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:- 1) In a club there are certain no. of males and females. If 15 females are absent then no. of males will be twice that of females. If 45 males are absent then female strength will be 5 times that of males. Find no. of males actually present.
Recent Answer : Added by Admin On 2020-05-17 12:01:55:
Let the number of males be given the name M.
Let the number of females be given the name F.
If 15 females are absent, then M will be twice that of
present females.
This means that M = 2 * (F – 15)
M = 2 * F – 30.
or 2 * F – M = 30.
Now if in addition to the 15 females being absent, we also
have 45 males being absent,
then this gives the equation,
(F – 15) = 5 * (M – 45)
which simplifies to
F – 15 = 5 * M – 225
5 * M – F = 210
Pulling the equations together, we get
5 * M – F = 210
-M + 2 * F = 30
Multiply the first equation by 2, and keep the second
equation as is.
10 * M – 2 * F = 420
– M + 2 * F = 30
Add the equations.
9 * M = 450
M = 50
Verify answer.
Calculate F
from – M + 2 * F = 30
-50 + 2 * F = 30
2 * F = 30 + 50
F = 40.
If 15 females are absent, then number of males will be twice
that of females.
40 – 15 = 25.
50 = 2 * 25. Confirmed.
If also 45 males were absent, then female strength would be
5 times that of males.
Female strength is 25 due to the 15 females being absent.
50 – 45 = 5.
25 = 5 * 5. Confirmed.
Ques:- In a series of six consecutive even numbers, the sum of the second and sixth numbers is 24. What is the fourth number?
Recent Answer : Added by Mogomotsi On 2021-11-24 16:45:04:
2a; 2a+2; 2a+4; 2a+6; 2a+8; 2a+10
t_2 + t_6 = 24 => 2a+2 + 2a + 10 = 24 => a = 3
therefore t_4 = 2(3) + 6 = 12
Ques:- About the number of pairs which have 16 as their HCF and 136 as their LCM, we can definitely say that
A. Only one such pair exists
B. No such pair exists
C. Only two such pairs exist
D. Many such pairs exist
Recent Answer : Added by Admin On 2022-09-23 18:49:19:
B. No such pair exists
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Ques:- The houses are numbered 1,2,3,? and reach the end of the street and backtracks toward s the starting point. The house numbered 10 is opposite to 23. there are even no of houses. Find the total no of houses.
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32
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Recent Answer : Added by Nandakumar Athiseshan On 2022-08-14 16:33:45:
P Q R S
TP -> UQ -> RV -> SW
Answers-
U sits opposite to S
Q sits in between S and R (after interchange)
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A. 13
B. 15
C. 16
D. 25
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Recent Answer : Added by Uday On 2022-03-11 17:00:33:
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Ques:- In 3.5 Kg rod , there is 74% silver. If it is alloyed with a 0.5 Kg rod, the % of silver goes p to 84%. The percent of silver in 0.5 Kg rod?
Recent Answer : Added by Munesh On 2022-08-14 16:29:31:
In 3500 g rod contains 74% of silver = 3500 * 74/100 = 2590 g
Then 3500 g + 500 g of rod contains 84% of silver
Let x be the silver contained in 500 g of silver
(2590/3500 * 100) + (x/500 * 100) = 84
74 + x/5 = 84
(370 + x) /5 = 84
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x = 50
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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