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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 S D
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mcaval
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    Welcome to MSE. You'll get a lot more help if you show that you have made a real effort to solve the problem yourself, even if you haven't made much progress. What are your thoughts? What have you tried? How far could you get? Where are you stuck? This question will likely be closed if you don't add more context. Please respond by editing the question body. Clarifications don't belong in the comments. – saulspatz Jul 22 '21 at 16:50
  • @saulspatz Thank you. I've updated the body. – mcaval Jul 22 '21 at 16:53
  • I guess $x$ stands for the number of students the first day, but your equation doesn't make sense. Let $t$ be the number of ticket that one student can prepare in one hour. You should be able to write down two equations involving $x$ and $t$. – saulspatz Jul 22 '21 at 17:01
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    Unless I'm missing something, this is a school algebra working together problem, not a logic problem. Instead of "logic", you should be using the tag "algebra-precalculus". – Dave L. Renfro Jul 22 '21 at 17:02
  • Let T be the number of tickets 1 student can make in 1 hour. Use that to solve the problem. – DanielV Jul 22 '21 at 17:08
  • On the off-chance that it could help, I answered in great detail a "working together problem" that explains an approach you can use for most any such problem without the use of charts and pictorial devices (often given in textbooks) that are easily forgotten unless you practice them a lot. – Dave L. Renfro Jul 22 '21 at 17:33

3 Answers3

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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 $t$.

TonyK
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2

$\require {cancel}$

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.)

fleablood
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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.

JMP
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