Expert: Sherry Wallin - 9/18/2009

Question
Leo is twice older than Martin and three times than Mike, if the sum of their ages is 33, what is Leo's age.

Answer
Hi Sly~
There are probably a gazillion ways to solve this problem but my first attempt doesn't even use simultaneous equations which means equations that occur at the same time. I can solve this problem with just one equation:

The most important part of this problem statement is the relationship between Leo and Martin and Leo and Mike and the fact that you are looking for Leo's age. Let x = Leo's age. Now if Leo is twice older than Martin this means that 2( ) = x. What is that something ()? Solve for it. So () is (1/2)x which is Martin's age. We will do the same thing with Mike's age. Leo is three times older than Mike so 3() = x or Mike is (1/3)x. Now sum them up: x + (1/2)x + (1/3)x = 33. Multiply all by 6 (because it will get rid of the fractions by multiplying by the GCD). 6x + 6(1/2)x + 6(1/3)x = 6(33)
which brings us to 6x + 3x + 2x = 198 or 11x = 198 which solves for x = 18, Leo's age.

Math Prof

You should always check that your answer makes sense in the problem, so what is Martin's age? (1/2)x = (1/2)(18) = 9 and Mike's age is (1/3)x = (1/3)(18) = 6. Is it true that 18 + 9 + 6 = 33????