GRE Quantitative Reasoning Percentile Problem

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 1

$$ 100-({45+\frac{45×4}{100}+35+\frac{35×6}{100}})$$

Answer 2

Let $m_1$ denote the percentage of men in $2012$

Let $m_2$ denote the percentage of men in $2013$

$m_2 = m_1+(0.04)(m_1) = m_1(1.04).$

$m_2 = 0.35(1.04)$

Let $w_1$ denote the percentage of women in $2012$

Let $w_2$ denote the percentage of women in $2013$

$w_2 = w_1+(0.06)(w_1) = w_1(1.06).$

$w_2 = 0.45(1.06)$

Let $c_2$ denote the percentage of children in $2013$

the sum of all percentages of population types represent $100%$ of the population, that is we use the number $1=100/100$ to state: $1=m_2+w_2+c_2$.

$c_2=1-(m_2+w_2)$.

$c_2=1-((0.35*1.04)+(0.45*1.06))$.

$c_2=0.159$

$c_2$ is approximately 16% of the population (which one?).

I personally hate such problems specially in crucial exam! Good luck!