How to find the number of pills from an assorted group taken at the end of a treatment?

Question

The problem is as follows:

Mike takes exactly each $6$ hours the following pills: $2$ of type $A$, $3$ of type $B$ and $4$ of type $C$. He began taking his treatment taking the three kinds of pills and finished when the sum of the amount from the pills of type $B$ and type $C$ was $42$. Find how many pills of type $A$, Mike took in total.

The alternatives given are as follows:

$\begin{array}{ll} 1.&14\\ 2.&18\\ 3.&12\\ 4.&16\\ 5.&10\\ \end{array}$

In this problem I'm totally lost as, because I'm confused exactly how should I account for the number of pills from type $A$. How can this be accounted?.

Can this be solved in an equation?. I'm confused at how to use the information of the time given in this problem, so an answer that would help me is one which may explain this part.

Answer 1

The answer is pretty much already given in the comments.

Each time Mike takes $3$ B and $4$ C pills. Lets say that he has take the $42$ B and C pills after n times. This means that $3n+4n=42$. Follows that he needs $n=6$ times to take the $42$ B and C pills. So he would have taken $6\times 2$ A pills.

Answer 2

Then in 36 hours $6$ times $6$ then is $42$ pills right?. But that's where I'm still stuck. ?

As you said: "I'm assuming that $3+4=7$ so in $6$ hours he has taken those." That´s right. $7$ pills ($B,C$) in $6$ hours. Now multiplying both numbers by $6$ gives $42$ pills in $36$ hours. Or you can look at these relations:

$$7 \textrm{ pills ($B,C$)} \ \hat = \ 6 \textrm{ hours} \\ 42 \textrm{ pills ($B,C$)} \ \hat = \ x \textrm{ hours}$$

Applying rule of three gives $x=6\cdot \frac{42}{7}=6\cdot 6=36 \ \color{grey}{ \textrm{hours}}$

And in $36$ hours you can take $36 \ \color{black}{ \textrm{hours}}\cdot \frac{2 \ \color{black}{ \textrm{pills A}}}{6 \ \color{black}{ \textrm{hours}}}=12 \ \color{black}{ \textrm{pills A}}$