Rs.265.80
b
– 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
Relative speed =65-45km/hr
Relative speed=20km/hr
Convert to m/s
S=20*5/18
S=5.555
Length of the faster train =5.55555*18seconds
=100meters
A control using ActiveX technologies. An ActiveX control
can be automatically downloaded and executed by a Web
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Java.
Assuming the youngest has X years, so we have
X+(X+3)+(X+6)+(X+9)+(X+12)=50
5X+30=50
5X=20
X=4
The youngest has 4 years
Ans = 8
Use simple box method
[1][2][2][2]= 1x2x2x2 =8
Logic >
once place have 2 number (2,6)
Tens place have also 2 number (7,3)
So that number is divisible by 4
Now, 2 numbers fixed at once and tens place left 2 numbers which is choice at hundredth place
Now 3 number fix left 1 number which is placed at thousands place
Total number form is 4×4 = 16
But divisible by 4 is 1x2x2x2 =8
They’ll never meet unless some of them stop and other continue to run
Let the total number of matches to played in the tournament be ‘x’.
Given that A county cricket team has won 10 matches and lost 4.
That means total number of matches played = 14.
So,
= > 70% of x = 14
= > (70/100) * x = 14
= > 70x = 1400
= > x = 20.
So, the number of matches to be played = 20.
They have a record of exactly 75% wins = 20 * 75/100
= > 15.
Therefore, the number of matches should the team win = 15 – 10 = 5.
Hope this helps!
360
1/6
15%
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)