ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Man's/Boat's Speed = X
Stream/Current/River Speed = Y

∴ Downstream speed = X + Y
Upstream speed = X - Y

X + Y = 18 km/hr and X - Y = 10 km/hr
Adding them we get,
X + Y + X - Y = 28 km/hr

∴ X = 14 km/hr = Speed of Motorist

Y = 18 - 14 = 4 km/hr = Speed of stream

Let speed upstream be x km/hr
Then, speed downstream = 3x km/hr.

Speed in still water = 1/2 (3x + x) kmph = 2x km/hr.

∴ 2x = 28/3
x = 14/3 km/hr;
Speed downstream = 14 km/hr

Hence, speed of the current
1/2 {14 - (14/3)} km/hr
= 14/3 km/hr
= 4(2/3) km/hr.

Let the speed of the stream be x km/hr
Upstream Speed = 15 + x
Downstream Speed = 15 - x

So,
{30/(15+x)} + {30/(15-x)} = 4(1/2)(4 hours 30 minutes)
⇒ {900/(225-x2)} = 9/2
⇒ 9x² = 225
⇒ x² = 25
⇒ x = 5.
Hence, the speed of the stream is 5 km/hr.

Suppose he moves 4 km downstream in x hours.
Then, Speed downstream = (4/x) km/hr
Speed upstream = (3/x) km/hr

∴ 48/(4/x) + 48/(3/x) = 14
⇒ 48x/4 + 48x/3 = 14
⇒ 48{(x/4) + (x/3)} = 14
⇒ 7x/12 = 14/48
⇒ 7x/12 = 7/24
⇒ 7x = (7/24) × 12
⇒ 7x = 7/2
⇒ 14x = 7
⇒ x = 7/14
⇒ x = 1/2

So, speed downstream = 7 km/hr
Rate of the stream = 1/2(8 - 6) km/hr = 1 km/hr.

Man's/Boat's Speed = X
Stream/Current/River Speed = Y

∴ Downstream speed = X + Y
Upstream speed = X - Y

∴ X+Y = 4.5+1.5 = 6 km/hr and X-Y = 4.5-1.5 = 3 km/hr
Let distance be D km

Downstream time = Distance/Speed = D/6
Upstream time = D/3

Average speed = Total distance/Time taken = (D + D)/(D/6 + D/3)
= (6 × 2D)/3D
= 4 km/hr.

Downstream speed = 55/(5/2) = 11 × 2
= 22 km/hours

Time taken in upstream = 2.2 × 5/2
= 5.5 hours

Upstream speed = 55/5.5
= 10 km/hour

∴ The speed of boat in still water

= (10 + 22)/2
= 32/2
= 16 km/hr.

Let PQ = Qr = x km
Let speed downstream = a km/hr.
and speed upstream = b km/hr.

Then,
x/a + x/b = 10
x = 10ab/(a + b) .........(i)

And,
2x/a = 4
x = 4a/2
x = 2a .............(ii)

From (i) and (ii) we have:
2a = 10ab/(a + b)
5b = a + b
a = 4b

Required ratio = Speed in the water/Speed of river
= {1/2(a + b)}/{(1/2) (a - b)}
= (a + b)/(a - b)
= (4b + b)/(4b - b)
= 5b/3b
= 5/3

Man's/Boat's Speed = X
Stream/Current/River Speed = Y

∴ Downstream speed = X + Y
Upstream speed = X - Y

X+Y = (45+Y) km/hr
1 hour 20 munites = 1 hour + 20/60 = (1 + 1/3) = (4/3) hours
Downstream speed = Distance covered/Time taken

∴ 45 + y = 80/(4/3)
∴ Y = 15 km/hr
X - Y = 45 - 15 = 30 km/hr

Time is taken to go against the stream = 80/30 hours = 2 Hours 40 minutes.

Given that,
time is taken to travel upstream = 2 × times taken to travel downstream
When the distance is constant, speed is inversely proportional to the time
Hence, 2 × speed upstream = speed downstream

Let speed upstream = x
Then speed downstream = 2x

we have,
1/2(x + 2x) = speed in still water
⇒ 1/2(3x)=7.5
⇒ 3x = 15
⇒ x = 5
∴ speed upstream = 5 km/hr

∴ Rate of stream = 1/2(2x - x)
= x/2
= 5/2
= 2.5 km/hr.

