A political party orders an arch for the entrance to the ground in which the annual convention is being held. The profile of the arch follows the equation $y = 2x - 0.1x^2$ where $y$ is the height of the arch in meters. The maximum possible height of the arch is

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Solution

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Official answer: B) 10 meters

Quantitative Aptitude

The height of the arch is given by the quadratic equation:

y = 2x - 0.1x^2

To find the maximum height, differentiate $y$ with respect to $x$ and set the derivative to zero:

\frac{dy}{dx} = 2 - 0.2x = 0 \implies x = \frac{2}{0.2} = 10

Verify that this is a maximum:

\frac{d^2y}{dx^2} = -0.2 < 0

Substitute $x = 10$ into the original equation:

y_{\text{max}} = 2(10) - 0.1(10)^2 = 20 - 0.1(100) = 20 - 10 = 10\text{ meters}

Thus, the maximum possible height of the arch is 10 meters.

Hence, the correct option is (B).

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This Quantitative Aptitude previous year question is Q62 of the GATE CSE 2012 paper. Find more GATE CSE PYQs on Quantitative Aptitude in General Aptitude, or browse all GATE CSE PYQs.

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Quantitative Aptitude has 53 GATE CSE PYQs across 13 papers from 2011 to 2026. 27 of them are 2-mark MCQs like this one, and GATE CSE 2012 has 5.

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.