Re: Worded Problem: At a doctor's convention, each doctor ma
minestrone said:
I'm stumped on this problem and have no clue how to solve it.
At a doctor's convention, each doctor made a full cup of tea with milk for himself or herself. After having finished his cup, Dr. Clive worked out that he had drank one fifth of the total amount of tea and one seventh of the total amount of milk used by all the doctors at the convention. Assuming that they all had cups of the same capacity, how many doctors were at the convention? Find all possible answers.
I fail to see a way of deriving a soution with the given information. I suspect you left our one important piece of information, the capacity of the cups. I assumed the most frequent capacity of cups of 8 ounces and proceeded as follows.
1--Let T = the number of ounces of tea in Dr. Clives cup and M = the numder of ounces of milk in the cup.
2--Then, T/5 += M/7 = 8 or 5M + 7T = 280
3--Dividing through by 5 yields M + T + 2T/5 = 56.
4--2T/5 must be an integer making T = 5k/2
5--Substituting back into (2) yields M = 56 - 7k/2.
6--From ascending values of k
...k....1....2....3....4....5....6....7....8....9....10....11....12....13....14....15....16
...T....-....5.....-...10...-....15...-...20....-....25.....-.....30.....-......35.....-.....40
...M...-...49....-....42..-.....35..-....28....-....21.....-.....14.....-.......7.....-......0
7--We also know that 4T/5 + 6M/7 = 8(n - 1), n being the number of doctors in attendance, or (30M + 28T)/280 = (n - 1).
6--Substituting the valid solutions of M and T into (7) we get the number of doctors, D, as
...k....1....2....3....4....5....6....7....8....9....10....11....12....13....14....15....16
...T....-....5.....-...10...-....15...-...20....-....25.....-.....30.....-......35.....-.....40
...M...-...49....-....42..-.....35..-....28....-....21.....-.....14.....-.......7.....-......0
...D.....6.75........6.5......6.25.......6.........5.75.........5.5...........5.25..........5
It would appear that there is only one valid value of D, the number of doctors being 6. Of course there could have been any multiple of 6 also.
If my assumption is incorrect, please let me know. If you ever get a valid solution with only the information you originally gave, please let me know too.
Thanks for a great problem.
