# Time and Distance Tutorial III: Boats & Streams, Escalators, Journey with Stoppages

This is a continuation of article on tricks to solve questions related to time and distance in quantitative aptitude examinations. In this part of tutorial we’ll learn about some of the important concepts for solving problems related to boats and streams, escalators and journey with stoppage.

## Objectives

• Boats and streams
• Escalators
• Journey with stoppages
• Speed of moving bodies as per the speed of sound
Aptitude questions related to boats and streams/escalator tests your skills in following areas;
• Understanding and visualizing the situation.
• Effective practical application of the concept of relative speed.

## Important Terminologies

Speed of boat in still water (b): Speed of the boat in the still water is the real speed of the boat without any additional supportive or resistive forces.

Speed of current/ stream / wave (c): This is the speed of water flow.

Downstream speed (v): If the boat is moving along with the direction of the flow then the effective speed of the boat is called downstream speed.
Downstream speed = sped of the boat in still water + speed of the stream; v = b + c

Upstream speed (u): If the boat is moving against the direction of the stream, then the effective speed of boat is called the upstream speed.
Upstream speed = speed of boat in still water - speed of the stream; u = b - c

Shortcuts and formula
(u + v)/2 = ((b - c) + (b + c))/2 = b
-> "Speed of boat in still water" = ("upstream speed" + "downstream speed")/2

(u - v)/2 = ((b - c) - (b + c))/2 = c
-> "Speed of stream" = ("downstream speed" - "upstream speed")/2

### A very useful illustration of the relationship between basic parameters

As per the above explanations, it is easy to observe that the Upstream Speed (u), Speed of the boat in still water (b) and the Downstream speed (v) are the three consecutive terms in an Arithmetic Progression in the mentioned increasing order, with a common difference which is the Speed of the Stream (c). image

"Upstream Speed " + " Speed of Stream " = " Speed of Boat in Still Water"
"Speed of Boat in Still Water " + " Speed of Stream " = " Downstream Speed"

## Escalators

This is most repeating type of questions in quant area of CAT, MAT and CMAT exams. Concept of escalators is same as the concept of downstream and upstream.

Let the speed of a person when he is walking with the escalator (both moving up or down)  = s and speed of the moving escalator = e

### Case I: An escalator is moving up at a speed of 'e'

Effective speed of a person who is moving upon escalator and walking with the escalator
 = "Normal speed of the person" + "speed of escalator"
= "Normal speed of the person" + "speed of escalator"
= s + e
If he is moving down against the movement of the escalator
"Then his effective speed"= "Normal speed of the person" - "speed of escalator"
= s - e

### Case II: An escalator is moving down at a speed of 'e'.

If the person is walking up against the direction of escalator, then his effective speed
= "Normal speed of the person" - "speed of the escalator"
= s - e
If he is walking down on escalator with the downward moving escalator, then his effective speed
 = "Normal speed of the person" + "speed of escalator"
= s + e

### Concept review questions

Speed of a boat in still water is 20 kmph. If it's upstream and downstream speeds are in the ratio of 4:5. Find the speed of the stream?
Speed of boat in still water = 20 kmph
Let the speed of stream = c kmph
Upstream speed = (20 - c) kmph
Downstream speed = (20 + c) kmph
"Upstream speed"/"Downstream speed" = (20 -c)/(20 + c) = 4/5
On solving this, we'll get c = 20/9 = 2.22 kmph

Alternate Method:
"Speed of boat in still water"/"Speed of stream" = (u + v)/(u - v)
-> 20/c = (5 + 4)/ (5 - 4)
-> c = 20/9 = 2.22 kmph
A boat goes 24 Km upstream in 48 minutes. If the speed of the stream is 5 kmph then find the downstream speed of the boat?
Upstream speed = 24 xx 60/48 = 30kmph
Speed of current = 5 kmph
Speed of boat in still water = 30 + 5 = 35 kmph
Downstream speed = 35 + 5 = 40 kmph
An escalator of length 32 meters is moving upward. When Abhijith is walking on the escalator at a constant speed, he covered escalator from bottom to top in 10 seconds. And he walks down against the direction of the escalator with the same speed, travelled from top to bottom of the escalator in 16 seconds. Find the speed of Abijith?
Length of escalator = 32 m
Let the speed of Abijith = s "m/s"
And the speed of escalator = e "m/s"
Time taken by Abijith for moving upward = 10 sec
s + e = 32/10 = 3.2 "m/s" -> (1)

Note that the speed of Abijith is more than that of escalator, and then only he can move against the escalator and reach down.

