GATE Questions

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All the above listed are the System calls for process management

Methods for selecting 4 questions out of 5 in the first section = 5 x 4 x 3 x 2 x 1/4 x 3 x 2 x 1 = 5. Similarly for other 2 sections also i.e 5 and 5

So total methods = 5 x 5 x 5 = 125.

In the given series 1 4 9 16 22 36

1 = 1 x 1 

4 = 2 x 2

9 = 3 x 3

16 = 4 x 4

25 = 5 x 5 (Not 22)

36 = 6 x 6

Hence, the odd man in the series is 22.

At least one black can be chosen in two ways from each box:
Now, probability of choosing at least one black ball from first box 

= 1/2 × [ (2C1× 2C1)/4C2 + 2C2/ 4C2 ]= 5/12

Probability of choosing at least one black ball from 2nd box
= 1/2 × [(4C1×12C1)/16C2 + 4C2/ 16C2] = 9/40

Final probability

= 5/12 + 9/40 = (50 + 27)/120 = 77/120

To find the LCM of 24, 36 and 40

24 = 2 x 2 x 2 x 3

36 = 2 x 2 x 3 x 3

40 = 2 x 2 x 2 x 5 

Now, LCM of 24, 36 and 40 = 2 x 2 x 2 x 3 x 3 x 5

= 8 x 9 x 5

= 72 x 5

= 360

It is used to get parent process identifier

Total number of possible ways = 9C3 = 84 ways

Required atleast one girl in the group of three = total possible ways - ways in which none is girl

None of the members in the group is girl = 6C3 = 20

Therefore, number of ways that at least one member is a girl in the group of three

= 84 - 20

= 64 ways.

We know that,

$$\begin{gathered} S.I. = PTR100 \\ ⇒ P = S.I x 100T x R \\ And also, \\ C.I = P1 + R100n- 1 \end{gathered}$$

From given data, P = Rs. 8625

Now, C.I  = $$\begin{gathered} 86251+41002-1 \\ = 862526252-1 \\ = 8625676625-1 \\ = 8625 x 51625 \\ C.I = Rs. 703.80 \end{gathered}$$