Math Core

Most algebra word problems on MATHCOUNTS and the AMC 8 boil down to one or two linear equations. The algebra itself is easy. What wins points is setting the problem up quickly and cleanly, and knowing a few shortcuts that skip the algebra entirely.

From words to an equation

Work in three steps:

Name the unknown. Choose the variable that makes the equations simplest, not necessarily the thing the question asks for.
Translate each sentence into an equation.
Solve, then check the answer against the original words and make sure you answer the question that was asked.

Choose your variable wisely

For consecutive numbers, let the middle one be $n$ . Five consecutive integers are $n - 2, n - 1, n, n + 1, n + 2$ , and their sum is simply $5n$ . Consecutive even (or odd) numbers step by $2$ : $n - 2, n, n + 2$ .

The "assume all" shortcut

Problems with two kinds of things (chickens and cows, nickels and dimes, adult and child tickets) can be done with two equations. But there's a faster way: pretend everything is the cheaper kind, see how far off the total is, and divide by the difference each swap makes.

The "add everything" trick

For a system like $a + b = 13$ , $b + c = 17$ , $a + c = 20$ , don't solve one variable at a time. Add all three equations: $2(a + b + c) = 50$ , so $a + b + c = 25$ . Then each variable is the total minus one equation: $a = 25 - 17 = 8$ , $b = 25 - 20 = 5$ , $c = 25 - 13 = 12$ .

Ages

In age problems, everyone ages at the same rate. If someone is $x$ now, they were $x - 5$ five years ago and will be $x + 12$ in twelve years. The difference between two people's ages never changes.

Common mistake

Read the last line of the problem again before you answer. It's very common to solve for $x$ correctly and then report $x$ when the question asked for $x + 2$ , "the larger number," or "the mother's age."

Worked example: Consecutive integers

The sum of five consecutive integers is $215$ . What is the largest?

With middle number $n$ , the sum is $5n = 215$ , so $n = 43$ . The largest is $n + 2 = 45$ .

Worked example: Assume all chickens

A farm has chickens and cows: $30$ heads and $84$ legs. How many cows are there?

If all $30$ animals were chickens, there would be $60$ legs. There are $24$ extra legs, and each cow adds $2$ extra legs compared with a chicken. So there are $24 \div 2 = 12$ cows. Check: $12$ cows and $18$ chickens have $48 + 36 = 84$ legs.

Worked example: Ages

A mother is three times as old as her son. In $12$ years, she will be twice as old as he is. How old is the mother now?

Let the son be $s$ , so the mother is $3s$ . In $12$ years: $3s + 12 = 2(s + 12)$ , so $3s + 12 = 2s + 24$ and $s = 12$ . The mother is $36$ . Check: in $12$ years they'll be $48$ and $24$ .

Worked example: Two ways to share

If each student in a class gets $3$ pencils, there are $8$ left over. If each gets $4$ , the teacher is $5$ pencils short. How many pencils are there?

Let $n$ be the number of students. The pencil count is both $3n + 8$ and $4n - 5$ . Setting them equal gives $n = 13$ , so there are $3 \cdot 13 + 8 = 47$ pencils.

The picture shows why: the two lines $y = 3x + 8$ and $y = 4x - 5$ cross at the one class size where both descriptions agree.

Pencils needed for each class size: the lines meet at 13 students and 47 pencils.Open in grapher →

Tip

On multiple choice, you can often plug in the answer choices. Start with the middle choice ( $C$ ); if it's too big or too small, you know which way to go.