Let,
the speed of the stream be = u kmph
the speed of the boat be = v kmph
speed upstream will be = v - u kmph
speed downstream will be = v + u kmph

30 km upstream in time duration = 30 / (v - u) hrs
44 km downstream in time duration = 44 / (v + u) hrs
44/(v + u) + 30/(v - u)
= 10 hrs .........(i)

Similarly,
40 /(v - u) + 55 / (v + u) = 13 hrs Multiply with 3/4:
30/(v - u) + 165 / 4(v + u) = 39/4 .......(ii)

Now (i) - (ii)
⇒ [44 - 165/4] / (v + u) = 10 - 39/4 = 1/4
⇒ v + u = 11 ..........(iii)
Substitute this in (1) to get: 44/11 + 30/(v - u) = 10
⇒ v - u = 30/6 = 5.........(iv)

Solving (iii) and (iv) ,
we get :
v = 8 kmph and
u = 3 kmph

Let the speed of the current be x kmph
Then speed of the boat in still water = 5x

∴ Downstream speed
= (5x + x) = 6x km/hr.
Upstream speed
= (5x - x)
= 4x km/hr

Now,
According to the question,
1.1 × 6x = 13.2 km
6.6x = 13.2
x = 13.2/6.6
x = 2 km/hr.

Upstream speed
= 4x = 4 × 2 = 8 km/hr

∴ 312 minutes
= 312/60 hours
= 5(1/5) hours

∴ Required Distance travelled upstream
= Speed × Time
= 8 × (26/5)
= 41.6 km

Speed of the boat in still water = 12 km/hr.
Speed of the stream = 4 km/hr.
Speed downstream = (12 + 4)
= 16 km/hr.
Time is taken to travel 68 km downstream
= 68/16
=17/4
= 4.25 hrs.

Man's/Boat's Speed = X
Stream/Current/River Speed = Y

∴ Downstream speed = X + Y
Upstream speed = X - Y

Downstream Speed = Distance covered/Time taken
= 20/1
= 20 km/hr
Upstream Speed = 20/2 = 10 km/hr

X + Y = 20 km/hr
and, X - Y = 10 km/hr

Adding them we get,
X + Y + X - Y = 30 km/hr
∴ X=15 km/hr = Speed of swimmer in still water

∴ Y = 20 - 15 = 5 km/hr = Speed of river.

Let the speed of the first train be x km/hr.
Then,
Sum of lengths of trains = (66 + 88)m = 154 m.

Relative speed of two trains = (x - 30) km/hr
= {(x - 30) × (5/18)} m/s

∴ 154/(x - 30) × (5/18) = 0.168 × 60
⇒ 5(x - 30) = (154 × 18)/10.08
⇒ 5(x - 30) = 275
⇒ x - 30 = 55
⇒ x = 85 km/hr.

Speed of the train relative to Person
= (125/25) m/s.
= 5 m/s.

∴ 5 × (18/5) km/hr
= 18 km/hr

Let the speed of the train be x km/hr.
then, relative speed = (x - 8) km/hr.

So, (x - 8) = 18
⇒ x = 26 km/hr.

Let the speed one train be x and the speed of the second train be y
Length of the first train = Speed × Time = 28x
Length of second train = Speed × Time = 18y

So, {(28x + 18y)/(x + y)} = 26
⇒ 28x + 18y = 26x + 26y
⇒ 2x = 8y

Therefore,
x : y = 4 : 1.

Let the length of the train be x metres and its speed be y m/sec.
Then,
x/y = 1.75
⇒ x = 1.75y

Since the train takes less time to pass a moving object than a stationary object,
it means that the cyclist is moving in a direction opposite to that of the train.

∴ x/(y + 10) = 1.5
⇒ x = 1.5 y + 15
⇒ 1.75 y = 1.5 y + 15
⇒ 0.25 y = 15
⇒ y = 15/0.25
⇒ y = 60.

Length of the train = 1.75 y = (1.75 × 60) m
= 105 m.

We know,
Distance(D) = Speed(S) × Time(T)
⇒ D = S × T
∴ S = D/T; T = D/S

Since the second train is 30 km/hr faster, it is moving at 120 km/hr
Now, let train A travel T hours before meeting train B.
Time for which train B travels = (T - 1/2) hrs.
This is because it starts half an hour late
Distance travelled is same.

