How many boys, girls, men and women are there? In a village, there are exactly $10$% more boys than girls; $15$% more women than men; $20$% more children than adults. The population is less than $6000$.

Solution:
$b = g + 0.1g$--------(i), where b and g represent the number of boys and girls respectively.
$w = m + 0.15m$-----(ii) , where w and m represent the number of men and women respectively.
$c = a + 0.20a$ -----(iii),  where c and a represent the number of children and adults respectively.
$c + a < 6000$------(iv)
$2a +0.20a  < 6000$ from (iii) and (iv)

So, I arrived at a decimal number for the number of adults, which doesn't seem to be right. Also, do I have to know the exact number of boys, girls etc. since I wasn't given the exact population?  
 A: You have $3$ equations with $4$ variables:


*

*$b=1.10g$

*$w=1.15m$

*$(b+g)=1.20(w+m)$


You have an additional inequity with the same variables:


*

*$b+g+w+m<6000$


So it's simply a matter of finding a single integer solution:


*

*$(b+g)=1.20(w+m)\implies$
$(1.10g+g)=1.2(1.15m+m)\implies$
$2.10g=2.58m$

*$b+g+w+m<6000\implies$
$1.10g+g+1.15m+m<6000\implies$
$2.10g+2.15m<6000\implies$
$2.58m+2.15m<6000\implies$
$4.73m<6000\implies$
$m<1268.5$
Now to the integer solution, which is possibly the hardest part:


*

*$m=700 \implies w=805,g=860,b=946$

A: Your final inequation looks like it should be $a + 0.2a + a < 6000$ and that simplifies to 
$$2.2a < 6000$$
which implies $a < 2727$ approximately.  That sounds about right since there are a little more children than adults.
If your concern is over the fact that you got a decimal, there are two possible answers.  One might be:  you shouldn't worry, it's an inequation, so the true population will be some number where, when you divide by the decimal, you get some whole number.
On the other hand, maybe this is how you're supposed to get the exact population.  Maybe you're supposed to find the population number so that all your numbers become whole numbers.
A: We have
$$
b=g+0.1g=1.1g = 11/10 g\\
w=m+0.15m= 1.15m = 23/20 m\\
c=a+0.20a= 1.2a = 6/5 a
$$
Starting with the third equation, we obtain
$$
 b+g = 6/5(m+w)\\
\iff \frac{b+g}{m+w} = \frac{6}{5}\\
\iff \frac{\frac{21}{10}g}{\frac{43}{20}m} = \frac{6}{5}\\
\iff g = \frac{43}{35} m
$$
Hence
$$
b = \frac{11}{10} g = \frac{11}{10}\frac{43}{35} m = \frac{473}{350} m\\
g = \frac{43}{35} m\\
w = \frac{23}{20} m\\
$$
So we get the smallest integer solution if we choose $m$ to be the lowest common denominator of $350$, $35$, and $20$, which is $700$.
