## Syntel Sample Train Questions

Dear Reader, Below are four sample questions based on train speeds and distances.

Question 1

Mumbai Rajadhani express left Delhi for Mumbai at 09.30 hrs, travelling at a speed of 60 kmph and Delhi Super Fast Express left Delhi for Mumbai on the same day at 11.30 hrs. travelling at a speed of 80 kmph. How far away from Delhi will the two trains meet ?

(a) 120 km    (b) 360 km   (c)  480 km    (d) 500 km

Solution:

Let us assume both the trains meet at x hours after 9.30 hours.
Then distance travelled by train travelling at 60Kmph for x hours should match the distance covered by train travelling at 80Kmph for x - 2 hours. (x - 2 hours is used for second train as it has started late by two hours.)

Applying the above logic we can form the equation :
60x=80(x - 2)
Simplifying we get, x=8
So Required distance = 60x = (60x8) = 480 km.
Both the trains meet 480 km from Delhi.

Question 2

Virendar travels by a train that leaves New Delhi, at 9.00 AM . The rate of the train was 30 km/hr. Virendar’s friend Shilaja leaves by another train that leaves Delhi at 40 km/hr at 2p.m. in the same direction. Shilaja could not get her ticket for the first train. How many kms from Delhi do they meet?

a)437 kms b)450kms c)550 kms d)600 kms

Solution:

Rate of Virendar’s train is 30 kmph. It starts at 9 am while Shilaja's train starts 5 hours late at 2 pm. Before Shilaja's train starts, Virendar's train would have travelled 30 Kmph X 5 hours = 150 km.
Since Virendar’s train and Shilaja’s train travels in the same direction, relative velocity of Shilaja's Train with respect to Virendar's Train = Shilaja's Train Speed - Virendar's Train Speed = 40-30 =10 kmph
Time taken to meet after Shilaja's train departure = Distance covered by Virendar's train before Shilaja's train starts / Relative velocity of Shilaja's Train with respect to Virendar's Train = 150/ 10 = 15 hours from 2.00 PM
15 X 40 (TIME X RATE) =600 KM
Virender’s train will meet Shilaja’s train at 600 km from New Delhi.

Question 3

Christiana drives 150km from Chennai to Sreekalahasti in 3 hours 20min to meet her uncle who is a business man there and returns to Chennai in 4 hours 10 min. What is the difference in average speed from Chennai to Sreekalahasti to that of the average speed for the entire trip?

(a) 5 km/hr    (b) 4.5 km/hr   (c) 4 km/hr   (d) 2.5 km/hr

Solution:

Average speed from Chennai to Sreekalahasti =150 / 3 1/3 hours = 150x3/10 =45 kmph ...(1)
Average speed from Sreekalahasti to Chennai =150/ 4 1/6hours = 150x6/25 =36 kmph ...(2)
Average speed for the entire trip can be calculated using the formula : 2xy / x+y
In this formula x represents the onward average speed from Chennai to Sreekalahasti while y represents return average speed from Sreekalahasti to Chennai.

Applying the calculated values from 1 and 2 in the formula we get :
2x45x36/ 45+36 = 40kmph
The difference between average speed from Chennai to Sreekalahasti to that of the average speed for the entire trip is 45 -40 = 5 kmph.

Question 4

Garden City Express, a super fast train that flies between Bangalore and Mysore. The average speed of Garden City Express was measured every 12 minutes and was found to have increased by 6 miles per hour during each successive 12 minutes interval. By half way through, assuming there were 5 intervals of 12 minutes, between Bangalore and Mysore the average speed was 60 miles per hour. How many miles did the super fast train travel in the first 12 minute interval?

(a)7.2 miles (b)12 miles (c) 10miles (d) 2 miles.

Solution:

We need to find the distance travelled by the super fast train in the first 12 minutes interval. We know, Distance= Rate x Time, and we already know the time is 12 minutes. All we need to do is to find the train’s speed (its rate) during the first interval in order to find the distance.

We are given that the train’s average speed increased by 6 miles per hour during each interval. During half way the average speed was 60 miles per hour. There were 5 intervals of 12 minutes. So, in the fourth interval the speed would be 60-6=54. Similarly, in the third interval it would be 54-6=48, in the second interval it would be 48-6=42 and in the first interval the speed must have been 42-6=36 miles per hour.

Now we have time (12 minutes) and speed (36 miles per hour). Since time is in minutes and speed is in units of miles per hour we need to convert the time to hours or the speed to miles per minute (to bring uniformity in units). An hour has 60 minutes so 12 minutes is 12/60 hours=1/5 of an hour. Therefore the distance is 1/5 hours x 36 miles per hour = 36/5=7.2 miles.

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