The Batlord

I can count to 20 just as quick as I can untie my shoes.

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Eleanor Rigby 14

Thanks for the link!!
Well, the problem is the following one:
A restaurant offers at each dinner three desserts and twice the number of first courses than seconds. Each dinner consists of a first course, a second course and a dessert. What is the least number of second courses that you have to offer so that a customer can have different dinners during the 365 days of a year?
Sorry if it's not very understandable.

Marie Monday

Just create a math thread, I'd love to help!

Edit: nvm just noticed you posted it here, that's fine too

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Marie Monday

OK so here's the deal. There is one independent quantity you don't know (number of second courses) and a second unknown quantity (first courses) but this one depends on the second courses. So basically you have only one unknown quantity, because if you know the number of second courses you know the number of firsts too. So you need to solve for the number of second courses, we'll call it x. Then we have:
Number of first courses: 2*x
Number of second courses: x
Number of desserts: 3
Now by the laws of statistics, if you want the total number of combinations (= number of different meals) in this case you multiply these:
2x*x*3 = 6x^2
This has to be equal to 365 if you want a different meal every day, so you solve 6x^2 = 365 for x.
Hope that helps, let me know if something about it isn't clear yet!

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Eleanor Rigby 14

Lol, yes, thank you very much!! It was easier than I thought. The correct answer is 8, right?

Marie Monday

yes, that's correct!

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Eleanor Rigby 14

Sorry, but I have another problem

Juan says to Luisa: If you give me 3 coins I will have "n" times as many as you'll have. Luisa responds:
True, but if you give me "n" coins, then I will have three times what you have left. For what values of "n" are these statements true?
I've done the following:
x+3=n (y-3)
3(x-n)=y+n
But it doesn't lead me anywhere.

Trollheart

You can afford shoes??

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Trollheart: Signature-free since April 2018

Marie Monday

The trick is to realise that you again have only one unknown number. Just start by writing out the story in mathematics. The initial situation is that Juan has 3 coins, and it's n times as much as Luisa, so:
Juan: 3
Luisa: 3/n
Then Juan gives n coins to Luisa:
Juan: 3 - n
Luisa: 3/n + n
But now we also know Luisa has 3 times as many coins as Juan:
Luisa: 3*(3-n)
We have two different expressions for the same amount (the amount of coins Luisa has) so we can say that they're equal:
3/n + n = 3*(3-n)
and then this equation can be solved.


edit: wait, I might actually have misunderstood the question. Does Juan start with 3 coins, or with x + 3?

Ok so if I understand the question, the solution is:
Juan: x + 3
Luisa: (x + 3)/n
But also:
Juan: x - n
Luisa: (x - n)*3
Now these are two equations with 2 unknowns, which you can solve. The equation for Juan is
x + 3 = x - n
this means n = -3.
Then we substitute n in the equation for Luisa:
(x + 3)/n = (x - n)*3
(x + 3)/-3 = (x + 3)*3
and you can solve for x

edit 2: lol I actually made a typo in the first solution too (I corrected it), but knowing how math problems usually are the second case is probably what is meant

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Marie Monday

ok scrap that last derivation, I was too hasty and I overlooked something. I'll do it properly here, wait a sec:

update: it's not solvable the way I understand the question. Maybe something is getting lost in translation. What is the original Spanish text?

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