Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1 cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2 i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61,P(B)=61,P(C)=65,P(A)=61,P(B)=65,P(C)=61 and P(A)=65,P(B)=61,P(C)=61
Hence the required probability = 61×61×65×6×3=21690=125
237.6 Kmh
– 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
45
b
55967
30/100*142.85 = approx. 100
so ans is 42.85
2years
Green bricks
let first digit be ‘X’
then 5th digit is ‘3X’
let 2nd digit be ‘Y’
then 3rd digit is ‘Y-3’
and 4th digit is ‘Y+4’
then the no is ‘(X)(Y)(Y-3)(Y+4)(3X)’
from the above we can say 3X<=9
so X<=3 and any of the digit in the number is <=9
and also given that 3 pairs sum is 11...
so make trial and error..
if X=1...any of the no is 10 which is wrong trial....
if X=2...then let Y+4=9 ==> Y=5
then no is 25296
first pair 2+9=11
second pair 2+9=11
third pair 5+6=11
now the answer is 25296
22.5