Finding Procedures

72 people are on a bus. At a stop, some get off and 15 get on — leaving 68. How many got off? The trap is option B (4), which comes from simply subtracting 72 − 68 = 4 and ignoring the 15 who got on. The correct method adds the 15 back: 72 + 15 − 68 = 19, or equivalently, work backwards from 68.

Bus/queue change problems — where a starting quantity changes by two steps (one unknown, one known) to reach a final quantity — are among the most common very-easy working-backwards questions in OC MR. They appear at the start of the paper and are designed to be solved quickly with a single equation.

The examiner tests whether students can translate a two-step change (subtract unknown, add known) into an equation and solve for the unknown. Option B (4) is the deliberate trap for students who compute start − end = 72 − 68 = 4 while ignoring the 15 who boarded.

A starting total is given. An unknown amount is removed, then a known amount is added (or vice versa). A final total is stated. Students must set up: start − unknown + known = end, then solve for the unknown.

Let's start with one of the most elegant shortcuts in OC MR — the supposition method. Instead of setting up two equations, you make one bold assumption ('suppose all animals are ducks'), measure how wrong you are, and use that gap to find the answer in two arithmetic steps.

Supposition (or "guess and adjust") problems appear in almost every OC MR paper, usually in the easy-to-medium range. They include legs-and-animals, tables-and-chairs, coin problems, and container questions — all sharing the same two-step shortcut.

The examiner is checking whether you know to assume everything is one type, compare the resulting total against the real total, and divide the gap by the per-swap difference — rather than setting up two simultaneous equations or trial-and-error across the answer options.

Two types of items each contribute a different fixed amount to a total (e.g. ducks have 2 legs, cats have 4 legs). You are given the number of items and the total contribution. You must find how many of each type there are.

Melissa has a 10-piece cake. She gives 2 to John, then gives Sue HALF OF WHAT IS LEFT, then eats 1 herself. What fraction remains? Three separate steps shrink the cake — and students must execute them in the correct order. The most common error is giving Sue half of the original cake (half of 10 = 5) rather than half of what's left after John (half of 8 = 4).

Multi-step distribution problems (give some away, then give half of the remainder, then use some more) are a staple of OC MR at the easy level. They require careful sequential tracking of a quantity. Students who draw a running tally (10 → 8 → 4 → 3) solve these reliably.

The examiner is testing whether students can follow three consecutive instructions in order without skipping or reordering any step. Option A (2/10) traps students who stop after step 1. Option E (3/4) traps students who skip the Sue step entirely. Option B (1/4) can arise from using the wrong denominator.

A total is divided through several sequential actions: give a fixed amount to one person, give half (or another fraction) of what remains to another, then consume or use one more. Students must track the running total at each step and finally express what remains as a fraction of the original total.

Hayden follows 3 instructions — choose a number, add 4, multiply by 2 — and ends up with 40. What is the digit sum of his starting number? The trick: you never need to guess! Work backwards step by step, undoing each operation in reverse.

Working-backwards (think-of-a-number) problems appear in every OC MR paper, usually rated easy to medium. They involve two or three operations applied to an unknown starting number. The final answer is given, and students reverse-engineer the starting number by undoing each step in the correct reverse order.

The examiner is testing whether students can identify the reverse of each operation (÷2 undoes ×2; −4 undoes +4), apply them in the right order, and then carry out the secondary task — here, finding the digit sum of the starting number rather than the number itself. Option A (6) traps students who take only the units digit of 16 instead of summing both digits.

A sequence of 2–3 arithmetic operations (add, subtract, multiply, divide) is applied to an unknown starting number. The final result is stated. Students work backwards through the operations in reverse order to find the starting number, then answer a question about it (e.g. digit sum, difference from another number).

Now for a different flavour of 'finding procedures': combined rate problems. When two pipes (or workers) contribute to the same task simultaneously, the shortcut is to think in rates per hour — not times. Once you see that, the question reduces to adding two fractions.

Pipe and worker rate questions appear in the medium difficulty band of almost every OC MR paper, often in the second half of the test. They are a reliable separating question — students who know the rate shortcut solve them in under a minute; those who guess or trial-and-error often choose 3 hours (just halving the faster pipe) and lose the mark.

The examiner is testing whether you know to convert time into rate (rate = 1 ÷ time), add rates to get the combined rate, and take the reciprocal to get the combined time. The most common wrong answer is 3 hours — the average of 6 and 12 halved — which is exactly what you'd get if you averaged times instead of adding rates.

Two pipes (or workers) can each complete a task independently in different amounts of time. Both work simultaneously. You must find how long the combined effort takes. Sometimes a drain or a slower agent is introduced to make the problem harder.