{3}{100} *(50000-x)$

                                            = $(50000-x)(1+\dfrac{3}{100} )$

                                            = $(50000-x)(\dfrac{100+3}{100} )$

                                            = $(50000-x)(\dfrac{103}{100} )$

Now, adding the two found values -

∵ New population adds up to 52,020 -

=> New no. of females + new no. of males = 52,020

Thus -

=> $(50000-x)(\dfrac{103}{100} )+\dfrac{21x}{20}=52,020$

↔ Opening the brackets, we get,

=> $\dfrac{50000*103-103x}{100} +\dfrac{21x}{20}=52,020$

=> $\dfrac{5150000-103x}{100} +\dfrac{21x}{20}=52,020$

↔ Taking LCM of 100 and 20 i.e. 100 in numerator in L.H.S .

=> $\dfrac{5150000-103x}{100} +\dfrac{5*21x}{5*20}=52,020$

=> $\dfrac{5150000-103x}{100} +\dfrac{105x}{100}=52,020$

=> $\dfrac{5150000-103x+105x}{100} =52,020$

=> $\dfrac{5150000+2x}{100} =52,020$

=> ${5150000+2x} = 5202000$

=> 2x = 5202000 - 5150000

=> 2x = 52000

=> x = 52,000/2

=> x = 26,000

Now -

∵ We assumed original number of males to be x.

=> x = 26,000

∴ The number of males in the previous year = 26,000 males

∵ We assumed original number of females to be (50,000-x).

=> x = 26,000

=> 50,000-x = 50,000-26,000

=> 50,000 - x = 24,000

∴ The number of females in the previous year = 24,000 females

Answer 2

Answer is in the attachment