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TheNightKing
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A boat can cover a certain distance of 180 km and come back in 12.5 ho 25 Feb 2020, 10:48
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C
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Difficulty:

55% (hard)

Question Stats:

71% (02:44) correct 29% (02:50) wrong based on 317 sessions

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A boat can cover a certain distance of 180 km and come back in 12.5 hours. If the ratio of speed of boat in still water to speed of stream is 5:1, then find out the speed of boat in still water (in km/h)?

(A) 25 km/h
(B) 28 km/h
(C) 30 km/h
(D) 26 km/h
(E) 32 km/h
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C
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A boat can cover a certain distance of 180 km and come back in 12.5 ho 25 Feb 2020, 21:06
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Speed of stream x

180/(x+5x) + 180/(5x-x) = 12.5

12.5x=30+45
x=6
5x=30

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A boat can cover a certain distance of 180 km and come back in 12.5 ho 28 Feb 2020, 05:46
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Solution – Using Options



Given
In this question, we are given that
    • A boat can cover a certain distance of 180 km and come back in 12.5 hours
    • The ratio of speed of boat in still water to speed of stream is 5:1

To find
We need to determine
    • The speed of the boat in still water (in kmph)

Approach and Working out
As the ratio of speeds is given as 5: 1, we can start with the options which are multiples of 5.

If speed of the boat in still water is 25, then speed of stream is 5
    • 180/(25 + 5) + 180/(25 – 5) = 180/30 + 180/20 = 6 + 9 = 15, which is not satisfying as the given total time is 12.5 hours

If speed of the boat in still water is 30, then speed of stream is 6
    • 180/(30 + 6) + 180/(30 – 6) = 180/36 + 180/24 = 5 + 7.5 = 12.5, which is satisfying as the given total time

Thus, option C is the correct answer.

Correct Answer: Option C
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A boat can cover a certain distance of 180 km and come back in 12.5 ho 28 Feb 2020, 05:50
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Solution – Using Equation



Given
In this question, we are given that
    • A boat can cover a certain distance of 180 km and come back in 12.5 hours
    • The ratio of speed of boat in still water to speed of stream is 5:1

To find
We need to determine
    • The speed of the boat in still water (in kmph)

Approach and Working out
If we assume the speed of stream as V, then speed of boat in still water is 5V

Hence, we can write
    • 180/(5V + V) + 180/(5V – V) = 12.5
    Or, 180/6V + 180/4V = 12.5
    Simplifying, we get V = 6

Hence, speed of the boat in still water = 5V = 5 x 6 kmph = 30 kmph

Thus, option C is the correct answer.

Correct Answer: Option C
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A boat can cover a certain distance of 180 km and come back in 12.5 ho 28 Feb 2020, 09:02
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TheNightKing
A boat can cover a certain distance of 180 km and come back in 12.5 hours. If the ratio of speed of boat in still water to speed of stream is 5:1, then find out the speed of boat in still water (in km/h)?
(a) 25 km/h
(b) 28 km/h
(c) 30 km/h
(d) 26 km/h
(e) 32 km/h
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total speed ;
180/(6x)+180/(4x)= 12/5
x = 6 hrs
speed still water ; 5x; 5*6 ; 30 kmph
IMO C
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A boat can cover a certain distance of 180 km and come back in 12.5 ho 16 Aug 2025, 23:37
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1. speed inversely proportional to time when distance from both side is same.

6x/4x = 12.5-t/t ==> 3/2=12.5-t/t = t=5 hours so time taken for side is 5 hours so

6x*t=180
6x*5= 180 so x= 6
Therefore, speed of boat in still water = 5x = 30. hence, C

2. avg speed = 180+180/12.5 = 360/12.5= Total distance/Total time
Avg speed = 360/12.5 = 36*4/5
Avg speed is HM of the two speeds when distance is same.
Let downstream speed be 6x and upstream =4x

2*6x*4x/10x=36*4/5
x=6
so 5x= 30
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A boat can cover a certain distance of 180 km and come back in 12.5 ho 02 Sep 2025, 04:09
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Here's an easier approach:

boat : river => \(5x:1x\)
combined speed => \(6x\)

ratio of speeds of boat in upstream vs downstream => \(6:4\)
Therefore, the time would be in the ratio of => \(2:3\)

So time taken to travel downstream would be \((\frac{2}{5})*(12.5)\)

now simply plug in speed*time = distance for downstream

\((\frac{2}{5})(12.5)(6x) = 180\)
\(x = 6\)

speed of boat = \(5x = 30\)
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A boat can cover a certain distance of 180 km and come back in 12.5 ho 10 Sep 2025, 08:28
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Could you please elaborate on how you moved from 5x:1x to a ratio of 6:4?
ZIX
Here's an easier approach:

boat : river => \(5x:1x\)
combined speed => \(6x\)

ratio of speeds of boat in upstream vs downstream => \(6:4\)
Therefore, the time would be in the ratio of => \(2:3\)

So time taken to travel downstream would be \((\frac{2}{5})*(12.5)\)

now simply plug in speed*time = distance for downstream

\((\frac{2}{5})(12.5)(6x) = 180\)
\(x = 6\)

speed of boat = \(5x = 30\)
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A boat can cover a certain distance of 180 km and come back in 12.5 ho 10 Sep 2025, 23:06
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user34745792203
Could you please elaborate on how you moved from 5x:1x to a ratio of 6:4?

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Let me clarify the transition from \(5x:1x\) to \(6:4\)

Understanding Relative Speeds in Water:

When a boat travels in a river, the stream affects its effective speed differently depending on direction:

Given:
  • Boat speed in still water = \(5x\)
  • Stream speed = \(1x\)

Key Concept - How Stream Affects Boat Speed:
  1. Downstream (with the current): The stream helps the boat, so we ADD the speeds Effective speed downstream = \(5x + 1x = 6x\)
  2. Upstream (against the current): The stream opposes the boat, so we SUBTRACT the speeds Effective speed upstream = \(5x - 1x = 4x\)

Therefore: Ratio of downstream to upstream speeds = \(6:4\)

Visual Analogy: Think of it like walking on a moving walkway at an airport:
  • Walking with the walkway → your speed + walkway speed
  • Walking against the walkway → your speed - walkway speed

Since time is inversely proportional to speed, the time ratio becomes \(2:3\)

This concept of adding/subtracting stream speed is fundamental to all boat-and-stream problems on the GMAT.
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