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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
The first 10 odd prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31.
Sum of the odd prime numbers = (3+5+7+11+13+17+19+23+29+31)
= 158
Number of odd prime numbers = 10
We know, Average = (sum of the 10 odd prime numbers ÷ Number of odd
prime numbers)
Average =
= 15.8
∴ The Average of first 10 prime numbers which are odd is 15.8
Yes answer is 18 days
4500 sal of jan
Might be using dual E1 in that case will be much more helpful.
15a=45
x:y=18:11
A certain sum amounts to Rs. 1725 in 3 years
and amounts to Rs.1875 in 5 years
so interest of 2 years = 1875 -1725
= 150
so interest of 1 year = 75
so interest of 3 years = 75 × 3 =225 rs
so , Principal = Amount – SI
= 1725 – 225
= 1500 rs
now ,
S.I. = P × N × R /100
75 = 1500 × 1 × R /100
R = 75 / 15
R = 5%
South-West
24kmph
time taken = x/40 + 2x/20
=> x=8
so, 3x = 24
The answer probably lies in finding the in-centre of the
traingle. Bisect all the angles of the triangles and the
point where these angle bisectors meet gives u the point P
190.85
required number =H.C.F of (73-25),(97-73) & (97-25)
=H.C.F of 48 , 24 and 72 = 4 (c)
Number of students behind Aruna in rank = (46 – 12) = 34. So, Arun is 35th from the last
A