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

- 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`

`-> "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

`"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'

`= "Normal speed of the person" + "speed of escalator"`

`= s + e`

`= s - e`

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

`= s - e`

`= s + e`

### Concept review questions

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:**

`-> 20/c = (5 + 4)/ (5 - 4)`

`-> c = 20/9 = 2.22 kmph`

Speed of current `= 5 kmph`

Speed of boat in still water `= 30 + 5 = 35 kmph`

Downstream speed `= 35 + 5 = 40 kmph`

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"`

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

`=("Difference between speeds"/"Speed without stoppage") xx 60 min`

### Concept review questions

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`

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

### Concept review questions

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`

Speed of train `= 1188[(8 - 36/5)/(36/5)] = 132 km`

Interval of shots `= 8 min`

Distance covered `= 132 xx (8/60) = 17.6km`