1:2:3 or 1:2:4
– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.
– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m
To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec
In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m
at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m
If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)
Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.
For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.
This will take an additional (9.6 + 2) seconds.
Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds
1km = 1000m
Xkm = 300m
X=0.30km
1hr = 3,600s
X = 40.5s
0.01125hr
Therefore; 0.3km/0.1125hr= 26.6666=26.67km/hr
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120 (5×4!)
30%
Avg = 2(X*Y)/X+Y
2*(60*40)/60+40
2*(2400)/100
2*24
48.
Total number of pairs is NC2^{N}C_2NC2. Number of pairs standing next to each other = N. Therefore, number of pairs in question = NC2^{N}C_2NC2 – N = 28/2 = 14. If N = 7,
7C2 – 7 = 21 – 7 = 14….
N =7
By 1hour both trains meet, so the distance travel by fly in
1hr is 120km.
1km is equal to 1000 meter
Therefore 225 meter is how many km
Then we have to cross multiply
1km – 1000 meter
? – 225 meter
0.225 km
Then we have to apply formula
Speed = Distance/Time
Speed = 0.225/12 sec
= 0.01875×3600sec
= 67.5
Therefore speed of train is = 67.5
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