Episode 8 — Aptitude and Reasoning / 8.13 — Boats and Streams
8.13.b Tips, Tricks, and Shortcuts -- Boats and Streams
Tip 1: The "Add and Subtract" Memory Device
The two most important formulas are the ones for B and S. Remember them as:
B = AVERAGE of downstream and upstream speeds
S = HALF the DIFFERENCE of downstream and upstream speeds
B = (D_s + U_s) / 2 (add, then halve)
S = (D_s - U_s) / 2 (subtract, then halve)
Quick mental check: B is always bigger than S (otherwise the boat cannot go upstream at all).
Tip 2: Stream Speed Cancels in Two-Boat Problems
This is the single most powerful shortcut for boats and streams:
When two boats are on the SAME river:
- Relative speed = difference of boat speeds (same direction)
- Relative speed = sum of boat speeds (opposite direction)
THE STREAM SPEED DOES NOT MATTER!
This means meeting and overtaking problems on a river reduce to simple Speed-Distance-Time problems using only the boats' still-water speeds.
Example:
Two boats (still-water speeds 10 km/h and 6 km/h) start 48 km apart
on a river, heading towards each other. When do they meet?
Time = 48 / (10 + 6) = 48/16 = 3 hours
Stream speed? Does not matter. Not needed.
Tip 3: The Ratio Shortcut for B and S
If you are given the ratio of upstream time to downstream time:
T_up : T_down = p : q (p > q since upstream takes more time)
Then speeds are in inverse ratio:
(B - S) : (B + S) = q : p
From this:
B/S = (p + q) / (p - q)
Example:
Upstream time : Downstream time = 7 : 5
B/S = (7 + 5) / (7 - 5) = 12/2 = 6
So the boat is 6 times faster than the stream.
Tip 4: Round Trip Time Shortcut
For a round trip of distance D each way:
Total time = 2DB / (B^2 - S^2)
If you know B and S, compute B^2 - S^2 first:
B^2 - S^2 = (B+S)(B-S) = downstream speed x upstream speed
So: Total time = 2DB / (D_s x U_s)
where D_s and U_s are downstream and upstream speeds.
Or: Total time = 2D / D_s + 2D is wrong... let me redo.
Actually simpler: just compute both legs separately.
T = D/(B+S) + D/(B-S)
Practical example:
B = 10 km/h, S = 2 km/h, D = 48 km
T_down = 48/12 = 4 hours
T_up = 48/8 = 6 hours
Total = 10 hours
Using formula: 2 x 48 x 10 / (100 - 4) = 960/96 = 10 hours (same)
Tip 5: The Floating Object Trick (Frame of Reference)
When a boat drops something in the river:
+--------------------------------------------------+
| |
| Time to return to the object = Time spent |
| moving away from the object. |
| |
| Stream speed is irrelevant. |
| |
+--------------------------------------------------+
Why? In the water's reference frame, the object is stationary. The boat moves at speed B relative to water in both directions. So time away = time back.
When the problem gives distance instead of time:
If the boat was X km upstream from the object when it turned back:
X is the distance relative to WATER (not ground).
Distance relative to ground is different, but you don't need it.
Time to return = X / B (in water's frame).
Tip 6: Converting Word Problems to B+S / B-S
Many problems disguise the boats-and-streams setup. Recognize the pattern:
"A rower takes 3 hours downstream and 5 hours upstream for 30 km each"
Downstream speed = 30/3 = 10 km/h = B + S
Upstream speed = 30/5 = 6 km/h = B - S
B = (10 + 6)/2 = 8 km/h
S = (10 - 6)/2 = 2 km/h
Always extract B+S and B-S first, then solve.
Tip 7: The "Still Water" Trap
"In still water" means S = 0.
If a problem says "a man can row 8 km/h in still water," then B = 8.
His actual speed on the river depends on the current.
If a problem says "a boat goes 8 km/h downstream," then B + S = 8.
B is NOT 8.
Tip 8: Use LCM for Finding Distance
When both upstream and downstream times are given but distance is not:
Assume distance = LCM of the two effective speeds.
