Quantitative Aptitude - MCQ Practice Questions

For equal distances, average speed = 3/($\frac{1}{40}$ + $\frac{1}{50}$ + $\frac{1}{60}$) = 3/((3+2.4+2)/120) = 3/(7.$\frac{4}{120}$) = $\frac{360}{7}$.4 = 48.65 ≈ 49.09 km/h. Using LCD: 3/((15+12+10)/600) = $\frac{1800}{37}$ ≈ 48.65 km/h.

Let current speed = c. Upstream speed = 15 - c, downstream = 15 + c. Distance = 4(15-c) = 2(15+c). 60 - 4c = 30 + 2c. 30 = 6c. c = 5 km/h.

Combined rate = $\frac{1}{30}$ + $\frac{1}{20}$ = $\frac{5}{60}$ = $\frac{1}{12}$. Time to fill full tank = 12 hours. Time to fill $\frac{2}{5}$ tank = 12 × $\frac{2}{5}$ = 4.8 hours. Average per pipe = 4.8 × 1.5 = 7.2 hours.

For equal distances, average speed = 2ab/(a+b) = (2 × 8 × 12)/(8 + 12) = $\frac{192}{20}$ = 9.6 km/h.

Let total distance = 400 km. Distance segments: 100 km at 50 km/h (2 hrs), 200 km at 60 km/h (3.33 hrs), 100 km at 75 km/h (1.33 hrs). Total time ≈ 6.66 hrs. Average speed = $\frac{400}{6}$.66 ≈ 60.8 km/h.

In ratio 3:2, total parts = 5. Cost = (3 × 40 + 2 × 60)/5 = (120 + 120)/5 = $\frac{240}{5}$ = ₹48 per kg.