6.3 Multi-Step Arithmetic Word Problems

6.3 Multi-Step Arithmetic Word Problems

Quick Answer: Multi-step problems are chains of simple operations. Plan → compute each intermediate → check → answer. Label every partial result. If the final option does not match a re-run of the chain, an intermediate step slipped.

Entrance-exam mathematics rarely needs advanced algebra. What separates high scores from mid scores is reliability on word problems that take three to six arithmetic steps. This section trains a methodical approach with Trinidad and Tobago civilian stories: markets, buses, school fees, household shopping, and shared trip costs. No firefighting technical calculations.

The four-phase method

Phase 1 — Read for structure, not just numbers

• Circle every number and its unit.
• Underline the question sentence (what is finally asked).
• Cross out distractors (extra numbers that are not needed).
• Note signal words:

Phase 2 — Write a short plan

Example plan (not numbers yet):

1. Find cost of all items.
2. Subtract discount.
3. Subtract amount paid to find change or add tax if required.

Plans prevent “doing the first numbers you see.”

Phase 3 — Execute with labelled intermediates

Write:

• Subtotal items = …
• After discount = …
• Change = …

Phase 4 — Check

• Reverse the last operation when possible.
• Re-estimate magnitude.
• Confirm the answer unit matches the question.

Worked example 1 — market shopping with change

Maya buys:

• 3 kg of potatoes at $8 per kg
• 2 kg of onions at $10 per kg
• 1 pack of seasoning for $12

She pays with a $100 note. How much change does she receive?

Step 1 — potatoes: 3 × 8 = $24.
Step 2 — onions: 2 × 10 = $20.
Step 3 — subtotal: 24 + 20 + 12 = $56.
Step 4 — change: 100 − 56 = $44.

Check: 56 + 44 = 100. Good.

Trap option: forgetting seasoning → change $56 (100 − 44 wrongly). Another trap: treating $8 as total for potatoes instead of per kg.

Worked example 2 — multi-leg bus fares and remaining cash

Keiran has $80. In order he:

1. spends $8 on a maxi-taxi to town,
2. spends $15 on lunch,
3. spends 25% of what remains after lunch on a book,
4. spends $6 on the return maxi-taxi.

How much money does he have at the end?

After transport to town: 80 − 8 = 72.
After lunch: 72 − 15 = 57.
Book = 25% of 57 = 0.25 × 57 = 14.25.
After book: 57 − 14.25 = 42.75.
After return: 42.75 − 6 = $36.75.

If someone applies 25% to the original $80, they get a wrong path. Intermediate labels prevent that. Always follow the story order when a percent applies to a remaining amount.

Worked example 3 — working backwards

Some stems give you the end of the story and ask for the start. Reverse the operations, working from the last event to the first.

After giving $40 to her brother and spending $25 on groceries, Ria has $95 left. How much did she have at first?

Work backwards (reverse operations in reverse order):

Before groceries she had 95 + 25 = 120.
Before giving her brother she had 120 + 40 = $160.

Check forward: 160 − 40 = 120; 120 − 25 = 95. Correct.

Rule: reverse of subtract is add; reverse of add is subtract; reverse of multiply is divide; reverse of divide is multiply—applied from the end of the story to the start.

Worked example 4 — sharing and leftover

Three friends share a $180 trip cost. Friend A also pays $24 for group snacks. Everyone should share both costs equally. Who owes whom?

Total group spend = 180 + 24 = 204.
Equal fair share = 204 ÷ 3 = $68 each.

Now settle who actually paid what. Assume the trip fare was paid $60 each up front, and A alone paid the $24 snacks:

B and C each pay A $8. After settlement everyone has effectively paid $68. Multi-step money problems reward a fair-share table like this.

Worked example 5 — combining percentages and totals

A school book costs $120. During a sale it is reduced by 15%. VAT-like final surcharge is not assumed unless stated—do not invent taxes. Question: sale price?

Discount = 15% of 120 = 0.15 × 120 = 18.
Sale price = 120 − 18 = $102.
Or 120 × 0.85 = 102.

Follow-up multi-step: buy 3 books at the sale price and pay with $400. Change?

3 × 102 = 306. Change = 400 − 306 = $94.

Worked example 6 — mixed units inside a story

A water cooler starts with 8 L. Each day the office uses 1,500 mL, and once midweek someone refills 2 L. After 4 full days of use and one 2 L refill on the morning of day 3 (before that day’s use), how much remains?

Convert everything to mL: start 8,000 mL; daily use 1,500 mL; refill 2,000 mL.

Day 1 end: 8,000 − 1,500 = 6,500.
Day 2 end: 6,500 − 1,500 = 5,000.
Day 3 morning refill: 5,000 + 2,000 = 7,000; end day 3: 7,000 − 1,500 = 5,500.
Day 4 end: 5,500 − 1,500 = 4,000 mL = 4 L.

Sequence and units both matter. Drawing a tiny day-by-day table is worth the time.

Worked example 7 — multi-step with average link

A cricketer scores 32, 45, and 28 in three matches. What must he score in a fourth match so that the mean of four matches is 40?

Required total for mean 40: 40 × 4 = 160.
Current total: 32 + 45 + 28 = 105.
Needed fourth score: 160 − 105 = 55.

This blends §6.1 averages with multi-step planning.

Intermediate checks (non-negotiable)

After each major step, ask:

1. Units still consistent?
2. Is this intermediate larger/smaller in the expected direction? (Discounts decrease price; adding items increases subtotal.)
3. Does mean × count recover a known total?
4. Can I reverse the last step?

If an intermediate looks impossible (negative money without debt being mentioned, or a mean outside the data range without explanation), stop and re-read.

Translating denser sentences

Break long stems into short lines:

“A trader buys 40 shirts at $25 each, sells 30 of them at $40 each and the remaining shirts at $20 each. Find profit.”

Lines:

• Cost: 40 × 25 = 1,000.
• Revenue high: 30 × 40 = 1,200.
• Remaining shirts: 40 − 30 = 10.
• Revenue low: 10 × 20 = 200.
• Total revenue: 1,200 + 200 = 1,400.
• Profit: 1,400 − 1,000 = $400.

Estimation first

Before exact work, rough bounds help eliminate MCQ options.

Example: 19 × 21 is near 20 × 20 = 400, so options near 40 or 4,000 are suspicious.
15% of 198 is near 0.15 × 200 = 30.

Common multi-step traps

Exam pacing tip

For a long word problem:

1. 20–30 seconds: parse and plan.
2. Compute with intermediates.
3. 10 seconds: reverse-check if time remains.
4. If stuck, skip and return—do not let one story problem burn the whole paper.

Bridge to rates

Many multi-step problems become easier once you rewrite them as rates (cost per item, km per hour, tasks per day). Section 6.4 specialises in time, distance, work, and rate structures that frequently appear as multi-step stories.