Example: Downstream speed = 12 km/h, Upstream speed = 8 km/h
LCM(12, 8) = 24 km
T_down = 24/12 = 2 hours
T_up = 24/8 = 3 hours
Tip 9: Speed of Current from Two Trips
If a boat covers D km downstream in T1 hours and D km upstream in T2 hours:
B + S = D/T1
B - S = D/T2
S = D/2 x (1/T1 - 1/T2) = D(T2 - T1) / (2 x T1 x T2)
B = D/2 x (1/T1 + 1/T2) = D(T1 + T2) / (2 x T1 x T2)
Tip 10: The Percentage Speed Reduction
Percentage speed reduction upstream vs still water:
Reduction = S / B x 100%
If S = B/4, the upstream speed is 3/4 of still water speed (25% reduction).
The downstream speed is 5/4 of still water speed (25% increase).
But the TIME increase upstream is 1/3 (not 1/4)!
(Because speed decreased by 1/4, time increases by 1/3 -- see Tip 3 of 8.11)
Tip 11: Quick Formula for "n Times" Problems
"A boat takes n times as long to go upstream as downstream"
T_up = n x T_down
(B-S) x n.T_down = (B+S) x T_down [same distance]
(B-S) x n = (B+S)
nB - nS = B + S
B(n-1) = S(n+1)
B/S = (n+1)/(n-1)
Ready table:
n (times longer) | B/S ratio
------------------|----------
2 | 3/1 = 3
3 | 4/2 = 2
4 | 5/3
5 | 6/4 = 3/2
7 | 8/6 = 4/3
Tip 12: If a Boat Cannot Go Upstream
If B <= S, the boat cannot make upstream progress.
B = S: Upstream speed = 0 (boat stays in place)
B < S: Boat is pushed backwards by the current
This is often tested as: "What is the maximum stream speed for
the boat to reach upstream?" Answer: S < B.
Tip 13: Average Speed for Round Trip (Shortcut)
Let downstream speed = a, upstream speed = b.
Average speed = 2ab / (a + b)
This is the SAME formula as the harmonic mean (from Section 8.11),
because the distances are equal.
Example:
Downstream speed = 15 km/h, Upstream speed = 5 km/h
Average for round trip = 2 x 15 x 5 / (15 + 5) = 150/20 = 7.5 km/h
Tip 14: Meeting After Departure from Same Point
If two boats leave from the same point on a river:
Boat A goes downstream, Boat B goes upstream.
They separate at rate: (B_a + S) + (B_b - S) = B_a + B_b
Time to be X km apart = X / (B_a + B_b)
(Stream speed cancels!)
Tip 15: The "Effective Speed" Visualization
Think of the stream as a conveyor belt:
DOWNSTREAM:
Boat walks forward on conveyor going forward.
=========================================>
--> boat on belt -->
Net speed = boat + belt
UPSTREAM:
Boat walks forward on conveyor going backward.
<=========================================
--> boat on belt
Net speed = boat - belt
Tip 16: Common Exam Patterns
Pattern 1: Given downstream and upstream speeds, find B and S.
--> Use: B = (a+b)/2, S = (a-b)/2
Pattern 2: Given B and S, find time for a given distance.
--> Compute a = B+S, b = B-S, then T = D/a or T = D/b
Pattern 3: Given times for same distance, find B and S.
--> Compute speeds first: a = D/T_down, b = D/T_up, then formulas.
Pattern 4: Round trip with total time given, find D.
--> D = T(B^2 - S^2) / (2B)
Pattern 5: Floating object dropped in river.
--> Return time = time spent going away. Stream speed irrelevant.
Summary of Top Shortcuts
1. B = (D_s + U_s)/2, S = (D_s - U_s)/2
2. Stream cancels in two-boat relative speed
3. T_up:T_down = (B+S):(B-S) inverted from speeds
4. n times longer upstream: B/S = (n+1)/(n-1)
5. Floating object: return time = departure time
6. Round trip avg speed = 2ab/(a+b) where a=B+S, b=B-S
7. Always extract B+S and B-S first, then add/subtract
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