Time taken by Abijith for moving downward = 16 sec
s - e = 32/16 = 2 "m/s" -> (2)
From equation (1) and (2), s = 2.6 "m/s"
Hence speed of Abhjit is 2.6 "m/s"
John can row up from a point A to point B in a river at a speed of 8 kmph. Once he reached B, immediately start his return trip from B to A. Find his average speed of the entire trip if the speed of the stream is 2 kmph.
Let the speed of John in still water = s kmph
Upstream speed = s - 2 kmph = 8 kmph
-> s = 10 kmph
Downstream speed = s + 2 = 12 kmph
Average speed = (2 xx 8 xx 12)/(8 + 12) = 9.6kmph

## Journey with stoppage

Time taken per hour for stoppage
If the speed of a car with a certain minutes of stoppages is q " kmph" and without stoppage the speed is p " kmph", then time taken per hour for stoppage = ((p - q)/p ) xx 60 min
=("Difference between speeds"/"Speed without stoppage") xx 60 min

### Concept review questions

A car travels at an average speed of 50 kmph without any stoppages and with stoppage the speed of the car reduced by 20 kmph. Find the time of the rest per hour?
Speed without stoppage = 50 kmph
50 km covered without stoppage in 60 minutes. Speed with stoppage = 30 kmph
-> 30 km covered with stoppage in 60 minutes. 50 km covered in effective time of 60 minutes.
:. 30 km is covered in (60/50 ) xx 30 = 36min, that is with stoppage the car is travelled in an effective time of 36 minutes per hour.
Hence the time for rest = 60 min - 36 min = 24 min

Alternate Method I:
Speed without stoppage = 50 kmph
Speed with stoppage = 30 kmph
Time spent for stoppage = ((50 - 30)/50) xx 60 = 24 min

Alternate Method II:
Speed without stoppage = 50 kmph
Speed with stoppage = 30 kmph
Decrease in speed = ((50 - 30)/50) xx 100 = 40%
Therefore stoppage = 40% of time
= 40% of 60 mins
= 24 min

Bhanu travels from A to B at a constant speed and B to A at the same speed, but in the return journey he took an average of 30 minutes per hour for rest. If his average speed for the entire trip is 40 kmph. Find his speed from A to B?
Speed from A to B = P kmph
Stoppage time = 30 min/hr = 1/2 of an hr
I.e. the variation happened in the return speed = 1/2 of the initial speed = 1/2 P kmph
Average Speed = (2 xx P xx 1/2P)/(P + 1/2P)
-> 2/3P = 40 kmph
P = 60kmph; ie: Speed from A to B = 60kmph

## Speed of moving bodies as per the speed of sound (Report of Guns- problem)

It is an old pattern of questions from time and distance. Initial period of GMAT exam GMAC (the counsel which conduct GMAT exam) started to ask it is in a 700+ range bin of questions, even though the concept is very simple.

### Concept:

Two guns were fired from same place at an interval of 'x' minutes. If a person who is approaching the shooting place by train/ car and hears the shooting sound at an interval of 'y' minutes, then it is possible to find the speed of the train/car.

Speed of the sound in air is 330 "m/s" in a fixed value.
Distance covered by the train in y minutes = Distance covered by the sound in (x-y) minutes
= 330 xx (x-y) xx 60 minutes
Speed of the train = (330 xx (x-y) xx 60)/y = (330 xx 60 xx (x - y))/1000 xx 60/y = 1188[(x-y)/y]  kmph

Speed of the train = 1188[(x-y)/y]  kmph

### Concept review questions

Two bullets were fired at interval of 14 minutes from the same shooting range. While in the journey towards the shooting range, Mr. Kennedy hears the sound of both the shots at an interval of 13 mins 30 secs. If the speed of the sound is 330 m/s then at what speed Mr.Kennedy was travelling?
Time difference = 14.00 - 13.30 = 30 s
In 30 seconds sound covers the same distance as Mr.Kennedy covers in 13 mins 30 sec 330xx30 = 9900 m = 9.9 km
Distance covered by Kennedy in 13.30 mins = 9.9 km
Speed of Mr.Kennedy = 90 xx (60/13.5) = 44km (13 min 30 sec = 13.5 min)
Alternate Method:
As per the above discussed rule, speed of Kennedy = 1188[(14 - 27/2)/13.5] = 44km
Two guns were fired from the same place at an interval of 8 minutes, but a person travelling in a train approaching the firing place hears the second shot 7 minutes 12 seconds after the first. Supposing the speed of sound travels at 330 m/s, find the distance covered by the train in the interval of two shots?
As per rule, X = 8 min and Y = 7:12 min = 36/5 min
Speed of train = 1188[(8 - 36/5)/(36/5)] = 132 km
Interval of shots = 8 min
Distance covered = 132 xx (8/60) = 17.6km

#### Popular Videos

How to improve your Interview, Salary Negotiation, Communication & Presentation Skills.

Got a tip or Question?
Let us know