ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: What is the length of a train that takes 12 seconds to pass a pole when it runs at a speed of 24 km/h?

Solution:
speed = 24 km/h
= (24 × 5/18) m/s
= 20/3 m/s

∴ The length of the train = (12 × 20/3) m = 80 m

Question: A boat sailing against a stream of river takes 6 hours to travel 36 km while sailing with the stream it takes 3 hours to travel the same distance. What is the speed of the stream?

Solution:
Let,
Speed of boat = x km/h and,
Speed of stream = y km/h

ATQ,
x - y = 36/6 = 6 ......... (i)
x + y = 36/3 = 12 ....... (ii)

From,
(ii) - (i)
x + y - x + y = 12 - 6
⇒ 2y = 6
∴ y = 3

So, Speed of the stream = 3 km/h

Question: An electric pump can fill a tank in 3 hours. Because of a leak in the tank it took  7/2 hours to fill the tank. If the tank is full, how much time will the leak take to empty it?

Solution:
pump fills 1/3 part in 1 hour

because of leak, pump fills 2/7 part in 1 hour

leak empty {(1/3) - (2/7)} part
= 1/21 part in 1 hour

∴ leak empty full tank in 21 hours

Question: A takes 3 hours more than B to walk d km, but if A doubles his speed, then he can make it in 1 hour less than B. How much time does A require for walking d km?

Solution:
Let,
B takes "t" hours to walk "d" km.

Then, A takes (t + 3) hours to walk d km.
With double speed, A will take = (1/2)(t + 3) hours.

ATQ,
t - (1/2)(t + 3) = 1
⇒ {2t - (t + 3)}/2 = 1
⇒ 2t - t - 3 = 2
∴ t = 5

∴ B takes 5 hours to walk d km.

∴ A takes (5 + 3) = 8 hours to walk d km.

Question: Four pipes can fill a tank in 15, 20, 30 and 60 hours respectively. The first pipe was opened at 8 a.m, second at 9 a.m, third at 10 a.m. and fourth at 11 a.m. When will the tank be full-

Solution:
Let,
the time be x hours after 8 am.
Then, the first pipe worked for x hours
Second pipe for (x - 1) hours;
Third pipe for (x - 2) hours;
Fourth pipe for (x - 3) hours.

ATQ,
(x/15) + {(x - 1)/20} + {(x - 2)/30} + {(x - 3)/60} = 1
⇒ (4x + 3x - 3 + 2x - 4 + x - 3)/60 = 1
⇒ 10x - 10 = 60
⇒ 10x = 70
∴ x = 7

So, the tank will be full 7 hours after 8 am = 8 + 7 = 15 = 3 p.m.

Question: A and B can do a piece of work in 45 and 40 days respectively. They began the work together but A leaves after some days and B finished the remaning work in 23 days. After how many days did A leave?

Solution:
Work done by B in 23 days {(1/40) × 23} = 23/40 part

Remaining work = (1 - 23/40) part
= 17/40 part

(A + B)'s 1 day's work = 1/45 + 1/40 = 17/360 part

Now,
17/360 part has done by (A + B) in 1 day
∴ 1 part has done by (A + B) in 360/17 day
∴ 17/40 part has done by (A + B) in {(360/17)(17/40)} days
= 9 days

Question: A long distance runner runs 8 laps of a 360 meters track everyday. His timings for four consecutive days are 78, 86, 79 and 77 minutes respectively. On an average, how many meters/minute does the runner cover?

Solution:
Total distance covered = (4 × 8 × 360) meters
= 11520 meters

Total time taken = (78 + 86 + 79 + 77) = 320 minutes

∴ Average speed = (Total distance ÷ Total time) m/min
= (11520 ÷ 320) m/min
= 36 m/min

Question: An outlet pipe can empty a cistern in 10 hours. In what time will it empty 3/5 part of the cistern?

Solution:
Given
outlet pipe can empty a cistern in 10 hours

∴ Time taken to empty 3/5 part of the cistern = (3/5 × 10) = 6 hours

Question: A man in a train notices that he can count 21 telephone posts in one minute. If they are known to be 25 meters apart, then at what speed is the train travelling?

