In the class of 10 boys and 5 girls boys have 4 balls each and girls have 1 ball each what will be the average number of balls in class?

={( number of boys X number of balls)+( number of girls X number of balls)}/ total students

={(10Ã—4)+(5Ã—1)}/10+5

=(40+5)/15

= 3 is the average number of balls in class.

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In following questions, a sequence of groups of alphabets and numbers is provided with one term missing , Choose the missing term from options. 2A11, 4D13, 12G17, ?

The first numbers in the terms follow the sequence x 2, x 3, x 4.The middle letter of each term is moved three steps forward to obtain the corresponding letter of the next term>The last numbers follow the sequence + 2, + 4, + 6.

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E is
the son of A.

D is the son of B.

E is married to C.

C is B's daughter.

How is D related to E?

Option (D)

C is B's daughter and D is B's son. So, D is the brother of C.

E is a male married to C. So, E is the husband of C,

Whose brother is D. So, D is the brother-in-law of E.

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All the 6 member of a family A .B .C .D .E and F are travelling together. B is the son of C but C is not the mother of B ,A and C are a married couple. E is the brother of C. D is the daughter of A. F is the brother of B.

Who is the mother of B?

B is the son of C but C is not the mother of B means C is the father of B. A is married to C means A is the mother of B. F is brother of B means F is son of A and C. D is daughter of A means D is daughter A and C. Clearly, A is the mother of B..

A man has a daughter and a son. The son is is three years older than the daughter. In one year the man will be six time as old as the daughter is now. In ten years the man will be fourteen years older than the combined ages of his children at that time. what is the man's present age?

Let's the age of daughter=x

age f son = x+3

age of father now=6x-1

condition:

M+0={(S+D)+ 10}+14]

6x-1=[{(x+3)+10}+14]

4x=28

x=7

Father's age=(6x-7)-1=41

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