Some chocolates were distributed equally among a few children and three chocolates were left. Had there been four times the number of children, then ten chocolates would have been left. Find the number of children.
Question
Some chocolates were distributed equally among a few children and three chocolates were left. Had there been four times the number of children, then ten chocolates would have been left. Find the number of children.
Solution
Let's denote the number of children as 'c' and the total number of chocolates as 'ch'.
From the problem, we have two equations:
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ch = c*x + 3, where x is the number of chocolates each child originally received.
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ch = 4c*y + 10, where y is the number of chocolates each child would receive if there were four times as many children.
Since the total number of chocolates doesn't change, we can set the two equations equal to each other:
cx + 3 = 4cy + 10
We know that if there were four times as many children, each child would receive a quarter of the chocolates they originally did. So, y = x/4.
Substitute y into the equation:
cx + 3 = 4c(x/4) + 10
Simplify the equation:
cx + 3 = cx + 10
Subtract c*x from both sides:
3 = 10
This is a contradiction, which means there must be a mistake in the problem. The number of chocolates left over should be the same in both scenarios if the total number of chocolates and the distribution are consistent. In this case, the problem states that there are 3 chocolates left in the first scenario and 10 chocolates left in the second scenario, which is not possible. Therefore, the problem as stated does not have a solution.
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