Ques:- A jogger running at 9 km/hr along side a railway track is 240 m ahead of the engine of a 120 m long train running at 45 km/hr in the same direction. In how much time will the train pass the jogger?
Recent Answer : Added by Ben Gulliano On 2023-05-23 15:53:20:
– 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
Ques:- There is a 5digit no. 3 pairs of sum is eleven each. Last digit is 3 times the first one. 3 rd digit is 3 less than the second. 4 th digit is 4 more than the second one. Find the digit.
Recent Answer : Added by Admin On 2020-05-17 11:59:40:
let first digit be ‘X’
then 5th digit is ‘3X’
let 2nd digit be ‘Y’
then 3rd digit is ‘Y-3’
and 4th digit is ‘Y+4’
then the no is ‘(X)(Y)(Y-3)(Y+4)(3X)’
from the above we can say 3X<=9
so X<=3 and any of the digit in the number is <=9
and also given that 3 pairs sum is 11...
so make trial and error..
if X=1...any of the no is 10 which is wrong trial....
if X=2...then let Y+4=9 ==> Y=5
then no is 25296
first pair 2+9=11
second pair 2+9=11
third pair 5+6=11
now the answer is 25296
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Recent Answer : Added by DK BOSS On 2021-07-30 16:26:33:
Employee Performance Standards. Employee performance measurements can determine an employee’s compensation, employment status or opportunities for advancement.
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Recent Answer : Added by SUNIL KUMAR MODI On 2021-08-17 14:47:48:
4) QEOKTWG
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Ques:- The product of two numbers is 2025 and their HCF is 15.Evaluate their LCM is?
Recent Answer : Added by RAKESH S On 2021-09-23 15:56:01:
number1*number2=lcm*hcf
so,
product of two numbers is ,
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Recent Answer : Added by Admin On 2020-05-17 12:03:27:
( c ) 10
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A. 72
B. 74
C. 75
D. 76
Recent Answer : Added by PRASHANT YADAV On 2022-09-28 17:42:49:
75
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Recent Answer : Added by Admin On 2020-05-17 12:03:16:
he removed one bolt in each three wheels
and fixed the bolts in stephiny wheel
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Recent Answer : Added by shivaprasad On 2021-10-12 12:52:39:
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X=20
So (20,22)
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119
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Reliance
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– 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