ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: Traveling at a speed of 50 kmph, how long is it going to take to travel 60 km?

Solution:
Time = Distance ÷ Speed 
= 60/50
= 1.2 hours
= 1 hour and 12 minutes.

Question: A train traveling at 60 kmph crosses a man in 6 seconds. What is the length of the train?

Solution:
Speed in m/sec = 60 × (5/18) = 50/3 m/sec.
Time taken to cross the man = 6 seconds.

∴ Distance traveled = (50/3) × 6 = 100 m = length of the train.

Question: A person travels from one place to another at 30 km/hr and returns at 120 km/hr. If the total time taken is 5 hours, then find the Distance.

Solution:
Here the Distance is constant, so the Time taken will be inversely proportional to the Speed.
Ratio of Speed is given as 30 : 120 = 1 : 4

So the ratio of Time taken will be 4 : 1. 
Total Time taken = 5 hours;
Time taken while going is 4 hours and returning is 1 hour. 

Hence, Distance = 30 ×  4 = 120 km

Question: A man travels from his home to office at 4 km/hr and reaches his office 20 min late. If the Speed had been 6 km/hr he would have reached 10 min early. Find the distance from his home to office.

Solution:
Let the Distance between home and office = d.
Suppose he reaches the office on Time, the Time taken = x minutes

Case 1: When he reaches office 20 minutes late,
Time taken = x + 20 minutes

Case 2: when he reaches office 10 minutes early,
Time taken = x - 10 minutes

As the Distance traveled is the same, the ratio of Speed in case 1 to the Speed in case 2 will be the inverse of the Time taken in both cases
Ratio of Speed in both cases = 4 : 6 = 2 : 3

Ratio of Time in both cases = 3 : 2

∴ (x + 20)/(x - 10) = 3/2
⇒ 2x + 40 = 3x -30
∴ x = 70

minutes Taking case 1, 
70 + 20 minutes
= 90 minutes

In 60 minutes he reached 4000 m
∴ In 1 minutes he reached 4000/60 m
∴ In 90 minutes he reached (4000 × 90)/60 m
= 6000 m
= 6 km

Question: A man decided to cover a distance of 6 km in 84 minutes. He decided to cover two thirds of the distance at 4 kmph and the remaining at some different speed. Find the speed after the two third distance has been covered.

Solution:
We are given that two thirds of the 6 km was covered at 4 kmph 
∴ 4 km distance was covered at 4 kmph.
Time taken to cover 4 km = (4 km)/(4 kmph) = 1 hr = 60 minutes

Time left = 84 - 60 = 24 minutes

Now,
The man has to cover remaining 2 km in 24 minutes = 24/60 hours = 0.4 hours

Speed required for remaining 2 km = 2/ 0.4 = 5 kmph

Question: While going to office, Salman travels at a speed of 30 kmph and on his way back, he travels at a speed of 45 kmph. What is his average speed of the whole journey?

Solution:
When distance travelled is same, then average speed = (2ab)/(a + b); (where a and b are two different speeds)

∴ The Average Speed = (2 × 45 × 30)/(45 + 30)
= 2700/75 kmph
= 36 kmph

Question: Two horses start trotting towards each other, one from A to B and another from B to A. They cross each other after one hour and the first horse reaches B, 5/6 hour before the second horse reaches A. If the distance between A and B is 50 km. what is the speed of the slower horse?

Solution:
If the speed of the faster horse be f and that of slower horse be s 
Then,
f + s = 50/1 = 50 km/h
∴ f = 50 - s

ATQ,
50/s - 50/f = 5/6
⇒ (50f - 50s)/(sf) = 5/6
⇒ 50(f - s) = (5/6)(sf)
⇒ 50(50 - s - s) = (5/6)(sf)
⇒ 6(2500 - 100s) = 5 × s × (50 - s)
⇒ 15000 - 600s = 250s - 5s²
⇒ 5s² - 600s - 250s + 15000 = 0
⇒ 5s² - 850s + 15000 = 0
⇒ s² - 170s + 3000 = 0
⇒ s² - 150s - 20s + 3000 = 0
⇒ s(s - 150) - 20(s - 150) = 0
⇒ (s - 150)(s - 20) = 0
∴ s = 150, 20 [ 150 not be acceptablr]

∴ The speed of the slower horse is 20 km/h

Question: The distance of the college and home of Rajeev is 80km. One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4km/h and thus he reached  to college  at the normal time. What is the changed (or increased) speed of Rajeev?

Solution:
Let the normal speed be x km/h
Then 
80/x - 80/(x + 4) = 1
⇒ x² + 4x - 320 = 0
⇒ x(x + 20) - 16(x + 20) = 0
⇒ (x + 20 ) (x - 16) =0
∴ x = - 20, 16 [- 20 is not acceptable]
∴ x = 16 km/h

Therefore (x + 4) = 20 km/h
Therefore increased speed = 20 km/h

Question: A man traveled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village?

Solution:
Average speed = (2ab)/(a + b) here a = 25, b = 4
= (2 × 25 × 4)/(25 + 4)
= 200/29 km/hr.

Distance covered in 5 hours 48 minutes
= Speed × time = (200/29) × (29/5)
= 40

Distance covered in 5 hours 48 minutes = 40 kms.
∴ Distance of the post office from the village = (40/2) = 20 km.

Question: Two trains starting at the same time from two stations 200 km apart and going in opposite directions cross each other at a distance of 110 km from one of the stations. What is the ratio of their speeds?

Solution:
In the same time, they cover 110 km and 90 km respectively.
Therefore, Ratio of their speeds = 110 : 90 = 11 : 9
 

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:
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