‘A’ starts his journey at 1:00 p.m. from a location P with a speed of 1 m/sec. ‘B’ starts his journey from the same location P and along the same direction at 1:10 p.m. with a speed of 2 m/sec. If ‘B’ meets ‘A’ at the location Q, then the distance PQ is || || ADRE 1.0 SLRC 2022 PAPER-III SOLVED QUESTIONS
‘A’ starts his journey at 1:00 p.m. from a location P with a speed of 1 m/sec. ‘B’ starts his journey from the same location P and along the same direction at 1:10 p.m. with a speed of 2 m/sec. If ‘B’ meets ‘A’ at the location Q, then the distance PQ is:
(A) 1.5 km
(B) 1.75 km
(C) 1.2 km
(D) 1.25 km
Solution:
Let's solve the problem step by step.
1. Determine the time distance between the starts of A and B:
- A starts at 1:00 p.m.
- B starts at 1:00 p.m.
- The time difference is 10 minutes, which is 600 seconds (since 1 minute= 60 seconds)
- A's speed= 1 m/sec
- Time= 600 seconds
- Distance traveled by A
= Speed × Time
= 1m/sec × 600 sec
= 600 meters.
- A's speed= 1m/sec
- B's speed= 2m/sec
- Relative speed of B with respect to A
= 2m/sec-1m/sec
= 1m/sec
- Distance to be covered by B to catch up with A= 600 meters
- Relative Speed= 1m/sec
Time taken= Distance/Relative Speed
= 600 meterss/1m/sec
= 600 seconds
5. Calculate the total distance PQ:
- In 600 seconds, B travels
= 2m/sec × 600 seconds
= 1200 meters
= 1.2 km
Therefore, the distance PQ is: (C) 1.2 km.
Importance of this question:
- Enhances logical reasoning skills.
- Strengthens time-distance concept understanding.
- Essential for solving real-world travel problems.
- Improves speed and accuracy in competitive exams.
- Previously asked in ADRE SLRC 2022 Paper III for the Grade-III Post.
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