Mathematics, three units · Grade 11

How do you turn a word problem into an equation without going wrong at the start?

Begin by defining the unknown and writing that definition down: x is Danny's age today. Then translate each sentence separately into an expression, join them into one equation, and only then solve. Most of the marks lost on a three-unit paper go missing not in the algebra but in this translation step, before a single equals sign is written.

Learning objectives

Define the unknown first

Before anything else, write one line: what x stands for. Not in your head — on the page. An examiner reading a solution without it cannot tell what you calculated, and three lines later neither can you.

The choice itself matters. In an age problem it is easier to let the unknown be the younger of the two, because then the other age is written as an addition rather than a subtraction, and there are fewer signs to get wrong.

If the problem has two unknowns, check whether one of the sentences lets you write one in terms of the other. Usually it does, and then you are left with a single equation instead of a system.

Translate one sentence at a time

Each sentence in the question is one expression. Do not try to write the whole problem in a single line — translate sentence by sentence, then join them.

A few translations recur. Older by 5 years means plus 5. Three times as many means times 3. In four years' time means plus 4 to every age, not just to one. Greater by 20 percent means times 1.2.

The word that creates the equals sign is usually the last one: altogether, in total, are equal. Up to that point you are building expressions; from it onwards you have an equation.

Check, and sometimes reject

After solving, substitute back into the word problem, not into the equation you wrote. If you mistranslated, checking against your own equation will simply confirm the mistake.

A quadratic gives two solutions and often only one is possible. A negative age, a non-whole number of pupils, a negative speed — each is rejected, in writing.

That writing is worth marks. A solution that says "the second value is rejected because an age cannot be negative" earns the point; one that silently ignores it may not.

Worked examples

  1. Danny is 6 years older than Yossi. In 4 years the sum of their ages will be 40. How old is Yossi now?

    1. Let x be Yossi's age today, so Danny is x plus 6
    2. In 4 years: Yossi is x plus 4, Danny is x plus 10
    3. Sum: x plus 4 plus x plus 10 equals 40
    4. 2x plus 14 equals 40, so 2x equals 26

    Answer: Yossi is 13 today and Danny is 19

  2. A price rose by 20 percent, then fell by 25 percent, and reached 90 shekels. What was the original price?

    1. Let x be the original price
    2. After the rise: x times 1.2
    3. After the fall: x times 1.2 times 0.75
    4. This gives 0.9x equals 90

    Answer: The original price was 100 shekels

  3. A rectangle has area 60 and its length exceeds its width by 7. Find the width.

    1. Let x be the width, so the length is x plus 7
    2. Equation: x times (x plus 7) equals 60
    3. Expand: x squared plus 7x minus 60 equals zero
    4. The solutions are 5 and negative 12

    Answer: The width is 5. Negative 12 is rejected because a width cannot be negative

Common mistakes

Writing an equation without defining the unknown
Without the definition there is no way to tell whether x is the age now or the age in four years, so the final answer addresses a different question. That line costs ten seconds and saves the whole solution.
Adding the years to only one of the ages
In four years everybody gets older. Adding to one side only is the commonest error in age problems, and it produces an answer that looks perfectly reasonable.
Presenting both roots of a quadratic without rejecting one
The question asks about a physical quantity — an age, a length, a count. A negative or fractional value is impossible, and a written, reasoned rejection is part of the answer rather than a decoration.

What to remember

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