Mathematics
Five years ago, Mohit was thrice of Manish. 10 years later, Mohit will be twice as old an Manish. Find their present ages.
Linear Eqns One Variable
65 Likes
Answer
Let the age of Manish be x.
According to the question, five years ago, Mohit was three times as old as Manish.
⇒ Mohit's age - 5 years = 3 x (Manish's age - 5 years)
⇒ Mohit's age - 5 = 3(x - 5)
⇒ Mohit's age - 5 = 3x - 15
⇒ Mohit's age = 3x - 15 + 5
⇒ Mohit's age = 3x - 10 ……………(1)
And,
Ten years later, Mohit will be twice as old as Manish.
⇒ Mohit's age + 10 years = 2 x (Manish's age + 10 years)
⇒ Mohit's age + 10 = 2(x + 10)
⇒ Mohit's age + 10 = 2x + 20
⇒ Mohit's age = 2x + 20 - 10
⇒ Mohit's age = 2x + 10
From equation (1),
⇒ 3x - 10 = 2x + 10
⇒ 3x - 2x = 10 + 10
⇒ x = 20
Mohit's age = 3x - 10
= 3 20 - 10
= 60 - 10
= 50
Hence, the present age of Mohit is 50 years, and the present age of Manish is 20 years.
Answered By
15 Likes
Related Questions
The numerator of a fraction is 5 less than its denominator. If 3 is added to the numerator and denominator both, the fraction becomes . Find the original fraction.
The difference between two numbers is 36. The quotient when one number is divided by other is 4. Find the two numbers.
The denominator of a fraction is 3 more its numerator. If the numerator is increased by 7 and the denominator is decreased by 2, the fraction is 2. Find the sum of the numerator and the denominator of the fraction.
The digits of a two-digit number differ by 3. If the digits are interchanged and the resulting number is added to the original number the result is 143. Find the original number.