A high school logical problem I'd like to know how to solve this problem.
"To raise money for prom, students hosted a raffle. A group of students spent 6 hours preparing n raffle tickets. The following day, with 2 more students helping the group,
they prepared the same amount of tickets as in the first day in only 4 hours. How long would it take 1 student to prepare n tickets?"
Options are $12$, $24$, $16$ or $20$.
I can only get as far as to think this:

Where x is the number of students in the group:
x = 6 hours
x + 2 = 4 hours

I can't think of a way to determine the right answer without knowing the amount of students in the group.
Thanks in advance.
 A: Let $t$ be the time it takes for one student to prepare $n$ tickets; and let $k$ be the number of students on the first day. Then you have

*

*$\dfrac{t}{k}=6;$

*$\dfrac{t}{k+2}=4.$
So $t=6k=4(k+2)$. From this, it is easy to derive $k$, and thence $t$.
A: $\require {cancel}$
Let $s$ be the number of students.
You have $s\text{ students}\times 6 \text{ hours} \to n\text{ tickets}$
So we need some conversion factor from $\text{student-hour} \to \text{tickets}$.
That is if we let $r\frac {\text{tickets}}{\text{student-hour}}$ be the amount of tickets that one student can make in one hour then
$s\text{\students}\times 6\text{ hours}\times r\frac {\text{tickets}}{\text{student-hour}}=$
$s\cancel{\text{\students}}\times 6\cancel{\text{ hours}}\times r\frac {\text{tickets}}{\cancel{\text{student}}\cancel{\text{hour}}}=$
$6sr\text{ tickets} = n\text{ tickets}$ and therefore the numeric valuse of $6sr = n$.
We maybe don't care what $r$ actually but we can assume it is positive and not zero (otherwise the students would never be able to do anything).
We also have
$(s + 2)\text{ students} \times 4 \text{ hours}\times r\frac {\text{tickets}}{\text{student-hour}}= n\text { tickets}$.
So We have $6sr \text{ tickets} = 4(s+2)r\text{ tickets}$ and from that we have the numeric values $6sr = 4(s+2)r$.  And we can solve for $s$ by dividing both sides by $r$ and getting $6s = 4(s+2)$
( so ..... $s = ... \text{what?}...$)
Now we need to figure out how long it takes one student to make $n$ tickes.  Call that time $t\text{ hours}$ and we have
$1 \text{ student}\times t\text{ hours} \times r\frac {\text{tickets}}{\text{student-hour}}=$
$1\cancel{\text{student}}\times t\cancel{\text{ hours}}\times r\frac {\text{tickets}}{\cancel{\text{student}}\cancel{\text{hour}}}=$
$tr\text{ tickets} = n\text{ tickets}$.
But remember $n \text{ tickets} = 6sr\text{ tickets}$.
So $tr \text{ tickets} = 6sr\text{ tickets}$ and dividing both sides by $r$ we get
$t = 6s$ and ... we know what $s$ is.
(... We can also figure out what $r$ is if we want.)
A: If $k$ students take $6$ hours, then $1$ student takes $6k$ hours.
If $k+2$ students take $4$ hours, then $1$ student takes $4(k+2)$ hours.
These must be the same, so $k=4$ and $1$ student takes $24$ hours.
