Mhl Interview Questions, Process, and Tips

145.2

Out of 10 persons, 4 are graduates; so, (10 – 4) = 6 are under-graduates.

If there is no restriction, any three can be chosen from the ten in (10C3) = 120 ways.

Now, if all three chosen are under-graduates; it can take place in (6C3) = 20 ways.

Therefore, the probability that there will be no graduate among the three chosen = (20 / 120) = (1 / 6).

Therefore, the probability that there will be at least one graduate among the three chosen = {1 – (1 / 6)} = (5 / 6) = 0.8333.

Maximum number of edges = 9. Start from one corner. Select any face including that corner. Complete a square (4 edges) around the face to reach at the starting corner point . Now, move to the opposite face through the edge joining them and passing through the starting point(1 edge). Now, complete the square of edges around this face(4 edges). Total = 4+1+4 = 9 edges

only if the results are greater than my input

c

D

c

11

9828

40

X = 1/2 * b*h
base incresed by 4 times & height
is devided by 2
X = 1/2 * 4b * h/2
X = h/4 * 4b
X = hb

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– 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

let d distance and s be speed.
d/7 = x and d/5 = x+12
solving we get d=210.

52.5

Suppose that total 100 employees are in company….. out of that 35 are man and remaining 65 are women.
20% of man 35 = 20*35/100
40% of women = 40*65/100
total employees = 7+26 = 33 out of 100
so, Ans = 33 %