Problems on Ages Questions

Let Ishika and her daughter present ages be p and q

 From the given data, 

p = 4/3(p - 9)

 => p = 36 years 

Hence, Ishika's present age = 36 years and

Her daughter's present age = 1/6 x 36 = 6 years 

Now, required her daughter’s age three years ago = 6 - 3 = 3 years.

12 year old Manish is three times as old as his brother Rajeev who must be 4 years old.
Diff in ages = 8 years.
when Rajeev will become 8 years old, Manish will be 16 yrs old and he will be twice as old as Rajeev.

Given sum of present ages of A, B, C & D

A + B + C + D = 176

and also given that,

(A-4) : (B-4) : (C-4) : (D-4) = 11 : 9 : 6 : 14

Let the ratio = x

=> A-4 + B-4 + C-4 + D-4 = 11x + 9x + 6x + 14x

=> (A + B + C + D ) - 16 = 40x

=> 176 - 16 = 40x

=> x = 4

=> A = 44 + 4 = 48

=> B = 40

=> C = 28

=> D = 60

Therefore, from the above conclusions

D is eldest with 60 years and C is youngest with 28 years of ages.

At the time of marriage, let Sudeer's age be 4x, then Swetha's age is 3x
12 years after:
Age of Sudeer = 4x+12
Age of Swetha = 3x+12
Now the equation is 3x+12=5/6(4x+12)
Solving equation we get x=6
Hence Swetha got married at the age of 18.

Let the daughter age be 'x'
Therefore, Son age = 5x, Mother age = 25x, Person age = 50x.

Thus we get, x+5x+25x+50x = 81
81x = 81
x = 1

Daughter's age = 1, Son's = 5, Mother's = 25, Person's age = 50.

Let Shilpa's present age is p years

According to question,

p = 4(x+4) – 4(x–4)

p = 4x+16 – 4x+16

p = 32

Shilpa's present age = 32 years

Rajeev is four years older than Shilpa

=> Shilpa's present age = Rajeev's age 4 years before.

=> Rajeev's age 4 years before = 32 years.