Question

The population of a town is $53000$. If in a year, the number of males was increased by $6\%$ and that of the females by $4\%$, the population would grow to $55630$. Find the difference between the number of males and females in the town at present.

• A. $3000$
• B. $4000$
• C. $2000$
• D. $5000$

Answer 1

The correct option is C $2000$

Let the population of men and women be $x$ and $y$ respectively.

Then according to the question ,

$x+y=53000$ ....(1)

($x+6\%$ of $x$ )+($y+4\%$ of $y$)$=55630$

$(x+0.06x)+(y+0.04y)=55630$

$1.06x+1.04y=55630$

$106x+104y=5563000$ ....(2)

Multiplying equation (1) by $104$ and subtracting from $\left(2\right)$

$106x+104y-104x-104y=5563000-5512000$

$2x=51000$

$x=25500$

Putting values of $x$ in (1)

$y=53000-25500$

$y=27500$

Difference between number of men and women

$x-y=27500-25500$=$2000$