MAT Exam 2021 Paper

Concept:When speed becomes $\frac{3}{4}$ times, time for the same distance becomes $\frac{4}{3}$ times.Explanation:Let usual speed be $s$ and total distance be $D$.Usual total time is $\frac{D}{s} = 1$ hour.First half distance is $\frac{D}{2}$.Time for first half at $\frac{3}{4}s$ is:$\frac{\frac{D}{2}}{\frac{3}{4}s} = \frac{2D}{3s} = \frac{2}{3}$ hour.Time remaining for the second half is:$1 - \frac{2}{3} = \frac{1}{3}$ hour.Required speed for the second half is:$\frac{\frac{D}{2}}{\frac{1}{3}} = \frac{3D}{2}$.Since usual speed is $s = D$ per hour, required speed $= \frac{3}{2}s$.Thus, she must travel the second half at $\frac{3}{2}$ times her usual speed.Answer:Option A: At $\frac{3}{2}$ times of her usual speed.