C
1/4E(E decreses by 4 times)
– 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
47
The work done by A in 8 days is = 8/ 12 = 2/3
Means A alone completes 2/3 part of work.
Remaining work which is (1–2/3) = 1/3 is completed by B in 8–2 = 6 days
So the complete work done by B in 6/(1/3) = 18 days.
B alone can complete the work in 18 days.
let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240
1/221
(130+x)/30=(45×5/18)
245 meters
50 paise coins= 400 = Rs 200
20 paise coins = 600 = Rs 120
10 paise coins= 800 = Rs 80
17
length of the base=3 times height
b=3h;
h=b/3;
so, 1/2bh=24
1/2b(b/3)=24
(b^2)/6=24
b^2=144
b=12
39/50