Question

The population of a town at present is 50,000. If during a year the number of males increases by 5% and that females by 3%, the population would grow to 52,020 in one year. Find the number of males and females in the town at present.

Answer 1

Answer:

No of male=100x=260*100=26000

female=100y=100*240=24000

Step-by-step explanation:

let male=100x,female=100y

then 100x+100y=50000

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

male increases by 5% =5x

so total male=105x

similarly femae=103y

so 105x+103y=52020-------(2)

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from (1) y=500-x

so from(2) 105x+103(500-x)=52020

105x+51500-103x=52020

2x=52020-51500=520

x=260

from(1) again

260+y=500

y=500-260=240

Hence

No of male=100x=260*100=26000

female=100y=100*240=24000