Word problems

A word problem never hands you a rational number ready-made — it hides one inside a sentence about a falling thermometer, a shrinking tank of water, or cloth being cut and shared. The skill this topic builds is translation: turning "rose," "fell," "shared equally" or "of what remained" into a signed fraction and the correct operation, then computing with the addition, subtraction, multiplication and division rules for rational numbers you already know. This sits inside the Rational Numbers chapter of ICSE Class 8 Mathematics. Every example below stays within that syllabus: no new number system, just careful reading, correct signs, and clean fraction arithmetic. By the end you should be able to look at a multi-step story problem, break it into labelled rational-number pieces, and combine them in the right order without losing a sign or a denominator along the way.

The core insight

Solving a rational-number word problem means labelling each quantity as a signed fraction, choosing addition, subtraction, multiplication or division based on the situation's signal words, and computing step by step in lowest terms.

The core idea

Every rational-number word problem follows the same three-part translation, no matter how it is dressed up in a story.

1. Identify the quantities and their signs. Gains, rises, and additions are positive; losses, falls, debts, and removals are negative. A fraction like 34\dfrac{3}{4} or a mixed number like 2122\dfrac{1}{2} describes a part of a whole, and it must be converted to an improper fraction before any arithmetic begins.

2. Choose the operation from the signal words. Addition or subtraction combines quantities or finds a change; multiplication handles "of" a quantity; division handles "shared equally," "each," or "per."

3. Compute using rational-number rules, always finding a common denominator for addition and subtraction, and simplifying the final answer. Two situations recur often enough to deserve their own template. For a running total built from a starting value and a sequence of signed changes:

final amount=starting amount+(sum of all signed changes)\text{final amount} = \text{starting amount} + (\text{sum of all signed changes})

where: the starting amount and each change are rational numbers with the correct sign attached from the wording ("rose" is positive, "fell" is negative), and the final amount is found by adding them all with a common denominator. For "sharing equally" problems, the operation is division: share of each=total quantitynumber of people\text{share of each} = \dfrac{\text{total quantity}}{\text{number of people}}. For "part of a part" problems — a fraction applied to what is left after an earlier fraction was already used — the operation is multiplication of the two fractions. A good habit that prevents the most common mistake — losing track of which sign belongs to which quantity in a long problem — is to write a short label for each rational number as you pull it out of the sentence, for example "start =−52= -\dfrac{5}{2}," "rise =+234= +\dfrac{23}{4}," before combining anything. Once every quantity in the story has a label and a sign, the rest of the problem is ordinary rational-number arithmetic, and there is nothing left to interpret. It helps to recognise four common families by their signal words:

  • Change/level problems (temperature, height above or below sea level, bank balance): "rose," "fell," "gained," "owes," "above," "below" fix the sign of each step.
  • Sharing/rate problems (dividing a quantity among people): "equally," "each," "per," "shared among" signal division.
  • Part-of-a-part problems (a fraction of what remains): "of," applied a second time after an earlier fraction has already reduced the quantity, signals multiplication.
  • Combination problems: many ICSE questions chain two or three of these in one scenario — work through it one sentence at a time, updating a running total after each step.

These four families are not rigid boxes; a single question often opens with a change/level sentence, moves through a part-of-a-part step, and finishes by sharing the result equally. Recognising which family a given sentence belongs to is what tells you which operation to reach for next, and that recognition comes from the signal words far more than from the numbers themselves — the same skill of reading for operation, not just for value, carries forward unchanged into algebraic word problems in later classes.

Figure
A flow diagram showing a word problem sentence being translated into a labelled signed fraction then combined by an operation into a final answer
Translating a sentence into a rational-number calculation "fell by 6 1/4 degrees" signal word: fell sentence in the story label = -25/4 signed rational number add to running total common denominator, then simplify the result Every word problem is this same pipeline: sentence, signed fraction, correct operation.
Derivation AThe running-total template for change problems

The goal is to show, with the temperature scenario as the running example, exactly how "starting amount plus signed changes" produces a correct final answer even when several changes of different signs and different denominators are chained together — this is the template a student should default to whenever a word problem describes a level that rises and falls in stages.

Step 1. Convert every quantity mentioned to an improper fraction and attach the correct sign from the wording. A starting temperature of −212 ∘C-2\dfrac{1}{2}\,^\circ\text{C} becomes −52-\dfrac{5}{2}; a rise of 534 ∘C5\dfrac{3}{4}\,^\circ\text{C} becomes +234+\dfrac{23}{4}; a fall of 614 ∘C6\dfrac{1}{4}\,^\circ\text{C} becomes −254-\dfrac{25}{4}. Writing a short label for each ("start," "rise," "fall") before combining anything is what prevents a sign from getting lost in a long problem.

