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A locomotive engine can go 40 km/hr. Its speed gets reduced by a quantity that varies directly as the square root of the number of wagons attached. It is known that its speed becomes 16 km/hr if 36 wagons are attached. What is the smallest number of wagons with which the engine is unable to move ?

This question was previously asked in
CDS 2 2026 Maths Question Paper (13-Sep-2026)
The correct answer is

100

Understanding the Speed Reduction Formula

The problem relates a locomotive engine's speed (\(S\)) to the number of wagons (\(w\)) attached. The base speed (no wagons) is \(S_0 = 40\) km/hr. The speed reduction (\(\Delta S\)) is directly proportional to the square root of the number of wagons: \(\Delta S = k \sqrt{w}\) where \(k\) is the constant of proportionality.

The speed of the engine with \(w\) wagons is given by: \(S(w) = S_0 - \Delta S\) \(S(w) = 40 - k \sqrt{w}\)

Calculating the Proportionality Constant (k)

We use the given information: when \(w=36\) wagons, the speed \(S = 16\) km/hr. Substitute these values into the equation: \(16 = 40 - k \sqrt{36}\) \(16 = 40 - k \times 6\)

Solve for \(k\): \(6k = 40 - 16\) \(6k = 24\) \(k = \frac{24}{6} = 4\)

Determining the Wagon Limit for Movement

With \(k=4\), the speed equation becomes: \(S(w) = 40 - 4 \sqrt{w}\)

The engine is unable to move when its speed \(S(w)\) is less than or equal to zero. We find the number of wagons where the speed is exactly zero by setting \(S(w) = 0\): \(0 = 40 - 4 \sqrt{w}\)

Solve for \(w\): \(4 \sqrt{w} = 40\) \(\sqrt{w} = \frac{40}{4} = 10\) \(w = 10^2 = 100\)

This means the engine stops moving when 100 wagons are attached. Therefore, the smallest number of wagons with which the engine is unable to move is 100.

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