Math Tip of the Week – Word Problems

I always loved foreign language in school, and like to think of math like just one more language, when it comes to (the dreaded!) word problems.   So in the Algebra Word Problems Section, I put together a Math-to-English Translation Table, which covers some of these conversions:



is, yields, will be


what number, how much (look at question)

“\(x\)”    (or any variable)

in addition to, added to, increased by


sum of \(x\) and \(y\)


difference of \(x\) and \(y\)


product of \(x\) and \(y\)

\(x\times y\) 

quotient of \(x\) and \(y\)

 \(\displaystyle x\div y\,\,\,\,\,\text{or }\,\,\,\frac{x}{y}\)

opposite of \(x\)


ratio of \(x\) to \(y\)

\(\displaystyle x\div y\,\,\,\,\,\text{or}\,\,\,\,\frac{x}{y}\) 

a number \(n\) less 3


a number \(n\) less than 3


a number \(n\) reduced by 3




\(p\) percent

\(\displaystyle \frac{p}{{100}}\), or move decimal left 2 places

half, twice

 \(\displaystyle \frac{n}{2},\,\,\,2n\)

consecutive numbers

Let \(n=\) first number, \(n+1=\) second number, \(n+2=\) third number…

odd/even consecutive numbers

Let \(n=\) first number, \(n+2=\) second number, \(n+4=\) third number… (Note: Even if you are looking for odd consecutive numbers, use \(n, n+2, n+4, …\)).

average of  \(x,y\) and \(z\)  (and so on)

\(\displaystyle \frac{{x+y+z+…}}{{\text{(how many}\,\,\text{numbers}\,\,\text{on}\,\,\text{top)}}}\) 

\(x\) per \(y\), \(x\) to \(y\), \(x\) over \(y\), \(x\) part of \(y\)

\(x\div y\)   or   \(\displaystyle \frac{x}{y}\)

Example: number of girls to total people can be represented by \(\displaystyle \frac{{\text{girls}}}{{\text{total}}}\).

\(x\) per \(y\), as in \(x\) “for every” \(y\)

Multiplication, or \(x\times y\). 

Example: if you drive 50 miles per hour, how many miles will you drive in 5 hours:  250 miles. 

\(y\) increased by \(x\%\)

 \(\displaystyle y+\left( {y\times \frac{x}{{100}}} \right)\)

\(y\) decreased by \(x\%\)

 \(\displaystyle y-\left( {y\times \frac{x}{{100}}} \right)\)

\(y\) is at least (or no less than) \(x\)

 \(y\ge x\)

\(y\) is at most (or no more than) \(x\)

 \(y\le x\)

\(y\) is between \(x\) and \(z\)

\(x\le y\le z\) (inclusive)    \(x<y<z\)  (exclusive)



An Example: “Find the Numbers” Word Problem (only using 1 variable):

The sum of two numbers is 18.  Twice the smaller number decreased by 3 equals the larger number.  What are the two numbers?


Ok, so we always have to define a variable, and we can look at what they are asking.  The problem is asking for both the numbers, so we can make “n” the smaller number, and “18 – n” the larger.


Do you see why we did this?  The way I figured this out is to pretend the smaller is 10.  (This isn’t necessarily the answer to the problem!)  But I knew the sum of the two numbers had to be 18, so do you see how you’d take 10 and subtract it from 18 to get the other number?   See how much easier it is to think of real numbers, instead of variables when you’re coming up with the expressions?

We don’t need to worry about “n” being the smaller number (instead of “18 – n“); the problem will just work out this way!

So let’s translate the English into math:



The sum of two numbers is 18. Twice the smaller number decreased by 3 equals the larger number. What are the two numbers?

\(\begin{array}{l}2n-3\,\,\,=\,\,18-n\\\underline{{+n\,\,\,\,\,\,\,\,\,\,\,=\,\,\,\,\,\,\,\,+n}}\\3n\,-3\,\,=\,\,\,18\\\underline{{\,\,\,\,\,\,\,\,\,+3\,\,=\,+3}}\\\,\,3n\,\,\,\,\,\,\,\,\,\,=\,\,\,21\\\,\frac{{3n}}{3}\,\,\,\,\,\,\,\,\,\,\,=\,\,\frac{{21}}{3}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n=7\,\,\,\,\,\,\text{(smaller number)}\\\,\,\,18-7=11\,\,\,\,\text{(larger number)}\end{array}\)

Let \(n=\) the smaller number, and \(18-n=\) the larger number.


The translation is pretty straight forward; note that we didn’t have distribute the 2 since the problem only calls for twice the smaller number, and then we subtract 3.


Let’s check our work:


The sum of 7 and 11 is 18.   √

Twice the smaller (\(2\times 7\)) decreased by 3 would be \(14-3=11\).   √

 See – not so bad, right? Happy Mathing 🙂 Lisa

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