Let the age of Rahul, Sagar and Purav be x, y and z respectively
According to the given information
Age of Sagar – Age of Rahul = Age of Rahul – Age of Purav

Also y + z = 66 years
From (i) x = 33 years
Also as per equation (i) we have Purav’s age + Sagar age = 66 years
by going through option (A) given Purav = 18, and Rahul = 33 years, Sagar = 48 years
Difference between Rahul’s and Purav’s age = 18 years

A’s age 10 Years ago = x years

 = 3/4

Let the present ages of Sameer and Anand be 5xyearsand4x5xyearsand4x years. respectiely

Then , 5x+34x+35x+34x+3  = 119119 ⇔⇔  9(5x+3)= 11(4x+3)9(5x+3)= 11(4x+3)    ⇔⇔    xx= 6

∴∴  Anand’s present age  = 4x4x  = 24 years.

Mother’s age when Virat’s brother was born = 36 years
Father’s age when Virat’s brother was born = (38 + 4) = 42 years
So, the difference between his parent’s age = (42 – 36) = 6 years

Suppose x years later the father will be twice as old as his son.
Then, x + 41 = 2 (x + 16)
⇒ x = 41 – 32 = 9 years

According to the question,
(36 + n) / (50 + n) = 3/4
⇒ 144 + 4n = 150 + 3n

Let son’s age 10 years ago be x years.
Then, man’s age 10 years ago = 7 years
Son’s present age = (x + 10) years,
Man’s present age = (7x + 10) years

Son’s present age = (x + 10) years
= (4 + 10) years
= 14 years

Let, today’s age of the daughter = ‘A’ years.
Mom’s age today = 3A years
After 1 decade,
3A + 10 = 29A + 10
=> A = 10 years

The ratio of present age of Meena and her son’s age = 3 : 1 (3x, x)

Meena’s present age = (2/5)*Meena’s father’s present age

Meena’s father’s present age = (15/2)x

According to the question,

x + 3x + (15/2) x = 138

(2x + 6x + 15x)/2 = 138

23x = 138*2

x = 12

Required difference = 90 – 12 = 78 years

Let age of the son before $10$ years $=x$ and
age of the father before $10$ years $=3x$

$(3x+20)=2(x+20)\Rightarrow x=20$

Age of the son at present $=x+10=20+10=30$
Age of the father at present $=3x+10=3\times 20+10=70$

Required ratio $=70:30=7:3$

Let the present ages of son and father be x and (60 – x) years respectively.

Total age of 10 students and their teacher = 11 x 15 = 165 years.

Total age of first 7 students = 7 x 15 = 105 years

Total age of 3 students = 3 x 11 = 33 years

∴ Total age of 10 students = 105 + 33 = 138 years

Let A’s age 10 years ago = x years.
Then, B’s age 10 years ago = 2x years.
(x + 10) / (2x+ l0) = 3/4
=> x = 5.
So, the total of their present ages =(x + 10 + 2x + 10)
= (3x + 20) = 35 years.

Currently, the ratio of their ages is 5: 3. Suppose, their ages are: 5x and 3x.
Sara’s age 4 years ago = 5x – 4
Vaishali’s age after 4 years = 3x + 4
Ratio of Sara’s age 4 years ago and Vaishali’s age after 4 years is 1 : 1
Therefore,
(5x – 4) / (3x + 4) = 1/1
Solving, we get x = 4
We are required to find the ratio between Sara’s age 4 years hence and Vaishali’s age 4 years ago.
Sara’s age: (5x + 4)
Vaishali’s age: (3x – 4)
Putting the value of x, we get:
(5x + 4) / (3x – 4) = 3/1 = 3:1

Let the ages of father and son 10 years ago be 3x and x years respectively.

Then, (3x + 10) + 10 = 2[(x + 10) + 10]

 3x + 20 = 2x + 40

 x = 20.

 Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.