How to Solve Speed and Distance Problems for NIFT Quant 2026

Speed and distance questions consistently appear in the NIFT 2026 General Ability Test, testing your ability to handle time-motion relationships quickly. The questions range from simple speed calculation to tougher ones involving multiple travellers through trains and relative motions. These problems can be cracked if you know the basic formula and identify some patterns in the problem statement, all in less than 90 seconds.

Speed and Distance Problems for NIFT Quant Section

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Logical thinking and formula application under pressure: Time and distance problems. Success depends on fast identification of question types and selection of the correct approach.

The fundamental relationship governing all motion problems is: Distance = Speed × Time. Every speed-distance question manipulates these three variables. When a car travels at 60 km/h for 3 hours, the distance covered is 60 × 3 = 180 km.

If someone covers 240 km in 4 hours, their speed is 240 ÷ 4 = 60 km/h. This triangular relationship forms the foundation - knowing any two variables instantly gives you the third.

Conversion of units is where most students lose many marks. Always ensure that speed, distance, and time are in compatible units. When speed is in km/h and time in minutes, convert minutes to hours by dividing by 60. To convert km/h to m/s, multiply by 5/18. To convert m/s to km/h, multiply by 18/5. Conversations like these appear in nearly all the NIFT papers.

Students are becoming quite confused about average speed. When a person travels for equal distances at two different speeds, it is important to emphasise that average speed is not the arithmetic mean of those two speeds. If you travel 120 km at 40 km/h and return the same 120 km at 60 km/h, don't just average 40 and 60 to get 50. Calculate total distance (240 km) divided by total time (3+2 = 5 hours) to get 48 km/h. For equal distances at speeds A and B, use the formula: Average Speed = (2AB)/(A+B).

Relative speed is the speed which determines how fast two objects are coming together or going apart. The speeds simply add when two people walk toward each other. If A walks at 5 km/h and B at 4 km/h toward each other from 18 km apart, they meet in 18 ÷ (5+4) = 2 hours. When moving in the same direction, speeds subtract. If A cycles at 15 km/h and B at 12 km/h in the same direction, with A behind by 9 km, A catches up in 9 ÷ (15-12) = 3 hours.

Train problems involve length considerations, usually. When a train crosses a stationary object like a pole, it covers its own length. A 150m train at 54 km/h (15 m/s after conversion) crosses a pole in 150 ÷ 15 = 10 seconds. When crossing a platform, the train covers its length plus the platform length. The same 150m train crossing a 250m platform covers 400m total, taking 400 ÷ 15 = 26.67 seconds.

When two trains cross each other, their lengths add, and speeds add (opposite direction) or subtract (same direction). A 120m train at 72 km/h and a 180m train at 54 km/h moving in opposite directions have a relative speed 72+54 = 126 km/h = 35 m/s. They cross in (120+180) ÷ 35 = 8.57 seconds.

However, in streams where boats move, the complexity is added to the speed of the current. When one rows downstream, the speed of the boat adds to that of the stream. When rowing upstream, stream speed opposes boat speed. If a boat rows at 12 km/h in still water and stream flows at 3 km/h, the downstream speed is 12+3 = 15 km/h, upstream speed is 12-3 = 9 km/h. To find boat speed when downstream and upstream speeds are known: Boat Speed = (Downstream + Upstream) ÷ 2. Stream Speed = (Downstream - Upstream) ÷ 2.

In a circular track problem, we need to consider the relative position held by the two. When two people run around a circular track in opposite directions, they meet after one complete round together on the track. If A runs at 8 m/s and B at 6 m/s on a 280m track, they meet every 280 ÷ (8+6) = 20 seconds. In the same direction, they meet when the faster person gains one full lap on the slower person, taking 280 ÷ (8-6) = 140 seconds.

Moreover, these time-speed-distance relations are in a proportional relationship. If speed is doubled, time is reduced to half for the same distance; if the distance becomes double at constant speed, time will also become double. Remembering these proportional relationships provides an easy way to solve some tough problems without going through tiresome calculations. When a journey normally takes 5 hours at 60 km/h, increasing speed to 75 km/h reduces time proportionally: (60 ÷ 75) × 5 = 4 hours.

Recognise whether a problem needs a basic formula application or the calculation of relative speed, or proportional reasoning. Most questions for NIFT cocoon a straightforward concept in a lot of words. Cut all that out and see the core relationship being tested.

Comprehending the basic formulations on speed and distance should go hand in hand with quick conversions of units and relative motion. The practice of different problem types is meant to form patterns and accuracy, after which speed would definitely improve. These become automatic tools for solving questions within 90 seconds and success in exams when practised frequently.

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