5/9
Circular Track sis of 11 km
Speed of Mens are
4 , 5.5 , 8 km/hr
Time at Which they are at starting Points again
11/4 , 11/5 , 11/8
11/4 , 2.2 , 11/8
We need to find LCM of these
to find at what time they Meet again at starting point
LCM 11 * 2 = 22
After 22 Hrs they will meet at starting point
12.5
The answer for this question is very simple: it is called Ceaser Cipher.
It consists of replacing a character by another located in (current character position + (key-1)), where key = how many positions you want to skip.
For Example: VICTORY -> YLFWRUB with the key = 3.
Then SUCCESS -> QXFFHQQ
Note: it should work like a circle(Z+1 = A).
I will keep my dog far away from that allergic employee,for that I chain down my dog didn’t leave dog open to avoid moving the dog here there.
4
In order to enhance my career, I asked my colleagues about my strength and weaknesses, the areas where I need to improve more, and acquired more knowledge from my superiors. I worked on my weaknesses to convert into a positive response.
Let the age of the man be x
Then age of his son becomes (x−24)
2 years later from now,
Age of man will be = x+2
and age of his son will be =(x−24+2)=x−22
According to question,
2(x−22)=x+2
i.e., 2x−44=x+2
i.e., x=46
Therefore ,
Present age of man=46 years
And present age of his son =46−24=22 years
– 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
66 seconds
11
Can someone confirm that it is : 5/9 * 4/8 * 3/7 = 11.9% that you choose 3 good bulbs in a row