Towns P and Q are 45 km apart on a straight road. Vikram starts from P towards Q, while Deepak starts from Q towards P, at the same time. They meet after 1 hour 30 minutes. Vikram reaches Q exactly 1 hour 15 minutes after Deepak reaches P. Find Vikram's speed in km/h.
Answer and solution
Correct answer: B
Solution
Answer: B — 12 km/h
1. Convert times: meeting after 1.5 hours, arrival gap 1.25 hours. Let speeds be v for Vikram and u for Deepak.
2. Meeting gives (u+v)×1.5=45, so u+v=30. Vikram arrives later, hence v<u.
3. Because both cover 45 km, 45/v−45/u=1.25. Substitute u=30−v.
4. 36(30−2v)=v(30−v), giving v²−102v+1080=0=(v−12)(v−90).
5. v=90 would make u negative. Thus v=12, u=18. Full times 3.75 and 2.5 hours differ by 1.25 hours. Answer B.
Why this method works:
The meeting-time equation uses the sum of the speeds because the travellers approach each other. Arrival times use the complete 45 km journeys measured from the common start; the difference of these full times is 1.25 hours.
Independent check and common trap:
The feasible speeds 12 and 18 sum to 30, giving 45/30=1.5 hours to meet. Their full durations are 15/4 and 5/2 hours; the difference is 5/4. The rejected root 90 would force the other speed−60, which is physically invalid.
Apply: Meeting gives u+v=30. Test v=12, u=18: full travel times 45/12 and 45/18 differ by 1.25 hours. Option B fits both clues.
Check: The feasible speeds 12 and 18 sum to 30, giving 45/30=1.5 hours to meet. Their full durations are 15/4 and 5/2 hours; the difference is 5/4. The rejected root 90 would force the other speed−60, which is physically invalid.
1. Convert times: meeting after 1.5 hours, arrival gap 1.25 hours. Let speeds be v for Vikram and u for Deepak.
2. Meeting gives (u+v)×1.5=45, so u+v=30. Vikram arrives later, hence v<u.
3. Because both cover 45 km, 45/v−45/u=1.25. Substitute u=30−v.
4. 36(30−2v)=v(30−v), giving v²−102v+1080=0=(v−12)(v−90).
5. v=90 would make u negative. Thus v=12, u=18. Full times 3.75 and 2.5 hours differ by 1.25 hours. Answer B.
Why this method works:
The meeting-time equation uses the sum of the speeds because the travellers approach each other. Arrival times use the complete 45 km journeys measured from the common start; the difference of these full times is 1.25 hours.
Independent check and common trap:
The feasible speeds 12 and 18 sum to 30, giving 45/30=1.5 hours to meet. Their full durations are 15/4 and 5/2 hours; the difference is 5/4. The rejected root 90 would force the other speed−60, which is physically invalid.
G Strategy shortcut
Spot: Time, Speed and Distance.Apply: Meeting gives u+v=30. Test v=12, u=18: full travel times 45/12 and 45/18 differ by 1.25 hours. Option B fits both clues.
Check: The feasible speeds 12 and 18 sum to 30, giving 45/30=1.5 hours to meet. Their full durations are 15/4 and 5/2 hours; the difference is 5/4. The rejected root 90 would force the other speed−60, which is physically invalid.
G Strategy: Spot: Time, Speed and Distance. Apply: Meeting gives u+v=30. Test v=12, u=18: full travel times 45/12 and 45/18 differ by 1.25 hours. Option B fits both clues. Check: The feasible speeds 12 and 18 sum to 30, giving 45/30=1.5 hours to meet. Their full durations are 15/4 and 5/2 hours; the difference is 5/4. The rejected root 90 would force the other speed−60, which is physically invalid.