Placement Papers - MCQ Practice Questions

15% of 80 = 0.15 × 80 = 12

Total distance = 60×2 + 40×3 = 120 + 120 = 240 km. Total time = 2 + 3 = 5 hours. Average = $\frac{240}{5}$ = 48 km/h

Let distance = d. Time = Distance/Speed. d/60 + d/40 = 10. LCM(60,40) = 120. (2d + 3d)/120 = 10. 5d = 1200. d = 240 km.

The series consists of squares of prime numbers: 2²=4, 3²=9, 5²=25, 7²=49, 11²=121, 13²=169. Next prime is 13, so 13²=169.

Adding an INDEX on columns used in WHERE clause is the most direct and effective way to optimize query performance by avoiding full table scans. This is a standard database optimization technique. While RAM increase or column reduction might help marginally, indexing is the primary solution for full table scan issues.

Average speed = Total Distance / Total Time. Distance one way = 240 km. Time from A to B = $\frac{240}{60}$ = 4 hours. Time from B to A = $\frac{240}{40}$ = 6 hours. Total Distance = 480 km, Total Time = 10 hours. Average Speed = $\frac{480}{10}$ = 48 km/h.

Let the 4th number be x. Then 5th number = previous number + x. If 5th = 20 and 6th = 32, then 6th = 5th + 5th_previous, so 32 = 20 + (5th_previous). Therefore, 5th_previous = 12. Since 5th_previous is the 4th number, the answer is 12.