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? ........
I Have given images above to solve ....... Try to help me ! .​

Answer 1

Answer:

No of males=100x=100*260=26,000

No of females=50000-26000=24,000

Step-by-step explanation:

I believe the following is easy ,Still I have attached the filled sheet

Let in the previous year males are 100x and female are 100y

Then 100x+100y=50,000

100(x+y)=50,000

x+y=500

y=500-x.....................(1)

Now increse in males=5% of 100x=100x*5/100=5x

Increase in females= 3% of 100y=100y*3/100=3y

Total population becomes

50000+5x+3y=52020

Putting value of y from(1)

5x+3*(500-x)=52020-50000

5x+1500-3x=2020

2x=2020-1500=520

x=520/2=260

Thus in the previous year

No of males=100x=100*260=26,000

No of females=50000-26000=24,000

Answer 2

Answer:

x = -260

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