SOLUTION: All coffee mugs in an office were placed into 4 boxes. The first box when divided by 5, the second box when divided by 4, the third box when divided by 3 and the fourth box when di
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-> SOLUTION: All coffee mugs in an office were placed into 4 boxes. The first box when divided by 5, the second box when divided by 4, the third box when divided by 3 and the fourth box when di
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Question 1039131: All coffee mugs in an office were placed into 4 boxes. The first box when divided by 5, the second box when divided by 4, the third box when divided by 3 and the fourth box when divided by 6 resulted in the same whole number. What is the least of number of mugs that could have been in the office? Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! let that same number be equal to x.
you get:
the first box contains 5x.
the second box contains 4x.
the third box contains 3x.
the fourth box contains 6x.
the total number of mugs in the 4 boxes is equal to 18x.
x can't be less than 1, so the smallest number of coffee mugs in the office has to be 18.
5 in the first box plus 4 in the second box plus 3 in the third box plus 6 in the fourth box = 18 total.
