# PROBLEMS ON TRAINS (PART-III)

## PROBLEMS ON TRAINS (PART-III)

#### QUERY 21

**A train running at ^{7}⁄_{11 }of its own speed reached a place in 22 hours. How much time could be saved if the train would run at its own speed?**

A) 14 hours

B) 7 hours

C) 8 hours

D) 16 hours

**MAHA GUPTA
**The train runs at 7/11 of its usual speed and takes 22 hours; it means 22 hours is 11/7 of the usual time

Now let the usual time that would have been taken by the trains at its usual speed = x hours

Now, therefore (11/7)x = 22

=> x = 14

Hence the time that would have been saved = Time taken – usual time

= 22 – 14 = 8 (option ‘C’)

#### QUERY 22

**One of the two trains whose speed is 120 km/h stops 3 minutes in every 75 km. The 2nd train runs at 50 km/h takes 1 minute stops in every 25 km. What distance 2nd train travelled when the first travel 600 km?**

A) 259^{1}⁄_{6} km

B) 250^{1}⁄_{6} km

C) 255 ^{1}⁄_{3 }km

D) 260 km

**MAHA GUPTA**

TRAIN-I:

For that we need to calculate the total time taken by the first train to travel 600 km.

Time taken by it to travel 600 km at 120 km/h = 600/120 = 5 hours = 300 minutes

Total 3 minutes stoppages that it made in completion of 600 km = ^{600}⁄_{75 }– 1 = 7

Therefore time taken by it in stoppages = 7*3 minutes = 21 minutes

So, total time taken by it = 300 + 21 = 321 minutes

TRAIN-II:

Time taken by the second train obviously the same that the first train took

So, total time taken by the second train = 321 minutes

You’ll observe that the second train will at least travel for without stoppages, so we just need to know how many km it travelled in 21 minutes after deduction of time for stoppages it made.

Well, distance travelled by it in 300 minutes at 50 km/h = 250 km

1 minute stoppages it made = 250/25 = 10 —- We did not deduct 1 here because the train has to still travel further in this case

So time taken by it in stoppages = 10*1 = 10 minutes

So remaining time in which it has to travel = 21 – 10 = 11 minutes

Distance travelled by it in 11 minutes at 50 km/h = 55/6 km = 9^{1}⁄_{6 }km

Therefore total distance travelled by it = 250 + 9^{1}⁄_{6} = 259^{1}⁄_{6 }km (option ‘A’)