Question: The population of a city A in 2012, consisted of 45% men, 35% women and the remaining children. Of the children, 40% were female and 60% male. Of the men, 10% were over the age of 60 years and 25% below the age of 40 years. Of the women, 20% were over the age of 60 years and an equal number were under 40 years of age. The number of men increased by 4% in 2013 and that of women increased by 6%. The population of the town in 2012 was 200,000.

What percentage of the population in 2013 consisted of children?

My Answer : 

2012 : 45%M+ 35%W + 20%C

2013 : 49%M(Increased by 4%) + 41%W(Increased by 6%) + (Remaining)10%C

My Expectation is 10% of children would be in 2013

Ans: 16%.

But i am not sure how this is 16%. can somebody please explain correct one?

Answer: 
You are correct that, for $x \in (0,1)$, $x > x^3$. However, recall what the sign function $\text{sgn}( x)$ is:

$$\text{sgn}(x) =\left\{ \begin{matrix}
-1 & x<0\\ 
0 & x=0\\ 
1 & x>0
\end{matrix} \right.$$

Thus, since $\text{sgn}(x-x^3) = 1$ for all $x \in (0,1)$,

$$\int\_0^1 \text{sgn}(x-x^3) dx = \int\_0^1 1 \; dx$$
You are correct that $x-x^3>0$ for $x\in(0,1)$, so $\mathrm{sgn}(x-x^3)$ is $1$ on that interval, so the answer is $1$.