IBA ফ্যাকাল্টি ভিত্তিক প্রস্তুতি

Question: A man covers half of his journey at 15 km/h and the remaining half at 5 km/h. His average speed is-

Solution: 
Here, x = 15 km/h and y = 5 km/h

We know, 
Average speed = 2xy/(x + y)
= (2 × 15 × 5)/(15 + 5)  
= 150/20
= 7.5 km/h

Question: A boatman goes 4 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will it take to go 7 km in stationary water?

Solution: 
Speed along current = 60/10 km/h
= 6 km/h
Speed against current  = 4 km/h

∴ Speed in still water = (Speed against current + Speed along current)/2
= (4 + 6)/2
= 10/2
= 5

∴ Required time = 7/5 h
= [(7/5) × 60] min
= 84 min
= 60 min + 24 min 
= 1 h 24 min 

Question: A train covers a distance in 30 minutes. If it runs at a speed of 56 km/h on average. The speed at which the train must run to reduce the time of the journey to 20 minutes is-

Solution:
Here,
Current speed = 56 km/h
Current time = 30 minutes
= 30/60 h 
= 1/2 hour
New time = 20 minutes
= 20/60 h
= 1/3 h

We know, 
Distance = Speed × Time
= 56 × (1/2)
= 28 km

∴ New speed = Distance/New time
= 28/(1/3)
= 84 km/h

Question: An outlet pipe can empty a cistern in 2 hours and 30 minutes. In what time will it empty 3/5 of the cistern?

Solution: 
The outlet pipe empties the one complete cistern in 2 hours and 30 minutes or 2.5 hours

∴ Time taken to empty 3/5 Part of the cistern = (3/5) × 2.5
= 1.5 hours

Question: An ambulance moving at 33 km/h is 45 m behind a school bus. After 15 s, it is 30 m ahead of the bus. Find the speed of the school bus.

Solution: 
Relative Speed = Total distance/Total time
= (30 + 45)/15
= 75/15 
= 5 m/s
= 5 × (18/5)
= 18 km/h

Now,
Relative Speed = Speed of ambulance - Speed of school bus
∴ Speed of school bus = Speed of ambulance - Relative speed
= 33 - 18
= 15 km/h

Question: The speed of a boat in standing water is 9 kmph, and the speed of the stream is 1.5 kmph. A man rows to a place at a distance of 105 km and comes back to the starting point. The total time taken by him is-

Solution: 
Speed upstream = (9 - 1.5) kmph
= 7.5 kmph
Speed downstream = (9 + 1.5) kmph 
=10.5 kmph
∴ Total time taken = (105/7.5) + (105/10.5) 
= 14 + 10
= 24 h

Question: A train, 150 m long, passes a pole in 15 seconds and another train of the same length, travelling in the opposite direction in 10 seconds. What is the speed of the second train? 

Solution: 
Given,
Length of the first train &  second train  = 150 m
Time to pass a pole = 15 seconds
Time taken by trains to cross each other = 10 sec

Speed of the first train = 150/15
 = 10 m/s

And, the relative speed of two trains = (150 + 150)/10 
= 30 m/s

Speed of the second train = (30 - 10) × (18/5)
= 20 × (18/5)
= 72 km/h 

Question: Three pipes A, B, and C can fill a tank from empty to full in 40 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all three pipes are opened. A, B, and C discharge chemical solutions P, Q, and R respectively. What is the proportion of the solution R in the liquid in the tank after 2 minutes?

Solution: 
Part filled by (A + B + C) in 2 minutes
= 2 [(1/40) + (1/20) + (1/10)]
= 2 × (7/40)
= 7/20

Part filled by C (solution R) in 2 minutes = 2/10
= 1/5

∴ Proportion of solution R = (1/5) × (20/7)
= 4/7

Question: Ashik and Ruby run a race with their speed in the ratio of 5 : 3. They prefer to run on a circular track of circumference 2 km. What is the distance covered by Ashik when he passes Ruby for the sixth time?

