You have 12 coins, one of which is fake. The fake coin is indistinguishable from the rest except that it is heavier. Can you det
ermine which is the fake coin heavier using a balance scale and only 3 weighings?
2 answers:
1) Divide 12 coins into 6 6
The heavier side contain the fake coin
2)Divide the 6 coins left by 3,you will get 2 coins per groups
Just put TWO groups of coins on the balance and keep the third group aside
**If two sides of the balance are balance,it means that the third group contain fake coin**
**If one side is heavier,that 2 coins contain fake coin**
3)Directly put the chosen 2 coins on each side and you can find out which is the fake one!
The heavier side contain the fake coin
2)Divide the 6 coins left by 3,you will get 2 coins per groups
Just put TWO groups of coins on the balance and keep the third group aside
**If two sides of the balance are balance,it means that the third group contain fake coin**
**If one side is heavier,that 2 coins contain fake coin**
3)Directly put the chosen 2 coins on each side and you can find out which is the fake one!
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I would start by setting up a chart like I did below.
Label one column age now and the other age in 5 years.
Since we don't know the son's age we use x.
We do know that the man's age is 4 times the son's age.
So the man's age will be 4x.
In the age in 5 year column, we add 5 to their current ages.
Now set up our equation.
Since it says "in five years" we use information in second column.
In 5 years time, he, "4x + 5", will be, equals,
3 times as old as his son, "3(x + 5)".
So we have 4x + 5 = 3(x + 5).
Solving from here, we find that x = 10.
So the son is 10 and the man is 4 times his age or 40.
