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500 = Total+50
Total(450) = only one paper(p) + 29+20+35 + all three (g)
285+212+127 = p + 2( 29+20+35 )+ 3g
solve above .. to get g = 45 …
( small corrctn .. i think .. questn shud be 20 read ONLY
hindu and
times of India and 29 read ONLY hindu and Indian express
and 35
read ONLY times of India and Indian express)
251
My greatest achievement is to avoid Giving up since it made me who l am today and l am able to walk fearless towards achieving my goals.
saying 20%alcohol in 15 litre solution.
It means (20/100)*15=3 litres
Now 5 litre of water is added not alcohol.
So it means only the volume of solution is increases by 5 not alcohol.
It means 15+5=20
Now % of alcohol is
(3/20)*100=15%.
Let d = 7r. And use distance is = rate × time
7r= ( r+12) 5
7r= 5r + 60
Subtract 5r from both sides
2r = 60
Divide out 2
Rate = 30 km/h
Original question 30 km/h × 7 = 210
30 +12 = 42× 5 = 210
Because he is very short for his height he can press upto 18th floor button only
(1/2 + 1/2) = 1
2(1 + 1/2) = 3
2(3 + 1/2) = 7
2(7 + 1/2) = 15
2(15 + 1/2) = 31
1^1,2^2,3^3,4:^4,5^5,6^6
1,4,27,256,3125,46656
50 each
3x -y = 100
4x-2y = 100
– 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
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