ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: A person crosses a 600 m long street in 5 minutes, What is his speed in km per hour?

Solution:
Speed = 600/(5 × 60) m/sec
= 2 m/sec
= 2 × (18/5) km/hr
= 7.2 km/hr

Question: A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes?

Solution:
Relative speed of the thief and policeman  =  (11 - 10) km/hr = 1 km/hr 

Distance covered in 6 minutes  = (1/60) × 6 km   = 1/10 km = 100 m

Therefore, Distance between the thief and policeman = (200 - 100) m = 100 m.

Question: A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is-

Solution:
Total time taken = (160/64 + 160/80)hrs
= 9/2 hrs.

Average speed = 320 × (2/9) km.hr
= 71.11 km/hr.

Question: Two trains A and B start simultaneously in the opposite direction from two points P and Q and arrive at their destinations 16 and 9 hours respectively after their meeting each other. At what speed does the second train B travel if the first train travels at 120 km/h?

Solution:
After meeting with each other the 1st train travels 120 × 16 km=1920km.
If the speed of the 2nd train be X km/h, it travels 1920/X hour before meeting.
After meeting, the 2nd train travels 9X km.
Before meeting, the first train covered this distance in 9X/120 hour.
As the two trains started simultaneously, before meeting their journey time was same.
So, 1920/X = 9X/120
⇒ 9X² = 1920 × 120
⇒ X² = (1920 × 120)/9
⇒ X = √[640 × 40]
∴ X=160.

∴ The train B travels at 160km/h.

Question: A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely?

Solution:
Clearly, pipe B is faster than pipe A and so,the tank will be emptied.

part to be emptied = 2/5

part emptied by (A + B) in 1 minute = (1/6 - 1/10) = 1/15

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

 so, the tank will be emptied in 6 min.

Question: A man reaches his office 20 min late, if he walks from his home at 3 km per hour and reaches 30 min early if he walks 4 km per hour. How far is his office from his house?

Solution:
Let distance = x km.
Time taken at 3 kmph = dist/speed = x/3  which is 20 min late.
Time taken at 4 kmph = x/4  which is 30 min earlier
Difference between time taken : 30 - (- 20) = 50 mins = 50/60 hours.
ATQ,
x/3 -  x/4 = 50/60
⇒ x/12 = 5/6
∴ x = 10 km.
 

Question: A man walking at the rate of 5 km/hr crosses a bridge in 15 minutes. The length of the bridge (in metres) is-

Solution:
speed = 5 × (5/18) m/sec
= 25/18 m/sec.

Distance covered in 15 minutes = (25/18) × 15 × 60 m
= 1250 m.

Question: If Deep had walked 20 km/h faster he would have saved 1 hour in the distance of 600km. What is the usual speed of Deep?

Solution:
Let the original speed  be x km/h,
Then
600/x - 600/(x + 20) = 1 
⇒ 600[{(x + 20) - x}/x(x + 20)] = 1
⇒ x² + 20x - 12000 = 0
⇒ x² + 120x - 100x - 12000 = 0
⇒ (x + 120)(x - 100) = 0
∴ x = - 120, 100 [- 120 not acceptable]
 
Therefore,  original speed = 100 km/h

Question: A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?

Solution:
Time taken by one tap to fill half of the tank = 3 hrs.
Part filled by the four taps in 1 hour = 4 × (1/6) = 2/3
Remaining part =(1 - 1/2) = 1/2

Four taps filled 2/3 part in 60 min
∴ Four taps filled full part in (60 × 3)/2 min
∴ Four taps filled 1/2 part in (60 × 3)/(2 × 2) min
= 45 min

So, total time taken = 3 hrs. 45 mins.

Question: A man takes 6 hours 15 minutes in walking a distance and riding back to the starting place. He could walk both ways in 7 hours 45 minutes. The time taken by him to ride both ways, is-

Solution:
Time taken in walking both ways = 7 hours 45 minutes ..............(i)
Time taken in walking one way and riding back = 6 hours 15 minutes ...................(ii)

By equation (ii) × 2 - (i), we have
Time taken to man ride both ways = 12 hours 30 minutes - 7 hours 45 minutes
= 4 hours 45 minutes

Question: An employee may claim Tk. 7.00 for each km when he travels by taxi and Tk. 6.00 for each km if he drives his own car. If in one week he claimed Tk. 595 for traveling 90 km. How many kms did he travel by taxi?

