A father decides to distribute $$65$$ candies in an inversely proportional way, i.e., to the hours that his three children, Patricia, Paula and Pablo, have taken to do their homework. If it has taken them $$2, 3$$ and $$4$$ hours, respectively: how many candies does each receive?
Development:
The first thing that it is necessary to see is that if the distribution is inversely proportional, the one that has taken less time will take more candies, and vice versa.
If we call $$x$$ the candies that Patricia gets, $$y$$ those that Paula deserves and $$z$$ those that Pablo will take, the relation of the distribution will be as follows:
$$\dfrac{x}{\frac{1}{2}}=\dfrac{y}{\frac{1}{3}}=\dfrac{z}{\frac{1}{4}}$$
Now we have to find a fraction comparable to the previous ones by the rule of the sum:
$$\dfrac{x}{\frac{1}{2}}=\dfrac{y}{\frac{1}{3}}=\dfrac{z}{\frac{1}{4}}=\dfrac{C}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}$$
If we operate the denominator with the sum of fractions we get:
$$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{6+4+3}{12}=\dfrac{13}{12}$$
With this fact and knowing that, as the statement specifies, the total amount of candies to be distributed $$C$$ is $$65$$:
$$\dfrac{x}{\frac{1}{2}}=\dfrac{y}{\frac{1}{3}}=\dfrac{z}{\frac{1}{4}}=\dfrac{65}{\frac{13}{12}}$$
So,
$$2x=3y=4z=\dfrac{12\cdot65}{13} \Rightarrow 2x=3y=4z=\dfrac{780}{13} \Rightarrow 2x=3y=4z=60$$
Therefore, Patricia will have:
$$2x=60 \Rightarrow x=\dfrac{60}{2}=30$$ candies
On the other hand, Paula will get:
$$3y=60 \Rightarrow y=\dfrac{60}{3}=20$$ candies
And Pablo gets:
$$4z=60 \Rightarrow z=\dfrac{60}{4}=15$$ candies
Solution:
$$30$$ candies to Patricia, $$20$$ to Paula and $$15$$ to Pablo.