Quote:
Originally Posted by 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.
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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!