ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: Rakib and Tanveer started running simultaneously towards each other with speeds in the ratio 3 : 4 if the initial separation between the two is 4.2 km and they meet in 3 minutes, what is the difference between their speeds?

Solution:
Given that,
Rakib and Tanveer have speeds in the ratio 3 : 4.
The initial distance between them is 4.2 km, and they meet in 3 minutes, which is 3/60 = 1/20 hours

Let,
Rakib's speed be 3x km/h and Josh's speed be 4x km/h. 
The relative speed when running towards each other is = 3x + 4x = 7x km/h.

Now we know,
Relative speed × time = distance
7x × (1/20) = 4.2
⇒ x = (4.2 × 20)/7
∴ x = 12

Thus, Rakib's speed is = 3 × 12 = 36 km/h.
And Tanveer's speed is = 4 × 12 = 48 km/h.

So the difference in speeds is = 48 - 36 = 12 km/h.

Question: If your car runs from A to B at 60 km/hr and on returning from B to A, it is 40 km/hr, then average speed in km/hr of the car will be-

Solution:
Given that,
The speed of the car is travelling and the returning are 60 km/hr and 40 km/hr

We know,
Average speed = Total distance/Total time

Time = Distance/Speed

Now,
Let the distance between A and B be x

∴ Total time = (x/60) + (x/40)
= (2x + 3x)/120
= 5x/120 = x/24
∴ Total time = x/24

And total distance = x + x = 2x 

∴ Average speed = 2x/(x/24) = 48 km/hr

Question: A car moves from Dhaka to Cumilla at the original speed at 60 kmph and reached Cumilla in 20 minutes late. If the car increased speed by one fourth of his original speed reached Cumilla on time, find the distance between Dhaka and Cumilla?

Solution:
Let distance = d km.
Original speed = 60 km/h. So time taken = d/60 hours.

But it is 20 minutes late. That means the scheduled time T (in hours) is such that d/60 = T + (20/60) = T + (1/3).

And
When speed is increased by one fourth. So new speed = 60 + (1/4) × 60 = 60 + 15 = 75 km/h.
Then time taken = d/75. And this equals T (on time).

So we have,
d/60 = T + 1(/3) .........(1)
d/75 = T ..........(2)

Now, subtract the second equation from the first,
(d/60) - (d/75) = 1/3
⇒ (5d - 4d)/300 = 1/3
⇒ d = 300/3
∴ d = 100 km

∴ Distance between Dhaka and Cumilla = 100 km

Question: A boy rides his bicycle 10 km at an average speed of 10 km/hr and again travels 12 km at an average speed of 10 km/hr. His average speed for the entire trip is-

Solution:
Given that,
Distance₁ = 10 km, Speed₁ = 10 km/hr
Distance₂ = 12 km, Speed₂ = 10 km/hr

∴ Total Distance = 10 + 12 = 22 km

And,
Time₁ = 10 ÷ 10 = 1 hr
Time₂ = 12 ÷ 10 = 1.2 hr

∴ Total Time = 1 + 1.2 = 2.2 hr

We know,
Average Speed = Total Distance ÷ Total Time
= 22/2.2 = 10 km/hr

∴ His average speed for the entire trip is 10 km/hr.

Question: A train moving at speed of 108 km/hr crosses a pole in 12 seconds. Find the length of the train.

Solution:
Length of the train is equal to the distance covered by train to cross the pole.
So, we will find the distance travelled by the train in 12 seconds.

Now,
Speed is given in Km/hr so we will convert it into m/s
Speed = 108 × (5/18) = 30 m/s
Time = 12 seconds

We know, 
Distance = Speed × Time
∴ Length of train = 30 × 12 = 360 meters

Question: A thief committed a crime and escaped from the spot at a speed of 12 m/s. A Security guard started chasing him 20 minutes after the thief started running and caught him in the next 20 minutes. What is the speed (in m/s) of the Security guard?

Solution: 
Given that,
Thief's speed = 12 m/s
Security guard starts 20 minutes or 1200 s later and catches thief in next 20 minutes or 1200 s

Now,
Distance covered by thief before guard starts,
d₁ = speed × time = 12 × 1200 = 14400 m ; [20 minutes = 1200 s]
So, thief has a 14400 m head start.

And
Guard catches thief in 20 minutes = 1200 s
d₂ = 12 × 1200 s = 14400 m 
∴ Guard covers total distance = head start + distance thief runs during chase = 14400 + 14400 = 28800 m
and Time taken by guard = 1200 s

∴ Speed of guard = Distance/Time = 28800/1200 = 24 m

Question: In a 500-meter race, Q starts 50 meters ahead of P, yet P defeats Q by a margin of 25 meters. What distance did Q cover when P reached the finish line?