Solution: 
Since the speeds of Ashik and Ruby are in the ratio 5 : 3 i.e., when Ashik covers 5 rounds, then Ruby covers 3 rounds
∴ Relative speed = 5 – 3 = 2 parts
 Now, the first time Ashik and Ruby meet, when Ashik completes (5/2 = 2.5) rounds, and Ruby completes 1/2 round.

∴ Ashik to pass Ruby for the sixth time, Ashik would have completed = (6 × 2.5) rounds
= 15 rounds

Since each round is 2 km,
Hence, the distance covered by Ashik = (15 × 2) km
= 30 km

Question: A boat while downstream in a river covered a distance of 50 miles at an average speed of 60 miles per hour. While returning, because of the water resistance, it took 1 hour 15 minutes to cover the same distance. What was the average speed during the whole journey?

Solution: 
Time taken to cover 50 miles downstream = (50/60)h
= 5/6 h
Time taken to cover 50 miles upstream = 1h 15m
= 5/4 h

Total time taken to cover 100 miles = (5/6) + (5/4)
= 25/12 h

∴ Average speed  = 100/(25/12)
= (100 × 12)/25
= 48 mph

Question: Walking 3/4 of his normal speed, Rafi is 12 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and office-

Solution:
Let s be Rafi's normal speed,  t be his usual time, and t' be his new time.

Since the distance to the office is constant, d = s × t
When Rafi walks at his normal speed his new time is expressed as, d = (3/4)s × t'

Again, since the distance is the same,
st = (3/4)s × t'
⇒ t' = (4/3)t 

ATQ,
t' - t = 12
⇒ (4/3)t - t = 12
⇒ (4t - 3t)/3 = 12
⇒ t/3 = 12
∴ t = 36 min

Question: A pipe can fill a tank in x hours, and another can empty it in y hours. In how many hours do they together fill it in (y > x)?

Solution:
Pipe fills the tank in x hours
∴ filling rate = 1/x tank/hour
Pipe empties the tank in y hours
∴ emptying rate = 1/y tank/hour
We are told y > x, so the filling pipe is faster than the emptying pipe

When both pipes are open, the net rate = (1/x) - (1/y)

∴ Time to fill the tank = Total work/Net rate
= 1/[(1/x) - (1/y)]
= 1/[(y - x)/xy]
= xy/(y - x) hours

Question: A takes 2 hours 30 minutes more than B to walk 40 km. If A doubles his speed, then he can make it in 1 hour less than B. What is the average time taken by A and B to walk a 40 km distance?

Solution: 
(40/A) - (40/B) = 5/2 . . . . . . . (i)
(40/B) - (40/2A) = 1 . . . . . . . (ii)

Equation (i) + Equation (ii)
(40/A) - (40/2A) = (5/2) - 1
⇒ (40/A) - (40/2A) = 7/2
⇒ 40/2A = 7/2
⇒ A = 40/7

Putting the value of A in equation (i)
7 - (40/B) = 5/2
⇒ 7 - (5/2) = 40/B
⇒ 9/2 = 40/B
⇒ B = 80/9

Now, 
Total time = [40/(40/7)] + [40/(80/9)]
= 7 + (9/2)
= 23/2

Required average time = (23/2)/2
= 23/4
= 5 hours 45 minutes. 

Question: A tap can completely fill a water tank in 8 hours. The water tank has a hole in it through which the water leaks out. The leakage will cause the full water tank to empty in 10 hours. How much time will it take for the tap to fill the tank completely with the hole?

