4 days
When none of the digits are repeated:
The hundred’s place can be filled by any of the digits: 2, 3, 5, 6, 7 or 8 except the one which has already been used at the thousand’s place, so it can be filled in 5 ways.
Similarly tens’ place can be filled in 4 ways: only those 4 numbers which have not been use either at hundred’s or thousand’s place.
Unit’s place can be filled in only 3 ways. So, total number of nos. Possible =4×5×4×3= 240
p = prob(1st digit not 7)*prob(2nd digit not 7)*prob(3rd
digit not 7)
=(8/9)*(9/10)*(9/10)
=0.72
Sandhya had it right except that there are 900 numbers
between 100 and 999 inclusive (not 899).
brother and sister
white
30 Sec
22
Soldier have to move 0.75 miles to West and then 0.375 miles to South to reach the camp.
How:-
Firstly soldier moving 1 mile to East from camp and then 1/2 miles= 0.50 miles to North .
Then it is moving 1/4 miles =0.25 miles to west
And then 1/8 miles =0.125 miles to South
Now we just have to count the difference and minus the value of camp to East with the value of North to west = 1mile – 0.25miles(1/4) = 0.75 miles
Same case with north and south = 0.50 miles(1/2) – 0.125 miles(1/8) = 0.375 miles
Hence proves to return camp the soldier have to move 0.75 miles to west and 0.375 miles to south
– 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
100
Ans : – 215
Explanation :-
1^3 = 1-1=0
2^3 = 8-1 = 7
3^3 = 27-1=26
4^3 = 64 – 1 = 63
5^3 = 125-1 = 124
Hence, Ans is
6^3 = 216-1=215
8/15 left.
1/15 * 4 + 1/20 * 4 = 7/15
1 – 7/15 = 8/15
EMNXDS
speed of the train respect to man
= (63 – 3) km/hr
= 60 km/hr
= 60 * 1000 / 3600 m/sec
= 50/3 m/sec
time
= distance/speed
= 500 * 3/ 50
= 30 sec
90 hours