Problems on Ages Test 1

Given that x+y = 48 + y+ z
x – z= 48 + y – y = 48
z is younger than x by 48 years

Let M’s age 12 years ago = x years.
Then, N’s age 12 years ago = 2x years.
(x + 12) / (2x+ l2) = 4/5
=> x = 4.
So, the total of their present ages =(x + 12 + 2x + 12)
= (3x + 24) = 36 years.

Zainab present age = x
Zainab age is 18 times her daughter’s age.
Daughter’s age = 6.
Therefore, 18 times of 6 = x
18 x 6 = x
x= 108 years = Zainab age

Let’s take the present age of the elder person = x
and the present age of the younger person = x – 32
(x – 12) = 7*(x-32-12)
=> x – 12 = 7x – 308
=> 6x = 296
=> x = 49years 3months

Let the son’s present age be x years.
Then, Michael present age = (x + 32) years
=> (x + 32) + 4 = 2(x + 4)
=> x + 36 = 2x + 8
So, x = 28

Let age of John be y
Smith is 18 years elder than John = (y + 18).
So Smith age 8 years ago = (y + 18 – 8)
John age before 8 years = (y – 8)
8 years ago, Smith is 6 times as old as John
(y + 18 – 8) = 6 (y – 8)
(y + 10) = (6y – 48)
5y = 58
y = 11.6
John age = 11.6years
Smith age = (y + 18) = (11.6 + 18) = 29.6years

Let sons age = x years. Then mothers age = (54 – x)years.
(x-8)(54-x-8) = 6(54- x – 8) hence (x-8) = 6 so x = 14
Their ages are 40 years and 14 years.

Son’s current age = x years.
Then father’s age = (84 – x)years
(84 – x) – 7 = 3(x – 7)
2x=56
x =28; after 7years it will be turn on 35

daughter’s present age = x
At present the father’s age is 7 times that of his daughter
i.e., Father’s present age = 7x.
‘Father’s age 7 years ago was 11 times the age of his daughter’
(7x – 7) = 11 * (x – 7)
x = 17.5 years,
The daughter’s present age = 17.5 years

Let John present age be x
John age before 15 years = (x – 15)
John age after 30 years = (x + 30)
We are given that,
John age after 30 years (x + 30) is 15 times his age 15 years back (x – 15)
Therefore,
(x + 30) = 15 (x – 15)
Solving the equation, we get
x + 30 =15x –225
14x = 255
x = 18.2 years