13 years ago, a father was five times as old as his son that time. at present, the father's age is 11 years more than two times his son's age. what is the sum of their present ages? (in years)
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Let the present age of son be x years
therefore, the age of father=2x+11 years
Son's age 13 years ago= x-13
father's age 13 years ago= (2x+11)-13
Acc.
(2x+11)-13=5(x-13)
or, 2x+11-13=5x-65
or, 2x-2=5x-65
or, 5x-65=2x-2
or, 5x-2x=-2+65
or, 3x=63
or, x=63/3= 21.
So, son's age= 21
Father's age=2*21+11=53
Therefore, the sum of their ages= 21+53 = 79 years (Ans)
therefore, the age of father=2x+11 years
Son's age 13 years ago= x-13
father's age 13 years ago= (2x+11)-13
Acc.
(2x+11)-13=5(x-13)
or, 2x+11-13=5x-65
or, 2x-2=5x-65
or, 5x-65=2x-2
or, 5x-2x=-2+65
or, 3x=63
or, x=63/3= 21.
So, son's age= 21
Father's age=2*21+11=53
Therefore, the sum of their ages= 21+53 = 79 years (Ans)
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