Solution:
Number of gaps between 21 telephone posts = 20

∴ Distance travelled in 1 minute or 60 sec = (25 × 20) meter
= 500 m

∴ Speed of the train = (500 ÷ 60) m/sec
= 25/3 m/sec
= {(25/3) × (18/5)} km/h
= 30 km/h

Question: Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is-

Solution:
1^st pipe fill in 1 minutes = 1/20 part
2nd pipe fill in 1 minutes = 1/24 part

∴ Work done by the waste pipe in 1 minute = {1/15 - (1/20 + 1/24)} part
= (1/15 - 11/120) part
= (8 - 11)/120 part
= - (3/120) part
= - (1/40) [ −ve sign means emptying]

ATQ,
1/40 part = 3 gallons

∴ Volume of whole=(3 × 40) gallons
= 120 gallons

Question: The speed of a car increases by 2 km after every one hour. If the distance travelled in the first one hour was 30 km, what was the total distance travelled in 10 hours?

Solution:
Total distance travelled in 10 hours =( 30 + 32 + 34 +...... upto 10 terms)
This is an A.P with first term, a = 30 ,
number of terms, n = 10
d = 2

Required distance = (10/2)[(2 × 30) + {(10 - 1) × 2}] km
= 5 (60 + 18) km
= (5 × 78) km
= 390 km

Question: A tap can fill a tank in 10 minutes and another can empty it in 6 minutes. If the tank is fill already two-fifths and both the taps are opened together, how long will it take to completely emptied?

Solution:
Given,
the outlet pipe is faster than the inlet pipe

Part to be emptied = 2/5 part

Net part emptied in 1 minute = (1/6 - 1/10) = (5 - 3)/30 = 2/30 = 1/15 part

ATQ,
1/15 part is emptied in 1 minute
∴ 2/5 part is emptied in (15 × 2/5) minute
= 6 minutes

Question: A bus left 40 minutes late due to bad weather and order to reach its destination 16 km away in time, it had increase its speed by 4 kmph from its usual speed. Find the usual speed of the bus?

Solution:
Let
usual speed be x kmph,
then new speed will be (x + 4) kmph.

∴ Time taken to cover 16 km with speed x kmph = 16/x 

∴ Time taken to cover 16 km with Speed (x + 4) kmph = 16/(x + 4)

ATQ,
(16/x) - {16/(x + 4)} = 40 min
⇒ {16(x + 4) - 16x}/x(x + 4) = 40/60
⇒ (16x + 64 - 16x)/x(x + 4) = 2/3
⇒ 64/(x² + 4x) = 2/3
⇒ 2(x² + 4x) = 192
⇒ x² + 4x = 96
⇒ x² + 4x - 96 = 0
⇒ x² + 12x - 8x - 96 = 0
⇒ x(x + 12) - 8(x +12) = 0
⇒ (x + 12)(x - 8) = 0
∴ x = -12 or 8
[But, speed cannot be negative]

So usual speed will be 8 km/h

Question: Two pipes A and B can fill a cistern in 24 and 36 minutes respectively. Both pipes are opened together, after how many minutes should B be turned off, so that the cistern be fill in 16 minutes?

Solution:
A can fill the cistern in 24 minutes
So in 1 min A can fill the cistern = 1/24 part
In 16 min, A can fill the cistern = 16/24 part
= 2/3 part

Remaining part = 1 - 2/3 = 1/3 rd

As B can fill full cistern in 36 minutes
So it will fill 1/3 rd part in = (1/3 × 36) minutes.
= 12 minutes.

Question: A salesman travels a distance of 45 km in 2 hours and 30 minutes. How much faster, in kilometers per hour, on an average, must he travel to make such a trip in 5/6 hour less time?

Solution:
Given,
Time = 2 hours and 30 minutes
= (2 + 30/60) hours
= (2 + 1/2) hours
= 5/2 hours

∴ Initial speed = (45 ÷ 5/2) km/h
= (45 × 2/5) km/h
= 18 km/h

New time to travel 45 km = (5/2 - 5/6) = 10/6 = 5/3 hours

∴ New speed = (45 ÷ 5/3) km/h
= (45 × 3/5) km/h
= 27 km/h

∴ Speed should increased = (27 - 18) = 9 km/h

Question: In covering a distance of 36 km, Hasan takes 4 hours more than Imran. If Hasan doubles his speed, he would take 2 hours less than Imran. What is Hasan’s speed?

