– 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
24*4= x(x+4)
solve this
answer is x=8
X+4= 12
12 sec
Pen
Prosperity for all
Prob(choosing two people who are a couple from 6 couples) = 1/6
We have 5 couple left (10 people)
Prob(choosing a person A from 10 people) = 1/10
Prob(choosing 1 person who is not a couple of A) = 1/8
Prob(choosing 4 people that has exactly one couple) = 1/6*1/8*1/10 = 0.0021
May 21
answer is 100
[(45 men*8hours)/30 meters]=12 (working rate)
[(x men*5 hours)/50 meters]=12 (working rate is same)
then x=100
The greatest number that will divide 187, 233 and 279 leaving the same remainder in each case.
To find : The number ?
Solution :
First we find the difference between these numbers.
The required numbers are
233-187=46
279-233=46
279-187=92
Now, We find the HCF of 46 and 92.
Therefore, The required largest number is 46.
3000(4x/100) + 5000(2x/100)=10
0987654321