Solution:
Let x and y be the respective km's travelled by man via taxi and by his own car.
Given x + y = 90
⇒ x = 90 - y
But according to the question,
7x + 6y = 595
⇒ 7(90 - y) + 6y = 595
⇒ 630 - 7y + 6y = 595
⇒ y = 630 - 595
∴ y = 35

∴ x = 90 - 35 = 55

Therefore, the distance travelled by taxi is 55 kms.

Question: The distance between two cities A and B is 330 Km. A train starts from A at 8 a.m. and travel towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 Km/hr. At what time do they meet?

Solution:
Suppose they meet x hrs after 8 a.m
then,
[Distance moved by first in x hrs] + [Distance moved by second in (x - 1) hrs] = 330.
Therefore, 60x + 75(x - 1) = 330
⇒ 60x + 75x - 75 = 330
⇒ 135x = 405
∴ x = 3

So,they meet at (8 + 3) = 11 a.m.

Question: A cistern is normally filled in 8 hours but takes two hours longer to fill because of a leak in its bottom. If the cistern is full, the leak will empty it in?

Solution:
If a tap can fill a cistern in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank.
Part emptied by the leakage in 1 hour
= 1/8 - 1/10
= (5 - 4)/40
= 1/40

∴ The leakage will empty the cistern in 40 hours.

Question: With an average speed of 40 km/hr, a train reaches its destination in time. If it goes with an average speed of 35 km/hr, it is late by 15 minutes. Find the length of the total journey?

Solution:
Let the time taken by train be 't' hrs.
Then,
40t = 35(t + 15/60)
⇒ 40t = 35t + 35/4
∴ t = 7/4 hrs

Therefore, Required length of the total journey d = v × t
= 40 × 7/4
= 70 kms.

Question: A man covers half of his journey by train at 60 km/hr, half of the remaining by bus at 30 km/hr and the rest by cycle at 10 km/hr. Find his average speed during the entire journey?

Solution:
Speed = 60 km/h (half journey), 30 km/h (one-fourth journey), 10 km/h (one fourth journey)

Average speed = Total distance covered/Total time taken

Let the total distance be 120 km
Time taken to cover distance at 60 km/h = 60/60 = 1 hr
Time taken to cover distance at 30 km/h = 30/30 = 1 hr
Time taken to cover distance at 10 km/h = 30/10 = 3 hr

∴ Average speed = 120/(1 + 1 + 3) = 24 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:
Work done by the waste pipe in 1 minute = 1/15 - (1/20 + 1/24)
= 1/15 - 11/120
= - 1/40 [- ve sign means emptying]

Volume of 1/40 part = 3 gallons
∴ Volume of whole = (3 × 40) gallons
= 120 gallons.

Question: A man whose speed is 4.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 kmph, find his average speed for the total journey?

Solution:
Speed of Man = 4.5 kmph
Speed of stream = 1.5 kmph
Speed in DownStream = 6 kmph
Speed in UpStream = 3 kmph

Average Speed = (2 × 6 × 3)/9 = 4 kmph.

Question: A person takes 20 minutes more to cover a certain distance by decreasing his speed by 20%. What is the time taken to cover the distance at his original speed?

Solution:
Let the distance and original speed be 'd' km and 'k' kmph respectively.
d/0.8k - d/k = 20/60
⇒ 5d/4k - d/k = 1/3
⇒ (5d - 4d)/4k = 1/3
⇒ d = (4/3)k

Time taken to cover the distance at original speed
= d/k = 4/3 hours = 1 hour 20 minutes

Question: Excluding stoppages, the average speed of a bus is 60 km/hr and including stoppages, the average speed of the bus is 40 km/hr. For how many minutes does the bus stop per hour?

Solution:
In 1hr, the bus covers 60 km without stoppages and 40 km with stoppages.
Stoppage time = time take to travel (60 -  40) km = 20 km at 60 km/hr.

∴ stoppage time = 20/60 hrs
= 1/3 hrs
= 20 min.

Question: Taps X and Y can fill a tank in 30 and 40 minutes respectively. Tap Z can empty the filled tank in 60 minutes. If all the three taps are kept open for one minute each, how much time will the taps take to fill the tank?

Solution:
Given taps X and Y can fill the tank in 30 and 40 minutes respectively. Therefore,
part filled by tap X in 1 minute = 1/30
part filled by tap Y in 1 minute = 1/40

Tap Z can empty the tank in 60 minutes. Therefore,
part emptied by tap Z in 1 minute = 1/60

Net part filled by Pipes X,Y,Z together in 1 minute = [1/30  + 1/40 - 1/60]
= 5/120
= 1/24

∴ The tank can be filled in 24 minutes.