Step 2. Add the changes to the starting amount in the order they occur, converting to a common denominator at each stage rather than trying to combine everything at once. Starting amount plus the first change: −52+234=−104+234=134-\dfrac{5}{2}+\dfrac{23}{4} = -\dfrac{10}{4}+\dfrac{23}{4} = \dfrac{13}{4}.

Step 3. Apply the next signed change to the new running total, again over a common denominator: 134−254=−124\dfrac{13}{4}-\dfrac{25}{4} = -\dfrac{12}{4}.

Step 4. Simplify the final fraction to lowest terms and attach the units and sign the question expects: −124=−3-\dfrac{12}{4} = -3, so the running total lands at −3 ∘C-3\,^\circ\text{C}. Because each step only ever combined two rational numbers with a common denominator, the method scales to any number of changes without new rules — it is the same addition performed repeatedly.

Worked example

A tank is 56\dfrac{5}{6} full of water. 25\dfrac{2}{5} of this water is used for washing, and the remaining water is shared equally among 3 buckets. What fraction of the (full) tank's capacity does each bucket receive?

Step 1. "25\dfrac{2}{5} of 56\dfrac{5}{6}" is a part-of-a-part statement, so multiply: 25×56=1030=13\dfrac{2}{5}\times\dfrac{5}{6} = \dfrac{10}{30} = \dfrac{1}{3}. So 13\dfrac{1}{3} of the tank's capacity is used for washing.

Step 2. Find the water remaining by subtracting from the full amount: 56−13=56−26=36=12\dfrac{5}{6}-\dfrac{1}{3} = \dfrac{5}{6}-\dfrac{2}{6} = \dfrac{3}{6} = \dfrac{1}{2}. Half the tank's capacity remains.

Step 3. "Shared equally among 3 buckets" is a rate/sharing statement, so divide: 12÷3=12×13=16\dfrac{1}{2}\div 3 = \dfrac{1}{2}\times\dfrac{1}{3} = \dfrac{1}{6}.

Each bucket receives 16\dfrac{1}{6} of the tank's full capacity.

Competitive stretch

If instead the remaining water were shared among 3 buckets in the ratio 1:2:31:2:3 rather than equally, the smallest share would be 12×16=112\dfrac{1}{2}\times\dfrac{1}{6} = \dfrac{1}{12} of the tank, since the parts sum to 66 shares and the first bucket gets 11 of those 66 shares of the remaining 12\dfrac{1}{2}.

Common traps

  • Missing or misplacing a negative sign for "loss," "drop," "below," "owes," or "removed" — these must be written as negative rational numbers before any calculation.
  • Confusing "of" (multiplication) with "more than" or "added to" (addition) — "14\dfrac{1}{4} of a quantity" is very different from "14\dfrac{1}{4} more than a quantity."
  • Forgetting to convert mixed numbers to improper fractions before adding, subtracting, multiplying, or dividing them.
  • Leaving the final answer as an unsimplified or improper fraction when the question calls for a mixed number, or the reverse — always reread what form is expected.

Ask the tutor: Stuck on turning a tricky word problem into the right expression, or want a fresh multi-step practice problem worked through in full?

How do I know whether a sentence needs addition or multiplication when both use the word of or more?

Look for whether the second quantity is a fraction being applied to the first (multiplication, signalled by "of") or a separate amount being combined with the first (addition or subtraction, signalled by "more than," "added to," "gained," or "lost"). "14\dfrac{1}{4} of 2020" scales 2020 down to a quarter of itself, giving 55; "14\dfrac{1}{4} more than 2020" adds a quarter of 2020 on top, giving 2525. Reading the sentence slowly and asking "is this fraction acting on the quantity, or being added to it" almost always settles the question.

Do I always need a common denominator, even for multiplication and division of rational numbers?

No — a common denominator is only needed for addition and subtraction, because you can only combine "like parts" directly. Multiplication and division of fractions work directly on the numerators and denominators (multiply straight across, or multiply by the reciprocal for division) without ever needing the denominators to match first. Mixing this up — trying to find a common denominator before multiplying two fractions — wastes time and does not change the correctness of the answer, but it is unnecessary work worth dropping.

What is the fastest way to check my answer to a rational-number word problem is reasonable?

Reread the story and ask whether your numeric answer makes sense in context before trusting the arithmetic. A "remaining fraction" answer should be a positive fraction less than the whole you started with, never negative and never larger than one whole; a temperature that "fell" repeatedly should end up lower than it started, not higher. If your final answer for a shared quantity comes out larger than the total that was being shared, or a loss comes out positive, that mismatch is a strong signal that a sign was dropped or an operation was chosen incorrectly somewhere in the working, and it is worth retracing your labelled steps rather than trusting the last line blindly.

Before this

Where this leads