ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: The speed of A and B are in the ratio 3 : 4. A takes 15 minutes more than B to reach a destination. Time in which A reach the destination?

Solution: 
Ratio of speed = 3 : 4
Ratio of time taken = 4 : 3 

Let time taken by A and B be 4x and 3x hour respectively.

Now 
4x - 3x = 15/60
x = 1/4

Hence, time taken by A = 4x hours
= 4 × 1/4
 = 1 hour

Question: A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is:

Solution: 
Let
the speed of the stream be x km/hr. Then,
Speed downstream = (15 + x) km/hr,
Speed upstream = (15 - x) km/hr.

Now,
{30/(15 + x)} + {30/(15 - x)} = 9/2
⇒ 30(15 - x) + 30(15 + x)/(15 + x)(15 - x) = 9/2
⇒ (450 - 30x + 450 + 30x)/(225 - x²) = 9/2
⇒ 900/(225 - x²) = 9/2
⇒ 100/(225 - x²) = 1/2
⇒ 200 = 225 - x²
⇒ x² = 225 - 200
⇒ x² = 25 
⇒ x² = 5² 
⇒ x = 5

the speed of the stream be 5 km/hr.

Question: A man covers half of his journey at 6 km/hr and the remaining half at 12 km/hr. His average speed is-

Solution: 
Let the total distance be 2x km. 

Then,
Time taken = (x/6​) + (x/12)
= (x + 2x)/12
= 3x/12
= x/4 ​hours

∴ Average speed= Total distance/Time taken
= 2x/(x/4) ​km/hr
=2x × (4/x) km/hr.
= 8 km/hr

Question: The speed of a boat in still water in 17 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 12 minutes is:

Solution: 
Speed downstream = (17 + 3) kmph = 20 kmph.
Distance travelled = 20 × (12/60) km = 4 km 

Question: Running at 5/4 of his usual speed, an athlete improves his timing by 10 minutes. The time he usually takes to run the same distance is:

Solution:
When the athlete walks at 5/4 of his usual speed then he takes 4/5 of his usual time and he saves 10 minutes.

Now
→ Usual time - (4/5) × usual time = 10 minutes
→ Usual time {1 - (4/5)} = 10 minutes
→ (1/5) × Usual time = 10 minutes
→ Usual time = 50 minutes.

Question: A man can row at 15 kmph in still water. If the velocity of current is 3 kmph and it takes him 1 hour to row to a place and come back, how far is the place?

Solution: 
Speed downstream = (15 + 3) kmph = 18 kmph
Speed upstream = (15 - 3) kmph = 12 kmph
Let the required distance be x km
Then,
(x/18) + (x/12) = 1
(2x + 3x)/36 = 1
5x/36 = 1
5x = 36
x = 36/5
x = 7.2 km

Question: If A travels to his school from his house at the speed of 3 km/h, then he reaches the school 7 minutes late. If he travels at the speed of 4 km/h, he reaches the school 8 minutes earlier than school time. The distance of his school from his house is:

Solution: 
Let the distance between school and home be x km.
The difference of time when A goes school to school with these two different speeds is = 8 - (- 7) = 15 min
=15/60 hour

Now
(x/3) - (x/4) = 15/60
(4x - 3x)/12 = 1/4
x/12 = 1/4
x = 12/4
x = 3

Question: The speed of a boat when travelling downstream is 42 km/hr, whereas when travelling upstream it is 38 km/hr, what is the speed of the stream?

Solution: 
Here, The speed of a boat in downstream = 42 km/hr.
The speed of a boat in upstream = 38 km/hr,
∴ The speed of stream = (Speed downstream - Speed upstream) / 2
= (42 - 38)km/hr / 2
= (4/2) km/hr
= 2 km/hr

Question: A train passes two bridges of lengths 700 m and 300 m in 100 seconds and 60 seconds respectively. The length of the train is

Solution: 
Let the length of the train be x m 
Speed s = (x + 700)/100.................(1)
Speed s = (x + 300)/60................(2)

Now 
(x + 700)/100 = (x + 300)/60
(x + 700)/5 = (x + 300)/3
5x + 1500 = 3x + 2100
5x - 3x = 2100 - 1500
2x = 600
x = 300

Question: A man can row 5 km/h in still water. If the speed of the current is 1 km/hr, it takes 3 h more in upstream than in the downstream for the same distance. The distance is-

Solution: 
Let the distance be = d
Speed of boat in upstream = 5 - 1= 4 km/h
Speed of the boat in downstream=5 + 1= 6 km/h
According to the question,

Now
d​/4 − d/6 ​= 3
(3d - 2d)/12 ​=3
d/12 = 3
d = 36 km 

Question: In a 120m race, Kamal defeats Jamal by 6 seconds. If the speed of Kamal is 18 kmph, then the speed of Jamal is:

Solution: 
Time taken by Kamal = 120/{18 × (5/18)}
= 24 seconds

Time taken by Jamal = 24 + 6 = 30 second

Jamal's speed=120/30
= 4 m/s
= (4 × 18)/5 kmph
=14.4 kmph

Question: A man can row 9 km/hr in still water and he finds that it takes him twice as long to row upstream as to row downstream the river. Find the rate of the stream.

Solution: 
Let, the speed of the current be x km/hr
Thus upstream speed = (9 - x) km/h and
downstream speed = (9 + x) km/hr

Let distance traveled be y

Then,
y/(9 - x) = 2y/(9 + x)
1/(9 - x) = 2/(9 + x)
9 + x = 18 - 2x
x + 2x = 18 - 9
3x = 9
x = 3 

the speed of the current be 3 km/hr.

