ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Speed upstream = (2/20)×60 km/hr
= 6 km/hr
Speed downstream = (2/15)×60 km /hr
= 8km/hr
Speed of the current
= 1/2(8-6) km /hr
= 1 km/hr

Let the speed in still water be x km/hr.
Then,
Speed downstream = (x+ 4) km/hr,
speed upstream = (x-4) km/hr.
6/(x+4) + 6 /(x-4) = 2
=>  1/(x+4) +1/(x-4)=2/6 = 1/3
=> (x+4)+(x-4)/x2-16= 1/3
=> x²-16= 6x
=>  x² -6x-16= 0
=> (x-8) (x+2) = 0
=>  x = 8.
∴ Speed of boat in still water = 8 km/hr.

Speed in still water =6 km/hr.
Speed against the current =6/3 km/hr = 2 km/hr
Let the speed of the current be x km/hr
so, 6-x = 2
=> x = 4 km/hr.

Let the distance be D km.
∴ Downstream Speed = D/4 km/hr
And Upstream Speed = D/5 km/hr
Given, Speed of current = 2 km/hr
Speed of the current = 1/2 ×(Downstream Speed - Upstream Speed)
2 = 1/2 ×(D/4 - D/5)
D = 80 km

Speed of boat in still water = 25 km/hr
Speed upstream
= 10/1
= 10 km/hr
Speed of the stream = (25-10) = 15 km/hr
Speed downstream = (25+15) = 40 km/hr
Time taken to travel 10 km downstream
= 10/40 hours
= (10×60)/40
= 15 minutes

Let the distance be x
Speed upstream = (40-10) = 30 kmph
Speed downstream = (40+10) = 50 kmph
Total time taken = 1 hr
⇒ x/50 + x/30 = 1
⇒ 8x/150 = 1
⇒ x = 150/8 = 18.75 km

Let the speed of the water in still water = x
Given that speed of the stream = 3 kmph
Speed downstream
= (x+3) kmph
Speed upstream
= (x−3) kmph
He travels a certain distance downstream in 4 hour and come back in 6 hour.
ie, distance travelled downstream in 4 hour = distance travelled upstream in 6 hour
since distance = speed × time,
we have
(x+3)4= (x−3)6
⇒ (x+3)2= (x−3)3
⇒ 2x+6= 3x−9
⇒ x= 6+9= 15 kmph

Let speed upstream = x
Then, speed downstream = 2x
Speed in still water
= (2x+x)/2
= 3x/2
Speed of the stream
= (2x−x)/2
= x/2
Speed in still water : Speed of the stream
= 3x/2 : x/2
= 3:1

Speed of the boat in still water = 22 km/hr
speed of the stream = 5 km/hr
Speed downstream = (22+5) = 27 km/hr
Distance travelled downstream = 54 km
Time taken = distance/ speed
= 54/27
= 2 hours

Let speed of the boat in still water = a and speed of the stream = b
Then
a+b = 14
a-b = 8
Adding these two equations, we get 2a = 22
=> a = 11
ie, speed of boat in still water = 11 km/hr

Man's speed with the current = 15 km/hr
=> speed of the man + speed of the current = 15 km/hr
Speed of the current is 2.5 km/hr
Hence, speed of the man
= 15-2.5
= 12.5 km/hr
Man's speed against the current = speed of the man - speed of the current
= 12.5-2.5
= 10 km/hr

Hint: If a boat moves to a certain distance downstream in 't1 ' hours & returns the same distance upstream in time 't2' hours, then
Speed of boat in still water = y{(t2+t1)/(t2–t1)} km/hr.
With the given parameters,
y = 6 km/hr, t1 = 3 hrs, t2 = 2 hrs
We can find, Speed of boat in still water
(x) = 6{(3+2)/(3–2)}= 30 km/hr

Let the length of the train be x metres.
Then, length of the platform = 2x metres.
Speed of the train = 90× (5/18) m/sec
= 25m/sec
∴(x+2x)/25 = 36
⇒ 3x= 900
⇒ x= 300
Hence, length of platform
= 2x= (2×300)m= 600m

Relative speed = (60+40) km/hr
= 100×(5/18) m/sec
= 250/9 m/sec.
Distance covered in crossing each other
= (140+160)m= 300m
Required time
= 300×(9/250) sec
= 54/5 sec
= 10.8 sec

Let the length of the train be x meters and its speed be y m/sec.
Then, xy = 20
⇒ y = x/20
∴ (x+100)/30 = x/20
⇒ 30x = 20x + 2000
⇒ 10x = 2000
⇒ x = 200 meters

Let the speed of the slower train be x m/sec.
Then, speed of the faster train = 2x m/sec.
Relative speed = (x+2x) m/sec = 3x m/sec.
So,(100 + 100)/8 = 3x
⇒ 24x = 200
⇒ x = 25/3
So, speed of the faster train =50/3 m/sec
= (50/3) x (18/5) km/hr
= 60 km/hr.

Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x-1) hours = 25(x-1) km.
20x+25(x-1) = 110
45x = 135
So, x = 3.
So, they meet at 10 a.m.

Let the speeds of the two trains be x m/sec and y m/sec respectively.
Then, length of the first train = 27x metres
and length of the second train = 17y metres
∴ 27x+17y/(x+y) = 23
⇒ 27x+17y = 23x+23y
⇒ 4x=6y
⇒ x/y= 3/2

Speed of the train relative to man
= 125/10 m/sec
= 25/2 m/sec
= (25/2)×(18/5) km/hr
= 45km/hr
Let the speed of the train be x km/hr. Then, relative speed=(x−5)km/hr
∴x−5= 45
⇒x = 50km/hr

Given: Speed of passenger train = 55 km/hr, length of goods train (P) = 250, length of passenger train (Q)= 200m
Hint:
Time = (P+Q)/(V1+V2) sec
Goods train and the passenger train move in opposite direction. Hence, the relative speed is the addition of two speeds.
Convert 55 km/hr into m/s
55 x (5/18) = 15.277 m/s
Therefore,
10 = (250+200)/(15.27+V2)
V2 = 29.73 m/s 

Given: Speed of the person = 5 km/hr, length of train = 100 m, speed of train = 60 km/hr
Speed of train relative to walking person = (60–5) = 55 km/hr
Convert km/hr into m/s
55 km/hr = 55 x(5/18) = 15.27 m/s
Distance to be covered by the train = 200 + 100 = 300 m
Therefore, time taken by the train to cross the person
= Distance over speed =300/15.27 = 19.64 sec

We know,
Speed =Distance/ Time
Speed =(10/15) 60 = 40×(5/18)m/sec
= 11.1 m/sec
Length of train = (Speed x Time)
= (11.11x10)
= 111.1 m