Speed and Time Calculation

Four boys, four bicycle colours, and a handful of positional clues — who wins the race? This question rewards students who slow down, match each rider to their bike, then build the finishing order step by step from the clues. The twist is one clue that hides a name behind a colour description.

Bicycle / colour-coded race ordering questions appear regularly in OC TS exams. They are a favourite because the stem deliberately uses both names and colours interchangeably, forcing students to track both. Missing the "level with" clue is the single most common source of errors.

The examiner is testing whether students can track two attributes (rider name and bicycle colour) simultaneously while building a linear finishing order. The "Marcel is level with the boy on the purple bicycle" clue is the deliberate trap: students must realise the purple bicycle belongs to Ralph, and therefore Marcel = Ralph in finishing position.

Four or five competitors each have a uniquely coloured item (bicycle, hat, bag). The stem gives relative finishing positions using both names and colour descriptions interchangeably. One clue uses a colour description for a rider whose name you already know, requiring you to mentally substitute the name before applying the clue.

A three-level structure holds food, water bowl, and a log. Two clues: food is below water, and food is below log. Which statement must be true? This is the simplest form of ranking deduction: when one item is lower than both others, it must be at the bottom — there is no other spot for it.

Three-item ranking questions with "must be true" phrasing are one of the most common OC TS question types. They test whether students understand the difference between "possible" and "necessary". Getting the right answer requires checking whether an arrangement is forced (must be true) or just one of several possibilities (could be true).

The examiner is testing whether students can deduce a forced position from two relative clues. Food is constrained to be lower than both other items — and in a three-level system, the only level lower than both remaining levels is the lowest. Options A and C are traps: the log and water can each be at middle or top, so neither is forced.

Three items must fill three distinct positions (e.g. levels, places in a race, floors of a building). Two clues each give a "higher/lower/before/after" relationship involving the same item. You must find which arrangement is forced — not just possible. The "must be true" phrasing is the key signal to look for forced, not optional, placements.

Four plants bloom at different times across a school: geranium in the canteen, guppy plant in the music room, wax plant in the school hall, bromeliad in the library. Three clues tell you the relative order. The key is Clue 2: the geranium bloomed on the same day as the school hall plant — which means you need to identify the school hall plant before you can place the geranium.

"Which one bloomed/finished/arrived first?" ordering questions with a tie-clue (two items happen on the same day) appear regularly in OC TS. Students who miss the tie-clue often place the geranium incorrectly, leading them to choose it instead of the bromeliad as the earliest. Always read every clue carefully before placing any item.

The examiner is testing two skills: (1) chaining relative order clues (bromeliad < wax plant < guppy plant), and (2) correctly applying the indirect same-day clue (geranium = wax plant = school hall). Students who read 'bloomed on the same day as the plant in the school hall' without identifying which plant is in the school hall will be confused. The question also tests careful reading of 'bloomed a day before any of the others' — this means first, not last.

Four items are placed in different named locations. Clue 1 gives a chain of relative order (A before B, B before C). Clue 2 uses a location name ("the plant in X bloomed the same day as Y") to indirectly tie two items together. You must identify what item is in location X, then chain all clues to produce the full order. The question asks for the first or last item.

Speed and Time problems are a core component of OC Thinking Skills — and the single most valuable insight in this entire unit is that if something travels at twice the speed, it takes exactly half the time. Once you hold that idea firmly, these questions become fast and reliable marks.

Relative speed questions — where two vehicles travel the same route at different speeds — appear consistently across OC TS tests. Students who apply the speed-time inverse relationship automatically solve them in under thirty seconds; students who try to calculate actual distances waste precious time.

The examiner is checking whether you understand the inverse relationship between speed and time: doubling the speed halves the journey time, and halving the speed doubles it. From there, the question asks you to count backwards from a fixed arrival deadline to find the departure time.

Two vehicles cover the same route, one faster than the other. You are told the speed ratio (e.g. "twice as fast"), given the timetable of one vehicle, and asked when the slower vehicle must depart to arrive by a specific time. Sometimes an extra number — like a specific named journey time — is included to anchor the scenario.

When the express is four times as fast, it covers the same distance in one-quarter of the time. The slow train took 6 hours, so the express takes just 1½ hours — count back from 7:00 pm and the departure is 5:30 pm.

Inverse speed-time problems with a speed ratio of 4× appear in the medium-to-difficult band of OC TS. Students who try to calculate actual speeds or distances waste time; recognising that 4× speed = ¼ time is the instant shortcut.

The examiner checks whether students understand that speed and time are inversely proportional over the same distance, and can then subtract the express journey time from the shared arrival time to find the departure time.