Solution: 
Tap alone fills the tank in 8 hours
⇒ Filling rate = 1/8 tank/hour
Leakage alone empties the full tank in 10 hours
⇒ Emptying rate = 1/10 tank/hour

∴ Net rate = Filling rate - Emptying rate
= (1/8) - (1/10)
= (5 - 4)/40
= 1/40

Time to fill the tank with the hole = 1 full tank/Net rate
= 1/(1/40) hours
= 40 hours 

Question: The speed of a train is 220% of the speed of a car. The car covers a distance of 950 km in 19 hours. How much distance will the train cover in 2.5 hours?

Solution: 
Speed of car = 950/19 
= 50 km/h

Speed of train = 220% × speed of car
= (220/100) × 50
= 110 km/h

Distance covered by the train = 110 × 2.5
= 110 × (25/10)
= 275 km 

Question: Two pipes can fill a tank in 36 and 40 minutes respectively, and a waste pipe can empty 3.5 gallons per minutes. All three pipes working together can fill the tank in 30 minutes. The capacity of the tank is-

Solution: 
Work done by the waste pipe in 1 minute = (1/30) - [(1/36) + (1/40)]
= (12 - 10 - 9)/360
= - (7/360) [Negative sign means emptying]

∴Volume of (7/360) part = 3.5 gallons
Volume of whole tank = (360 × 3.5)/7 gallons
= 180 gallons

Question: A man travelled a distance of 1260 km in 17 hours. He travelled partly by car at a speed of 40 km/h, and partly by train at a speed of 80 km/h. What is the distance traveled by the train?

Solution: 
Given,
 Total distance = 1260 km
Total time = 17 hours
Speed of car = 40 km/h
Speed of train = 80 km/h

Let,
Distance travelled by car = x km
Distance travelled by train = (1260 - x) km

ATQ,
(x/40) + [(1260 - x)/80] = 17
⇒ 2x + 1260 - x = 17 × 80
⇒ x = 1360 - 1260
⇒ x = 100

∴ Distance traveled by the train = (1260 - 100) km
= 1160 km

Question: A train travelling at the speed of x km/h crossed a 300 m long platform in 30 seconds, and overtook a man walking in the same direction at 6 km/h in 20 seconds. What is the value of x?

Solution: 
Train speed = x km/h
Length of train = L
Length of platform = 300m
Man's speed = 6 km/h

∴ (x - 6) × (5/18) = L/20
⇒ (5x - 30)/18 = L/20
⇒ 100x - 600 =18L . . . . . . (i)

And, x × (5/18) = (L+ 300)/30
⇒ 150x = 18L + 5400
⇒ v150x - 5400 = 18L . . . . . . (ii)

From equations (i) & (ii)
⇒ 100x - 600 = 150x - 5400
⇒ 50x = 4800
∴ x = 96 

Question: Two boats are travelling towards each other at speeds of 46 km/h and 62 km/h, respectively. What is the distance between the two boats half a second before they collide?

Solution: 
Relative speed = (46 + 62) km/h
= 108 km/h
= [108 × (5/18)] m/s
= 30 m/s

We know,
Distance = Relative speed × Time
= [30 × (1/2)] m
= 15 m 

∴ Distance between the two boats half a second before collision = 15 meters 

Question: A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time as the tank is filled by the third pipe alone. The second pipe fills the tank 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 time taken by the first pipe = x hours

Then,
The second pipe fills the tank 5 hours faster than the first pipe
∴ Time = x - 5 hours
The third pipe fills the tank 4 hours faster than the second pipe
∴ Time = (x - 5) - 4 = x - 9 hours

ATQ,
(1/x) + {1/(x - 5)} = 1/(x - 9)
⇒ (x - 5 + x)/{x(x - 5)} = 1/(x - 9)
⇒ (2x - 5)(x - 9) = x(x - 5)
⇒ 2x² - 23x + 45 = x² - 5x 
⇒ x² - 18x + 45 = 0
⇒ x² - 15x - 3x + 45 = 0
⇒ (x - 3)(x - 15) = 0
∴ x = 3 or x = 15
Since time cannot be less than 9 hours
Hence, x = 15