How to solve number word problems

What are number word problems in Algebra?

The purpose of number word problems is to give you practice in translating back and forth from words to numbers and vice versa.

For some people these are fun number games, but they do appear on tests and in math classes from time to time, so it’s good to be comfortable with them.

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In a number word problem you’re given information about a pair or group of numbers and you usually need to translate the information into equations to solve for the numbers.

Some helpful vocabulary:

Consecutive numbers are numbers that are in order, like $4$ and $5$.

Consecutive even numbers then would be even numbers that are in order, like $4$ and $6$.

Consecutive odd numbers are odd numbers that are in order, like $5$ and $7$.

And of course “sum” means add, “difference” means subtract, and “product” means multiply. Remember that place value can also be used in these problems.

How to solve number word problems

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Examples of number word problems

Example

The sum of two consecutive integers is $25$. Find the numbers.

“Consecutive” means numbers are in order.

Let $X$ be the first number. Then the next integer would be $X+1$.

Let’s write the sum.

$X+(X+1)=25$

Now let’s solve for $X$.

$X+X+1=25$

$2X+1=25$

$2X+1-1=25-1$

$2X=24$

$\frac{2X}{2}=\frac{24}{2}$

$X=12$

The first integer is $12$, the next one is $12+1=13$. Therefore, the integers are $12$ and $13$. We can double check the answer and see that $12+13=25$.

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Let’s look at another style of problem.

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In a number word problem you’re given information about a pair or group of numbers and you usually need to translate the information into equations to solve for the numbers.

Example

If you add the digits of a certain two digit number, the sum is $17$. Reversing the two digits gives a number $9$ smaller than the original number. What is the original number?

Let $T$ be the tens digit and $U$ be the units digit in the original number. Then the sum of the digits is $T+U=17$.

The value of the original number is

$10T+U$

Reversing the digits gives us a number whose value is

$10U+T$

The second number is $9$ smaller than the original number so we can write

$\text{Original number}-9=\text{Second number}$

That is

$10T+U-9=10U+T$

$10T-T+U-10U-9=0$

$9T-9U-9=0$

Dividing through by $9$ gives

$T-U-1=0$

$T-U=1$

Add the sum of the digits equation, $T+U=17$, to this one.

$T-U+(T+U)=1+(17)$

$T+T-U+U=1+17$

$2T=18$

$T=9$

Plug $T=9$ into $T+U=17$ to find $U$.

$9+U=17$

$U=8$

The original number is $98$.

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