A and B run a 1 km race. Case 1: A gives B a start of 50 m and still wins by 14 s. Case 2: A gives B a start of 22 s and then B wins by 20 m. Find the time taken by A to run 1 km.

Introduction / Context:
This blends two handicap conditions (distance head-start and time head-start) to solve the underlying speeds. With vA and vB extracted from the pair of equations, compute A’s 1 km time as 1,000/vA.

Given Data / Assumptions:

• Scenario 1: A runs 1,000 m; B runs 950 m; B’s time − A’s time = 14 s.
• Scenario 2: A starts 22 s late; when B finishes 1,000 m, A is 20 m short (at 980 m).

Concept / Approach:
Let vA = a, vB = b (m/s). From Case 1: 950/b − 1000/a = 14. From Case 2: a*(1000/b − 22) = 980. Solve simultaneously to obtain a, then tA = 1000/a.

Step-by-Step Solution:

Verification / Alternative check:
Check Case 1: tA = 100 s; tB = 950/b. Using a = 10 in the relations reproduces the 14 s gap. Check Case 2: A’s distance at B’s finish equals 980 m with a 22 s delay.

Why Other Options Are Wrong:
Other times mismatch one or both scenarios when substituted back.

Common Pitfalls:
Treating both handicaps as distances or both as times; keep them distinct and set equations carefully.

Final Answer:
100 sec