Solution:
Let
Hasan's speed be x km/h

ATQ,
36/x - 36/2x = (4 + 2)
⇒ (72 - 36)/2x = 6
⇒ 36/2x = 6
⇒ 12x = 36
∴ x = 3

∴ Hasan's speed be 3 km/h

Question: Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?

Solution:
Let,
the cistern be filled by pipe A alone in x hours.
Then, pipe B will fill it in (x + 6) hours

ATQ,
{1/x + 1/(x + 6)} = 1/4
⇒ (x + 6 + x)/{x(x + 6)} = 1/4
⇒ (2x + 6)/(x² + 6x) = 1/4
⇒ (x² + 6x) = 8x + 24
⇒ x² + 6x - 8x - 24 = 0
⇒ x² - 2x - 24 = 0
⇒ x² - 6x + 4x - 24 = 0
⇒ x(x - 6) + 4(x - 6) = 0
⇒ (x - 6)(x + 4) = 0
⇒ x = 6 or - 4
∴ x = 6 [neglecting the negative value of x]

∴ the cistern be filled by pipe A alone in 6 hours

Question: A truck covers a distance of 540 meters in 1 minute where as a bus covers a distance of 56.7 km in 45 minutes. The ratio of their speeds is-

Solution:
Speed of truck = 540/60 = 9 m/s
Speed of bus = (56.7 × 1000)/(45 × 60) = 21 m/s

∴ Ratio of speed = 9 : 21 = 3 : 7

Question: A thief steals a car at 1.30 pm and drive it off 40 km/hr. The theft is discovered at 2 pm and the owner sets off in another car at 50 km/hr he will catch the thief at-

Solution:
Distance covered by thief in (2 pm - 1.30 pm) = 1/2 hours

1/2 hours at speed of 40 km/h = 40 × (1/2) = 20 km

Their relative speed in same direction = (50 - 40) km/h
= 10 km/h

ATQ,
20 km, is the distance that has to be covered by owner to catch the thief.

Required time = (20/10) hours
=2 hours

Therefore, he will over take the thief at :
= 2 pm + 2 hours
= 4 pm

Question: A swimming pool is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the pool in the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is?

Solution:
Let,
the first pipe alone takes x hours to fill the swimming pool.
Then, the second and third pipes will take (x - 5) and (x - 9) hours respectively to fill the swimming pool.

ATQ,
1/x + 1/(x - 5) = 1/(x - 9)
⇒ (x - 5 + x)/x(x - 5) = 1/(x - 9)
⇒ (2x - 5)/(x² - 5x) = 1/(x - 9)
⇒ 2x² - 18x - 5x + 45 = x² - 5x
⇒ 2x² - 23x + 45 - x² + 5x = 0
⇒ x² - 18x + 45 = 0
⇒ x² - 15x - 3x + 45 = 0
⇒ x(x - 15) - 3(x - 15) = 0
⇒ (x - 15)(x - 3) = 0
∴ x = 15 [neglecting x = 3 ]

∴ the first pipe alone takes 15 hours to fill the swimming pool.

Question: A train travels a distance of 600 km at a constant speed. If the speed of the train is increased by 5 km/h, the journey would take 4 hours less. Find the speed of the train.

Solution: 
Let,
the speed of the train be x km/h

ATQ,
600/x - 600/(x + 5) = 4
⇒ {600(x + 5) - 600x}/x(x + 5) = 4
⇒ (600x + 3000 - 600x)/(x² + 5x) = 4
⇒ 3000/(x² + 5x) = 4
⇒ 4x² + 20x = 3000
⇒ 4x² + 20x - 3000 = 0
⇒ x² + 5x - 750 = 0
⇒ x² + 30x - 25x - 750 = 0
⇒ x(x + 30) - 25(x - 30) = 0
⇒ (x + 30)(x - 25) = 0
⇒ x = - 30 or 25
∴ x = 25 [neglecting the negative value of x]

∴ Speed of the train 25 km/h