Job Roles

Number of students behind Manoj in rank = (31 - 7) = 24.

So, Manoj is 25th from the bottom.

Number of students behind Satya in rank = (31 - 11) = 20.

So, Satya is 21st from the bottom.

A+7B=112

it is clear that A = 14, then it becomes A>=B, but A is smallest angle) given
so, range of A is 0.0001 to 13.9999 ( I am taking upto 4 decimal places)
so, range of B becomes 14 to 16 ( after rounding off to 4 decimal places)
so, range of C becomes 152 to 164 ( after rounding off to 4 decimal places)

OKG
Remaining all are not in series
only OKG is in series of(15,11,7)

Here it follows the random difference of '4'.

For the given problem there are three possiblities.
Father's age can be = 36, 31, 29
Mother's age = 32, 27, 25
Son's age = 8, 3, 1.

By using only B it can be calculated.

For the first time since Nagaland officially gained recognition as a state, a state Police Officer AtuZumvu has been awarded one of the nations' military honour, ‘Shaurya Chakra’.

From the given information; 

Members   Sex    Profession  

     A         M          Advocate
     Q         F           Doctor
     B         M          Advocate
     R         F           Advocate
     C         M          Doctor
     S         F           Doctor
     D         M          Doctor
     T         F           Teacher
     E         M          Teacher
     U         F           Teacher
     P         F           Doctor

 Answer (D)

 From option (D) A and B are two male advocates, P and S are two female doctors and u is female teacher.

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.

Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.

Hence, we can select 1 black ball from 4 black balls
or 2 black balls from 4 black balls.
or 3 black balls from 4 black balls.
or 4 black balls from 4 black balls.

Hence, number of ways in which we can select the black balls

= 4C1 + 4C2 + 4C3 + 4C4
= 24-1 ........(A)

Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.

Hence, we can select 1 red ball from 3 red balls
or 2 red balls from 3 red balls
or 3 red balls from 3 red balls

Hence, number of ways in which we can select the red balls
= 3C1 + 3C2 + 3C3
=23-1........(B)

Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)
or 1 blue ball from 5 blue balls
or 2 blue balls from 5 blue balls
or 3 blue balls from 5 blue balls
or 4 blue balls from 5 blue balls
or 5 blue balls from 5 blue balls.

Hence, number of ways in which we can select the blue balls
= 5C0 + 5C1 + 5C2 + … + 5C5
= 25..............(C)

From (A), (B) and (C), required number of ways
=  2524-123-1

The series follows the following pattern

5 X 5 – 5 = 20

20 X 4 – 4 = 76

76 X 3 – 3 = 225

225 X 2 – 2 = 448

Hence, the term 225 denoted by (d) completes the series