The quadratic equation whose roots are reciprocal of 2x2+5x+3=0 can be obtained by replacing x by 1/x.

Hence, 2(1/x)(1/x)+ 5(1/x) + 3 = 0

=> 3x2+5x+2=0

The chemical element with the atomic number 35, Bromine is a non metal that remains liquid at room temperature. Mercury also remains liquid at room temperature but it is a metal. Bromine is the third lightest halogen.

Yielding the floor :  The speaker of the Lok sabha can ask a member of the house to stop speaking and let another member speak.

Crossing the floor:  Passing between the member addressing the House and the Chair which is considered breach of Parliamentary etiquette.

Point of Order:  A Member of Parliament may raise a point of order if he feels that the proceedings of' the House do not follow the normal rules. The presiding officer decides whether the point of order raised by the member should be allowed.

Calling Attention Motion:  With the prior permission of the Speaker, any member of the Parliament may call the attention of a Minister to a matter of urgent public importance. The Minister may make a brief statement about the matter or he may ask for time to make a statement later.

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.

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

Since floor starts form the 1 therefore 2 will be the 1st floor.
Form 1 to 10 only 4,6,8,9,10 are not prime.
Form 11 to 20 only 14 ,16,18 are not prime also non of their digits are prime.
From 21 to 30 each number has a prime number so it must no be consider.
In between 40 to 49 we have to consider only 40,44,46,48,49.
50 to 59 follow either of the condition so we have too neglect it.
In between 60 to 68 60,64,66,68.
total option = 17.

Let us assume the two persons who can speak two languages speak Hindi and Tamil. The third person then speaks all the three languages.

Tamil – Number of persons who can speak is 6. Only Tamil 6 – 2 – 1 = 3

Hindi - Number of persons who can speak is 15. Only Hindi 15 – 2 – 1 12

Gujarati – Number of persons who can speak is 6. Only Gujarati 6 – 1 = 5

Thus the number of persons who can speak only one language is 3 + 12 + 5 = 20

Number of persons who can speak two languages = 2

Number of person who an speak all the languages = 1

Total number of persons = 23.