Two years ago, Dilip was three times as old as his son and two years hence, twice his age will be equal to five times that of his son. Find their present ages.
Answers
Step-by-step explanation:
2 years ago,
Son's age = x
Dilip's age = 3x
Present age of Son = x+2
Present age of Dilip = 3x+2
After 2 years,
Son's age = x+2+2 = x+4
Dilip's age = 3x+2+2 = 3x+4
A/Q 2(3x+4) = 5(x+4)
6x+8 = 5x+20
x = 12
∴ Present age of son = x+2 = 12+2 = 14
Present age of Dilip = 3x+2 = 3x12+2 = 38
Answer:
Let Dilip's son's age 2 years ago be x years.
Then, Dilip's age 2 years ago = (3x) years.
.·. the son's age 2 years hence = ( x + 4 ) years.
Dilip's age 2 years hence = ( 3x + 4 ) years.
.·. 2 ( 3x + 4 ) = 5 ( x + 4 )
=> 6x + 8 = 5x + 20
=> 6x - 5x = 20 - 8
=> x = 12.
.·. Dilip's son's age 2 years ago = 12 years.
And Dilip's age 2 years ago = 3 × 12 = 36 years.
.·. son's present age = 14 years, and
Dilip's present age
=> 3x + 4
=> 3 × 12 + 4
=> 38 years.
Step-by-step explanation:
As obtained the son's present age is 14 years and Dilip's present age is 38 years.
(i) Son's age 2 years ago
=> 14 - 2
=> 12 years.
Dilip's age 2 years ago
=> 38 - 2
=> 36 years.
.·. 2 years ago , Dilip's age= 3 × (son's age)
Thus, the first condition is verified.
(ii) Son's age 2 years hence
= 14 + 2
= 16 years.
Dilip's age 2 years hence
= 38 + 2
= 40 years.
.·. 2 years hence,
2 × (Dilip's age)= 2 × 40 = 80 years.
5 × (Son's age) = 5 × 16 = 80 years.
.·. 2 years hence,
2 × (Dilip's age) = 5 × (Son's age) .
Thus, the second condition is also verified