Question

The population of a town increases by 10.5% every year. If its present population is 795000
it a year ago?
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Answer 1

let the population of town 1 year ago be x.

according to given condition,

x + 10.5x/100 = 795000

100x + 10.5x = 79500000

110.5x = 79500000

x = 79500000/ 110.5

x =719457

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Answer 2

$\huge{\boxed {\mathfrak \pink {\fcolorbox {red}{blue}{AnSwEr}}}}$

Let the population of town a year ago be x.

Therefore,

= x+(10.5%of x) = 795000

= x+105x/1000 = 795000

= 1105x/1000 = 795000

= 1105x = 795000×1000

= x = 795000000/1105

= 719457

Hence, the population a year ago was 719457

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