Solution:
Total distance Q needed to cover = 500 - 50 = 450 meters
Distance covered by P = 500 meters
But P defeats Q by 25 meters

∴ Distance covered by Q when P reaches the finish line = 450 - 25 = 425 meters

∴  Q covered a distance of 425 meters when P reached the finish line.

Question: A man travels from A to B at a speed of 36 km/hr in 74 minutes and he travels a distance from B to C with a speed of 45 km/hr in 111 minutes. Find the average speed of whole journey.

Solution:
Given that,
A man travels from A to B at a speed of 36 km/hr in 74 minutes and he travels a distance from B to C with a speed of 45 km/hr in 111 minutes. 

We know,
Average speed = Total distance/Total time taken

Now,
Time taken = 74 min : 111 min   [given]
Ratio of Time taken = 2 : 3

∴ Average Speed = {(36 × 2) + (45 × 3)}/(2 + 3) = 207/5
 = 41.4 km/hr

∴ So the average speed of whole journey is 41.4 km/h

Question: A biker travels at 60 km/h. If instead, he had traveled at 80 km/h for the same duration, he would have covered 100 km more. How far did he actually travel?

Solution:
Let, the actual distance travelled be x km.
Then,
x/60 = (x + 100)/80 
⇒ x/6 = (x + 100)/8
⇒ 6(x + 100) = 8x
⇒ 6x + 600 = 8x
⇒ 8x - 6x = 600
⇒ 2x = 600
⇒ x = 600/2
⇒ x = 300 km

Question: Kobita runs 5/2 times as fast as Babita. In a race, if Kobita gives a lead of 40 m to Babita, find the distance from the starting point where both of them will meet (correct up to two decimal places).

Solution:
Given that,
Kobita runs 5/2 times as fast as Babita
Kobita gives a lead of 40 m to Babita 

We know,
Distance = Speed × Time

Let the speed of Babita be = 2x
∴ Speed of Kobita = (5/2) × 2x = 5x

And,
Let the distance covered by Kobita be y meters
∴ Distance covered by Babita = (y - 40) meters

As time is constant, distance is directly proportional to speed,
2x/5x = (y - 40)/y
⇒ 2/5 = (y - 40)/y
⇒ 2y = 5y - 200
⇒ 3y = 200
⇒ y = 200/3
∴ y = 66.67 m

∴ The distance from the starting point where both of them will meet is 66.67 m.

Question: P does one-third as much work as Q in one-fourth of the time. If together they take 24 days to complete a work, how much time shall Q alone take to do it?

Solution:
Let Q takes x days to do the work.
P takes 1/4 of x time to do 1/3 of the work.
∴ the work will be done by P in (x/4) × 3 days
= 3x/4 

ATQ,
(1/x) + (4/3x) = 1/24
⇒ 7/3x = 1/24
⇒ x = 56
∴ Q alone will take 56 days

Question: Walking 3/4 of his normal speed, Rabi is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and office.

Solution:
Let,
Total time = x minutes
So, when it is late then required time = x + 16
If actual speed = d metre/min
Then reduced speed = (3d/4) metre/min

ATQ,
dx = 3d(x + 16)/4
⇒ dx = (3dx + 48d)/4
⇒ 4dx = 3dx + 48d
⇒ 4dx - 3dx = 48d
⇒ dx = 48d
∴ x = 48

∴ Total time = 48 minutes

Question: A man completes a certain journey by a car. If he covered 30% of the distance at the speed of 20kmph. 60% of the distance at 40km/h and the remaining of the distance at 10 kmph, his average speed is-

Solution:

Question: 45 toymakers can prepare 30 toys per day. Rifat wants 360 toys. How many toymakers should he employ to get the job done in 12 days?

Solution:
Let, the required number of toymakers x
45 toymakers make 30 toys per day
So, 1 toymaker makes = 30/45 = 2/3 toys per day
Each toymaker in 12 days makes = (2/3) × 12 = 8 toys
So, x toymakers will make = 8x toys

ATQ,
8x = 360
⇒ x = 360 × (1/8)
∴ x = 45

Question: A man covers half of his journey at 6 km/h and the remaining half at 3 km/h. His average speed is-

Solution:
Let the total distance be 2d km.
First half at 6 km/h.
Second half at 3 km/h.

Now,
Time for first half, t₁ = d/6 hours
Time for second half, t₂ = d/3 hours

∴ Total Time = (d/6) + (d/3) = 3d/6 = d/2

∴ Average speed = Total distance​/Total time = 2d/(d/2) = 4 km/h

 ∴ Average speed = 4 km/h