MAT Exam 2020 Paper

Concept:The basic relation is time = distance ÷ speed. Use the given changes in speed and time to form two equations in distance and original speed.Explanation:Let the distance be $z$ km and the original speed be $x$ km/h.When speed increases to $(x + 4)$ km/h, the time saved is 30 minutes = $\frac{1}{2}$ h.So, the time difference is given by:$$\frac{z}{x} - \frac{z}{x+4} = \frac{1}{2}$$On simplification, this gives:$$8z = x^2 + 4x \quad \text{... (1)}$$When speed decreases to $(x - 2)$ km/h, the time lost is 20 minutes = $\frac{1}{3}$ h.So, the time difference is given by:$$\frac{z}{x-2} - \frac{z}{x} = \frac{1}{3}$$On simplification, this gives:$$6z = x^2 - 2x \quad \text{... (2)}$$From equations (1) and (2), equate $z$:$$\frac{x^2 + 4x}{8} = \frac{x^2 - 2x}{6}$$Cross-multiplying:$$6x^2 + 24x = 8x^2 - 16x$$This simplifies to:$$2x^2 - 40x = 0$$$$\Rightarrow 2x(x - 20) = 0$$Since speed cannot be zero, $x = 20$ km/h.Substitute $x = 20$ in equation (2):$$6z = 20^2 - 2(20) = 400 - 40 = 360$$$$z = \frac{360}{6} = 60 \text{ km}$$Answer:The distance covered is 60 km, i.e. option B.