Rates and Ratios

Let's start with the most fundamental ratio question type: given a ratio between two quantities, one quantity changes — find the other. The key move is finding how many 'groups' the new total contains, then scaling the other side by that same number of groups.

Ratio and proportion questions appear in every OC MR paper, spread from easy to medium difficulty. The scaling format shown here is the most common variant — it is an easy mark for students who use the unit-groups method and a common error for those who try to cross-multiply fractions under time pressure.

The examiner is checking whether you can identify the matching ratio part (pupils = 25), divide the given total by it to find the number of groups, and then multiply the other ratio part (teachers = 3) by that number — rather than setting up a proportion equation that is slower to solve.

A ratio between two quantities is stated (e.g. 3 teachers : 25 pupils). One quantity is scaled to a new value (1050 pupils). You must find the corresponding value of the other quantity. The total is always chosen to divide exactly by the ratio part, so no decimals are needed.

A zoo feeds 8 penguins with 24 kg and 4 sea lions with 24 kg per week. The question asks for fish needed for 4 penguins AND 8 sea lions — both quantities have changed from the given values. The key is the unitary method: find the per-animal rate first, then scale to the new count.

Unitary method / scaling ratio problems appear frequently at the easy end of OC MR. A common trap is treating the given totals as fixed and adding them (24 + 24 = 48), forgetting that the animal counts have changed. Students who convert to per-animal rates first never fall for this.

The examiner tests whether students can apply the unitary method across two different animal types simultaneously. Option C (48) traps students who only compute the sea lion fish (8 × 6 = 48) and forget the penguins. Option A (12) traps students who only compute the penguin fish.

Two groups are given with their total consumption (or cost, etc.). The question then changes the group sizes and asks for a new combined total. Students must find the per-unit rate for each group, scale to the new size, then add the two results.

2 identical blocks raise a jug from 100 mL to 150 mL — a rise of 50 mL total, so 25 mL per block. 5 more blocks add 5×25=125 mL. New level = 150+125 = 275 mL. Option A (225 mL) traps students who add 125 mL to the original 100 mL instead of the current 150 mL.

Proportional displacement problems appear at the easy level of OC MR. Students find the rate per item from an initial experiment, then scale to find the new total. The key error is using the wrong base level (100 instead of 150) when adding the extra displacement.

The examiner tests proportional reasoning in a real-world context (block displacement). Option A (225 mL = 100 + 125) is the classic trap — students who correctly find 25 mL/block but forget to start from 150 mL (not 100 mL) after the first experiment.

An initial experiment tells you how much N identical items change a measurable quantity. Students find the per-item rate, then calculate the new total after M more items are added to the current (not original) state.

Now for the most instructive speed question type: the same distance, two different speeds, and you need the time difference. The setup looks simple, but students who rush consistently choose the wrong answer because they subtract speeds instead of computing both times separately.

Same-distance, different-speed comparison questions appear regularly in the medium difficulty section of OC MR. They test whether you apply T = D ÷ S twice rather than taking a shortcut with the speeds — the shortcut never works.

The examiner is testing whether you can apply the T = D ÷ S formula to each leg of the journey independently, convert the result to the same time unit (minutes or hours), and subtract — rather than trying to work with the speeds directly.

A vehicle travels the same distance twice at two different speeds. You are asked how much longer (or shorter) one leg takes than the other. The distance is chosen to divide evenly by both speeds, making the arithmetic clean once you commit to the right method.

4 bananas = 6 apples → 1 banana = 1.5 apples. Convert each bag: Bag1=10×1.5+9=24, Bag2=6×1.5+12=21, Bag3=16×1.5+3=27. All different. The key insight: without converting to a common unit, students might think bags with 'similar' counts weigh the same.

Equivalent weight/value ratio problems appear at the medium level of OC MR. Students must use a given equivalence (A weighs the same as B) to express everything in one unit, then compare totals. This pattern also appears with money exchange rates and sticker values.

The examiner tests whether students can apply a weight equivalence to convert mixed quantities to a single unit. Option A (all same) traps students who don't bother calculating. Options B–D trap students who make arithmetic errors in one of the three bags.

A weight equivalence between two items is given. Three containers each hold a different mix of the two items. Students express each mix in one unit and compare.

Squigs, zots, grozzles — made-up money with two conversion rates! To find how many squigs equal 20 grozzles, you must chain two conversions together: grozzles → zots first, then zots → squigs. Miss either step and you land on a wrong answer.

Two-step chain-conversion problems (A→B→C) appear in every OC MR paper at medium difficulty. They use either real units (cm→m→km, cents→dollars→hundreds) or invented units to prevent pattern-matching. Students must identify which direction each conversion runs and apply them in sequence.

The examiner is testing whether students can chain two separate ratio relationships without short-circuiting the middle step. Option A (24) and Option B (36) trap students who apply only one of the two conversions. Option D (108) traps students who double the correct answer by miscounting the groups.

Two exchange rates are given: X units of A = Y units of B, and P units of B = Q units of C. A target amount of C is given. Students convert C→B using the second rate, then B→A using the first rate, working in groups to avoid fractions.

Romesh takes 40 minutes to chop the vegetables alone. Lucinda also takes 40 minutes alone. Working together on Saturday, starting at 8:35 am — when do they finish? The key insight: when two workers have the same individual time, together they take exactly half that time. 40 ÷ 2 = 20 minutes → finish at 8:55 am.

Work rate problems (two workers combining to finish a shared job) appear in OC MR at medium difficulty. When both workers have the same individual time T, the combined time is T÷2. When the times differ, use the 1/T₁ + 1/T₂ = 1/T formula. This question uses the symmetric case, making the arithmetic straightforward once students extract both individual times correctly.

The examiner is testing whether students can extract individual completion times from start/finish pairs (converting the timestamps to durations), combine fractional rates, and then apply the combined time to a new start time. Options B (9:05) and D (9:25) are classic traps — those are the individual finish times for Lucinda and Romesh respectively.

Two workers complete the same job independently, each in a stated amount of time (given as start and finish times). On another day they work together from a new start time. Students calculate each individual time, add the rates (1/T each), find the combined time, and add it to the new start time.