A round trip is one stretch of length d covered once downstream and once upstream. The fast downstream leg and the slow upstream leg take different times, so you must handle them separately β never divide the total distance by the still-water speed.
Total time, distance from time, and the true average speed
Net speeds first: D=b+s and U=bβs. The total there-and-back time is Ddβ+Udβ.
β’ When the round-trip time T is given and you need the one-way distance, skip adding fractions and use d=D+UTΓDΓUβ (cancel T against D+U before multiplying).
β’ The round-trip average speed is bDΓUβ, not the plain average of D and U β the slow upstream leg drags the average below the midpoint.
β’ Speed-doubling puzzles fall out of the round-trip time formula T=b2βs22bdβ: take the ratio of the two cases and the distance cancels.
β΅Boats & Streams SimulatorSet speeds Β· enter a distance Β· see the full round trip
Boat speed (B) in still water
km/h
Stream speed (S / current)
km/h
Downstream (β)B + S = 14 km/h
Upstream (β)B β S = 6 km/h
Round trip β one-way distance dkm
LEG 1 Β· DOWNSTREAM (fast)60 Γ· 14 = 4.29 h
LEG 2 Β· UPSTREAM (slow)60 Γ· 6 = 10 h
Total time = d/D + d/U
4.29 + 10 = 14.29 h
Round-trip average = DΓU Γ· B
14Γ6 Γ· 10 = 8.4 km/h
below B = 10 β the slow leg drags it down
Shortcut check (time given β distance):d = TΓDΓU Γ· (D+U) = 14.29Γ14Γ6 Γ· 20 = 60 km β
With U=5, D=15 and a round trip of 8 h, the one-way distance 208Γ15Γ5β = km.
πPractice Questions
Q1A boat goes 4 km/h upstream and 12 km/h downstream. A round trip takes 8 h. The one-way distance is:
Q2A boat rows at 10 km/h in still water on a 2 km/h stream. It goes 48 km and returns. Total time is:
Q3A boat has downstream speed 12 km/h, upstream speed 6 km/h and still-water speed 9 km/h. Its round-trip average speed is:
π
Real exam questions β Round Trip & Return Journey
2 question types Β· 7 solved examples from real SSC papers
Solve any boats-and-streams round trip, from same-distance up-and-back legs to return-journey time-and-speed puzzles, using clean net-speed shortcuts.
How to solve this type
First write the net speeds: D=b+s and U=bβs. The total round-trip time is Ddβ+Udβ. When the total time T is given and you need the distance, skip adding fractions and use d=D+UTΓDΓUβ. The round-trip AVERAGE speed is bDΓUβ, NOT the plain average of D and U β the slow upstream leg drags it down.
A boat goes 5 km/h upstream and 15 km/h downstream. It makes a round trip (one way and back) in 8 hours. Find the one-way distance.
A30 kmB40 kmC45 kmD50 km
A boat's speed in still water is 6 km/h and the stream flows at 2 km/h. The boat rows to a point 24 km away and comes back. Find the total time.
A8 hB10 hC12 hD9 h
A boat goes 25 km/h upstream and 35 km/h downstream. The total round-trip time is 12 hours. Find the one-way distance.
A160 kmB150 kmC175 kmD180 km
A man rows to a place 48 km away and returns, taking 14 hours in all. His downstream and upstream speeds are in the ratio 4:3. Find the speed of the stream.