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# [Smart Math] Time Speed Distance Problem 12

Here’s and example of a SMART MATH problem for TIME SPEED DISTANCE. ### Problem

A train traveling at 36 km/hr completely passes another train at 50% more speed but 50% of its length in the opposite direction in 12 secs. It takes 90 secs for the same train to pass a platform. Find the length of the platform.

1. 700 meters
2. 900 meters
3. 950 meters
4. 1000 meters
5. 1100 meters

### The Usual Method

Speed of slower train = 36 km/hr = $36\times \frac{5}{18}=$10 mt/sec

Speed of faster train = 36 x 1.5 = 54 km/hr = $54\times \frac{5}{18}=$15 mt/sec

Let the length of the slower train be 2l. $\therefore$length of the faster train = $\frac{2l}{2}=l$

Using the concept of relative speed, we get: $\frac{2l+l}{10+15}=12$ $\therefore l=\frac{12\times 25}{3}=100$meters

Let the length of the platform be ‘x’ meters. $\therefore$ Using relative speed concepts, we get: $\frac{2\times 100+x}{10}=90$ $\therefore 200+x=900$ $\therefore x=900-200=700$meters

(Ans: 1)

Estimated Time to arrive at the answer = 45 seconds.

### Using Technique

Note that it takes 90 secs for the train traveling at the speed of 10 mt/sec (36 km/hr) to traverse a length equal to its own plus that of the platform.

i.e. length of train + length of platform = 90

aaaaaaaaaaaaaaa10 $\therefore$ length of train + length of platform = 90 x 10 = 900 meters.

Hence, the length of platform has to be < 900 meters. Hence 700 meters (option ‘1’).

(Ans: 1)

Estimated Time to arrive at the answer = 10 seconds.

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