Cut the knot: learn to enjoy mathematics
A math books store at a unique math study site. Shopping at the store helps maintain the site. Thank you.
Learning Math Online
Sites for teachers
Sites for parents
Terms of use
Awards
Interactive Activities

CTK Exchange
CTK Wiki Math
CTK Insights - a blog
Math Help

III Millennium Olympiad

Games & Puzzles
What Is What
Arithmetic
Algebra
Geometry
Probability
Outline Mathematics
Make an Identity
Book Reviews
Stories for Young
Eye Opener
Analog Gadgets
Inventor's Paradox
Did you know?...
Proofs
Math as Language
Things Impossible
Visual Illusions
My Logo
Math Poll
Cut The Knot!
MSET99 Talk
Other Math sites
Front Page
Movie shortcuts
Personal info
Privacy Policy

Guest book
News sites

Recommend this site

Sites for parents

Education & Parenting

Manifesto  |  Bookstore  |  Contents  |  Amazon store  |  Term index  |  What changed?  |  Contact  |  Recommend
RSS Feed: Recent changes at CTK

A Typical Age Problem

Here is a word problem the likes of which make frequent appearances in textbooks and contests:


Jon's father is 3 times as old as Jon. Four years ago he was 4 times older. How old is Jon?


Solution

Copyright © 1996-2010 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution

The problem admits several solutions. None is better than the other. You should not worry about picking up the "right" one.

The question is about Jon's present age, so it is natural to consider it unknown and give it a name. The name can be anything, x for example, but to better remember the significance of the name, I'll use the letter J:


Let J be Jon's age.


The problem's data can be reformulated as follows:


Four years ago Jon was J - 4,Jon - 4,J - 4,J + 4
Jon's father is 3 times as old, meaning that he is 3J,3J,3 + J,J - 3,J/3
but 4 years ago he was 3J - 4.


The only piece of data that we have not used so far is the fact that four years ago Jon's dad was 4 times older,younger,nicer,taller,olderthan was Jon at the time. This leads to an equation:


4(J - 4) = 3J - 4.


This is the equation we now need to solve. The idea is to collect the like terms. But first use the distributive law to carry out multiplication on the left:


4J - 16 = 3J - 4,
4J - 3J = 16 - 4,
J = 12.


So Jon's age is 12. Just to check, his father is 12×3 = 36,12×4 = 48,12 + 4×3 = 24,12×3 = 36,12×(3 + 4) = 84. Four years ago they were 12 - 4 = 8 and 36 - 4 = 32. We see that indeed four years ago Jon's dad was 4 (= 32/8) times as old as Jon was at the time.


We managed to solve the problem introducing just one variable, J. Some will find that using two variables, one for Jon's age, the other for his dad's age, makes the solution more transparent.


So, let J be Jon's age and F the age of his father. What we know is this:


F = 3J,
F - 4 = 4(J - 4).


This is a straightforward translation of the problem into the language of algebra. What we got is a system of two linear equations with two unknowns. Again, there is no single "right" way to solve such a system. Here's one possible way.


We may plug the value of F from the first equation into the second:


3J - 4 = 4(J - 4),


which directly leads to the equation in one variable we have already handled. We may also subtract, say, the second equation from the first:


F - (F - 4) = 3J - 4(J - 4),


which simplifies to


4 = -J + 16, or
J = 16 - 4 = 12,


as before.

Copyright © 1996-2010 Alexander Bogomolny

35690934Page copy protected against web site content infringement by Copyscape

Search:
Keywords:

Google
Web CTK