Join 60,000+ competitive exam aspirants
or with email
Your Cart
Subtotal₹0
Total₹0
Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
Join 60,000+ competitive exam aspirants
Two cyclists P and Q start from the same point. P cycles at 12 km/h. After 't' hours Q starts and catches P after 3 hours of his own cycling. If Q's speed is 16 km/h, what is the value of 't'?
0.5
1
1.5
2
1
Use the ratio of speeds: 16:12=4:3. Since times are inversely proportional to speed, the ratio of time taken is 3:4. Q takes 3 hours, so P takes 3×(4/3)=4 hours. P was moving for t+3=4 hours, so t=1.
Speed of P = 12 km/h, Speed of Q = 16 km/h, Q starts t hours later than P, Q catches P in 3 hours of his own cycling.
DistanceP=DistanceQ⟹SpeedP×(t+TimeQ)=SpeedQ×TimeQ
Use the ratio of speeds: 16:12=4:3. Since times are inversely proportional to speed, the ratio of time taken is 3:4. Q takes 3 hours, so P takes 3×(4/3)=4 hours. P was moving for t+3=4 hours, so t=1.
Students often add the time t to Q's time instead of P's time, forgetting that P has been traveling for a total duration of t+3 hours.
Calculate distance covered by Q
Q travels at 16 km/h for 3 hours.
DistanceQ=16×3=48 km
Relate distance to P
Since Q catches P, P must have covered the same 48 km in t+3 hours.
TimeP=SpeedPDistance=1248=4 hours
Solve for t
We know the total time for P is t+3=4.
t=4−3=1
B is correct because the total time P traveled is 4 hours, and since Q started 3 hours after P, the delay t is 4−3=1 hour.
This concept of equal distance for unequal start times is identical to problems involving 'Trains overtaking' or 'Boats and Streams' relative velocity scenarios.
Get instant step-by-step AI explanations, full shift Previous Year Papers (SSC GD, RRB, SSC JE), and regional rank analysis on the official Examoogle App.
Free Download • 100K+ Aspirants