How to Solve Word Problems
It seems that one of the biggest challenges many people have
in math, is how to solve basic word problems. After a recent
newsletter I was reminded of this when a reader emailed me about
a riddle in that issue, which was in the form of a word problem.
He had used the "guess and test" method to arrive at
the solution, but was wondering how to solve problems like this
one that were more difficult.
With that in mind, here's the riddle/word problem, followed
by a refresher lesson on algebra for those of us that have been
out of school for a couple decades.
An Example of a Word Problem
A pound of onions plus a pound of carrots costs $2.70 total.
A pound of carrots plus a pound of potatoes costs $2.30 total.
A pound of onions plus a pound of potatoes costs $2.60.
What is the price per pound for each vegetable?
To solve word problems, start by replacing the statements
with symbols, to create a simple equation. In this case, we can
replace the vegetables with the first letter of each, with each
letter representing a pound of that vegetable. Now the riddle
looks something like this:
o + c = $2.70
c + p = $2.30
o + p = $2.60
o = ?
c = ?
p = ?
Now, one method (there are others) to solve this, is to just
combine the first two statements and make an equation with the
last statement on the other side:
o + c + c + p = o + p + $2.40
We arrived here because we know that $2.70 plus $2.30 (o +
c = $2.70 and c + p = $2.30) equals $5.00. To make the right
side of the equation equal then, we added $2.40 to the cost of
a pound of onions plus a pound of potatoes, which we know costs
$2.60 ($2.60 + $2.40 = $5.00). If you want to break that down
further: $2.60 + x = $5.00, so we subtract $2.60 from each side
and we get x = $2.40.
If you recall your algebra, you know you can also remove an
"o" from each side and a "p" from each side,
and both sides will remain equal, leaving us with:
c + c = $2.40
or
2 x c = $2.40
To find the value of c now, just divide each side by 2, and
you get:
c = $1.20
Now you can easily solve for the other costs:
o + $1.20 = $2.70
Subtract $1.20 from each side and you get :
o = $1.50
And
$1.20 + p = $2.30
Subtract $1.20 from each side and you get :
p = $1.10
So:
Onions cost $1.50 per pound.
Carrots cost $1.20 per pound.
Potatoes cost $1.10 per pound.
This may not be exactly how an algebra teacher would do it,
but then I'm not an algebra teacher. This method works for me,
but in any case, it is more important to understand the principles
involved than remember the exact forms and steps of the algorithms.
To solve word problems then, replace the words with an equation
and work out a solution for each unknown. Then test your solution.
Does $1.50 (onions) plus $1.20 (carrots) equal $2.70? If so,
we have a correct solution.
When I was in school, I learned how to solve word problems
more easily than I learned other math, because word problems
at least seemed related to real life. More abstract math, while
used in life by engineers, astronomers and others, was never
presented in such understandable contexts, and so was more difficult
for me. But then, I usually had my own obscure and difficult
to explain algorithms for such problems.
In fact, like the reader who emailed me, I would have probably
used the "guess and test" method to solve the problem
above. Part of using our minds most efficiently is knowing intuitively
when to use one method or another. Intuition comes into play
because it isn't always clear which way will be quicker, especially
with word problems.
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