∴ D = D
∴ 90 km/hr × T hrs = 120 km/hr × {T - (1/2} hrs
∴ 90T = 120T - 60
∴ T = 2 hours

Distance travelled by train A = 90 km/hr × 2 hours = 180km.
Thus they meet 180 km from starting point.

Let the length of the tunnel = x metre
Then, distance = (800 + x) metre.
Time = 1 minute = 60 seconds.

∴ Speed = 78 km/hr
= 78 × (5/18)
= (65/3) m/s

According to the question,
800 + x = {60 × (65/3)}
⇒ 800 + x = 1300
⇒ x = 500 metre.
Hence, The length of the tunnel is 500 metre.

Let the speeds of the faster and slower trains be x m/sec and y m/sec respectively.
Then,
240/(x - y) = 60
⇒ x - y = 4 ........(i)

And, 240/(x + y) = 3
⇒ x + y = 80 ........(ii)

Adding (i) and (ii), we get
2x = 84
⇒ x = 42
Putting x = 42 in (i), we get:
42 - y = 4
⇒ y = 38

Hence, speed of the faster train = 42 m/sec.

Let the speed of slower train = S km/hr
Speed of faster = (S + 14) km/hr
Trains meet after 12 hours.

Distance travelled by slower train in 12 hrs. = 12S
Distance travelled by faster train in 12 hrs. = 12(S + 14)

The total distance to be travelled between the two stations is given.
So, 12S + 12(S + 14) = 240
2S + 14 = 20
S = 3 km/hr.

Hence, The speed of the slower train is 3 km/hr.

Let the length of the train is x m. and its speed is v. m/s.

Distance = Speed × time [S = V × T]
x = v × 9 .........(i).
(x+700) = v × 30 ........(ii).

Dividing the eqn. (i) by (ii).

x/(x+700)= 3/10.
⇒ 10x=3x + 2100.
⇒ 7x=2100.
⇒ x= 2100/7.
⇒ x= 300. m.

putting x = 300 in eqn. (1).
300 = v × 9
⇒ v = 300/9
⇒ v = 100/3 m/s.

Let the train crosses a 800 m. long platform in t seconds.
(x + 800) = v × t .........(iii) [ S = V × T]
⇒ (300 + 800) = (100/3) × t. [putting x= 300. and v= 100/3.]
⇒ t = (1100×3)/100
⇒ t = 33 seconds.

Relative speed
= (80 + 55) km/hr
= 135 km/hr.
= 135 × (5/18) m/sec
= (75/2) m/sec.

Distance covered = (120 + 90 + 90)
= 300 m.

∴ Required time = {300 × (2/75)}
= 8 sec.

Speed of the first train = 48 km/hr.
Let the length of the first train = 2x metre.

Speed of the second train = 42 km/hr.
Let the length of the second train = x metre.

Distance = (2x + x) = 3x metre.
Time = 12 seconds

Relative speed = 48 + 42 = 90 km/hr.
= 90 × (5/18) = 25 m/s.
3x = 25 × 12
⇒ x = 100 metre.

Length of the first train = 200
Time is taken to cross the platform = 45 seconds.
Speed of first train = 48 km/hr
= 48 × (5/18)
= 40/3 m/s.

Let the length of the platform = y metre.
Distance = 200 + y metre.
⇒ 200 + y = 45 × (40/3)
⇒ 200 + y = 600
⇒ y = 400 metre.

The difference in speed due to stopping = Speed without stoppage - Speed with stoppage
∴ Difference = 40-35 = 5km/hr
Thus, in 1-hour train covers 5 km less.

Time taken to cover 4km = 5km/(40 km/hr)
= 1/8 hours.
= (1/8 × 60) minutes.
= 7.5 minutes.

Hence, The train halts 7.5 minutes per hour.

Speed of the first train
= 120/10 m/sec.
= 12 m/sec.

Speed of the second train
= 120/15 m/sec.
= 8 m/sec.

Relative speed = 12 + 8 = 20 m/sec.

∴ Required time
= (120 + 120)/20 sec.
= 240/20 sec.
= 12 sec.

Let the length of the train be x metres and its speed be y m/s.
Then,
x/(y - a) = b and x/{y - (a + 1) = (b + 1)
⇒ x = b (y - a) and x = (b + 1)(y - a - 1)
⇒ b (y - a) = (b + 1) (y - a - 1)
⇒ by - ba = by - ba - b + y - a - 1
⇒ y = (a + b + 1).