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# Funny problem to practice solving systems of equations.

Many students I worked with tend to think that many notions they learn in mathematics are just that, notions and have little or no utility in real life situations.

So, I like to give them problems, where they need first to derive their own equation or system of equations and then solve it.

Here is a funny problem that never fails to surprise students: at first they feel lost for it sounds like a pseudo tongue-twister, but then they see the elegance of the solution emerge from within what sounds a hopeless situation.

Problem: Mark is twice Bob's age and, when Bob will be Mark's age, together they will be 60 years old. What are Mark and Bob's ages right now?

Solution: The "difficult" part is to write this problem in the language of a system of equations, actually a system of 2 equations.

Let x be Mark's age and y Bob's age, respectively. Then we know that x=2y.

For Bob to be Mark's age, we will need x-y years. At that point Bob will be x-year old and Mark will be x+(x-y) = 2x-y years old. Thus, from the second part of the statement of the problem we must have that 2x-y =+ x = 3x-y =60.

Thus we have the following simple system of equations:

x=2y
3x-y =60

If we replace x in the second equation with 2y we get 6y-y=5y = 60 => y =12. Hence x=2y=2*12=24.

Hence we have shown that Bob's age is 12 and Mark's age is 24.

\$50p/h

Maurizio T.