Two vehicles travel the same route and arrive at the same time. One is a fixed multiple faster than the other. A timetable is given for the slower vehicle; students must find when the faster vehicle departed.

Six differently-coloured lunch boxes, three positional clues, and the question: which two boxes sit at each end? This is a classic constraint-satisfaction problem where one powerful clue (purple has only one neighbour) pins down the whole arrangement. Find that anchor clue first and everything else falls into place.

Six-item linear ordering questions appear frequently in OC TS, often using coloured objects on a shelf or in a row. The "only one neighbour" clue is a favourite examiner device because it is the strongest possible positional constraint — it forces an item to an end — and students who miss it often waste time on dead-end arrangements.

The examiner is testing whether students can identify the most constraining clue first and use it to anchor the arrangement. The "purple has only one neighbour" clue uniquely fixes purple to an end position, which then combines with the "three boxes between red and blue" clue to lock in both ends as red and purple in all valid arrangements.

N uniquely named items are placed in a row. Each clue gives a relative distance or adjacency constraint. One clue will say an item has only one neighbour (or is adjacent to only one other), which forces it to an end. Subsequent clues fill in the remaining gaps. The question usually asks for the items at the ends or a specific position.

Four friends in a car with two front seats and two back seats. Three constraints: Katelyn drives or sits beside Talia; Talia is behind the driver; Jimmy is not in the back and not beside Andrea. Who drives? The key is to try both interpretations of the first clue — one works, one creates an unavoidable contradiction with the third clue.

Two-row seating arrangement questions with an "either/or" clue appear regularly in OC TS. The "either/or" clue always generates exactly two cases to test. Students who test only one case — or who do not check all clues against their final arrangement — frequently arrive at wrong answers. Case-elimination is the core skill here.

The examiner is testing whether students can apply an 'either/or' clue as a branch-point (Case A vs Case B) and then eliminate the invalid branch using a second constraint (Jimmy not beside Andrea). Students who stop after finding one valid-looking arrangement without checking all clues, or who assume Katelyn must be beside Talia (ignoring the driving option), will pick the wrong answer.

A grid of positions (seats, desks, rooms) needs to be filled. One clue gives an "either/or" choice for a person, creating two cases. A second clue is very specific and anchors one person immediately. A third clue eliminates one of the two cases by creating a contradiction. The remaining case gives the unique answer.

Four sandwich ingredients — chicken, mayonnaise, cheese, lettuce — must be stacked in four layers. Three rules constrain the order: lettuce can't touch mayo, cheese can't touch lettuce, and mayo goes directly above chicken. The direct-adjacency rule (mayo on top of chicken) is the strongest starting point — use it to anchor the pair and then test where the pair can go.

Four-item linear stacking questions with adjacency and non-adjacency constraints appear regularly in OC TS. The "directly on top of" rule is always the best starting clue because it locks two items into a fixed relative position. Students who start with the "must not touch" rules and try all permutations waste time; starting with the forced pair and testing three positions is much faster.

The examiner is testing whether students can systematically use an anchor clue (mayo directly above chicken) to reduce the search space to three cases, then eliminate two cases using the remaining constraints. Students who guess or try all 24 permutations are likely to make errors. The diagram-building skill (writing layers 1–4 and filling in known pairs) is the key technique.

Four items must be placed in a strict vertical order. One clue forces two items into direct adjacency ("directly above/below"). Two clues forbid adjacency between specific pairs. The anchor clue creates three possible positions for the adjacent pair. One of the three positions satisfies all remaining constraints; the other two fail. The question asks for the item at a specific position (top, bottom, etc.).

Now for the most demanding type in this unit — a multi-leg international flight problem that stacks three calculations on top of each other: adding multiple flight segments, including a layover, and then converting the result into a different time zone. On top of that, the question deliberately hides a red herring inside the numbers to trip you up.

Multi-leg flight problems with time zone conversions appear regularly in the middle-to-hard section of OC TS tests. They reward students who work in a clear, structured sequence and who have trained themselves to spot irrelevant numbers immediately.

The examiner is testing three things simultaneously: whether you can add flight times and layover time in sequence without mixing up the order, whether you can identify and discard an irrelevant piece of information (the total distance in km), and whether you can apply a time zone offset correctly — adding or subtracting based on which city is ahead.

A traveller flies from City A to City C via City B, with two given flight durations and a layover time. The total route distance in kilometres is also stated — this is always a red herring when flight durations are already given. You are told the time zone relationship between origin and destination, then asked what the local clock at the destination shows when the plane lands.