Question: A cyclist moving on a circular track of radius 105 meters completes one revolution in 3 minutes. What is the average speed of cyclist (approx.)?

Solution: 
Distance covered in one complete revolution
= 2πr
= (2 × 105 × 22)/7
= 660 m

Average speed = 660/3 = 220 m/min

Question: An aeroplane covers a certain distance at a speed of 260 kmph in 5 hours. To cover the same distance in 5/3 hours, it must travel at a speed of:

Solution: 
Distance = (260 × 5) = 1300 km.
Speed = Distance/Time


Speed = 1300/(5/3) km/hr.
           = 1300 × (3/5) km/hr.
           = 780 km/hr.

Question: Thrice the speed of a boat downstream is equal to four times the speed upstream. The ratio of its speed in still water to its speed in current is-

Solution: 
Let,
Speed in still water = x km/hr.
Speed of current = y km/hr.
Speed downstream = (x + y) km/hr.
Speed upstream = (x - y) km/hr.
so, 
3 (x + y) = 4 (x - y)
⇒ 3x + 3y = 4x - 4y
⇒ 4x - 3x = 3y + 4y
⇒ x = 7y
⇒ x/y = 7/1
∴ x : y = 7 : 1

Question: A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:

Solution: 
Let speed of the car be x kmph.
Then, speed of the train = 150x/100 = 3x/2

Now
(75/x) - {75/(3x/2)} = 125/(10 × 60)
(75/x) - (50/x) = 5/24
(75 - 50)/x = 5/24
25/x = 5/24
5x = 24 × 25
x = ( 24 × 25)/5
x = 120 kmph

Question: A train 165 metres long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?

Solution: 
Speed of train relative to man =(60 + 6)km/hr = 66km/hr
=66 × 5/18m/sec
=55/3 m/sec

∴Time taken to pass the man =165 × 3/55 sec = 9 sec

Question: An individual is cycling at a speed of 37.5 km per hour. He catches his predecessor who had started earlier in two hours. What is the speed of his predecessor who had started 3 hours earlier?

Question: 
The distance covered in two hour,
= 2 × 37.5= 75 km
Time taken by first individual = (3h + 2h) = 5h

Then, the speed of predecessor,
= 75/5
= 15 kmph.

Question: Two ladies simultaneously leave cities A and B connected by a straight road and travel towards each other. The first lady travels 2 km/hr faster than the second lady and reaches B one hour before the second lady reaches A. The two cities A and B are 24 km apart. The speed of second lady?

Solution: 
Let the speed of the second lady be x km/hr.
Then, speed of first lady = (x + 2) km/hr.

Now 
(24/x) - 24/(x + 2) = 1
{24(x + 2) - 24x} / {x(x + 2)} = 1
(24x + 48 - 24x) / {x(x + 2)} = 1
48/(x² + 2x) = 1
x² + 2x = 48
x² + 2x - 48 = 0
x² + 8x - 6x - 48 = 0
x(x + 8)  - 6(x + 8) = 0
(x + 8)(x - 6) = 0
 x = 6

Speed of second lady = 6 km/hr.

Question: A certain distance is covered at a certain speed. If half the distance is covered in double the time, the ratio of the two speeds is-

Solution: 
Let x km be covered in y hours.
Then, speed = x​/y km/hr

In the second case, x/2​ km is covered in 2y hours.
∴ New speed = (x/2)/2y km/hr
= x/4y

∴ Ratio of speeds = x/y : x/4y
= 1 : 1/4
= 4 : 1 

Question: The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:

Solution: 
Speed = (45 × 5)/18 m/sec = 25/2 m/sec
Time = 30sec
Let the length of bridge be x metres

Now
(130 + x)/30 = 25/2
260 + 2x = 750
2x = 750 - 260
2x = 490
x = 490/2
x = 245 

Question: A motor boat takes 2 hours to travel a distance of 9 km downstream and it takes 6 hours to travel the same distance against the current. The speed of the boat in still water-

Solution: 
Downstream speed,
= 9/2 kmph
= 4.5 kmph

Upstream speed,
= 9/6 kmph
= 1.5 kmph


Thus, speed of boat in still water,
= (4.5 + 1.5)/2
   = 3 km/h

Question: A man can row three-quarters of a kilometer against the stream in 45/4 minutes and down the stream in 15/2 minutes. The speed (in km/hr) of the man in still water is:

Solution: 
three - quarters of a kilometer = 750 meters
45/4 minutes = (45/4) × 60 = 675 sec
15/2 minutes = (15/2) × 60 = 450sec


Rate upstream = 750/ 675 m/sec
                         = 10/9 
Rate downstream =750/450 m/sec
 =5/3 m/sec


∴Rate in still water= (1/2){(10/9) + (5/3)} m/sec
= (1/2){(10 + 15)/9}
= (1/2)(25/9)
= 25/18
= (25/18)(18/5) km/hr.
= 5 km/hr

Question: A train covers a distance between two stations A and B in 45 minutes. If the speed of the train is reduced by 5 km/hr, then it covers the distance in 48 minutes. The distance between the stations A and B is :

Solution: 
Let
the distance between the stations A and B be x km
Time taken = 45 min = 45/60 = 3/4 hour
 
∴ Original speed =(x × 4/3)km/hr = 4x/3 km/hr

New speed = {(4x/3) - 5} km/hr.
= (4x - 15)/3 km/hr.

Now 
x/{(4x - 15/3} =48/60
3x/(4x - 15) = 4/5
16x - 60 = 15x
16x - 15x = 60
x = 60 km