Question

Ramu's age is 1/6 times of his father's age. The present sum of the ages of Ramu, father and mother is 70. When Ramu's age gets doubled their sum is twice the present sum. Then what is the father's age? Plz explain​

Answer 1

Father =x
son=y

x+2y=70
2x+y=95
solving we get
y=15
x=40


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Answer 2

Data given has some mistakes , Father Age = 35 years with corrected data

Step-by-step explanation:

Let say Ramu Age  = R Years

Ramu's age is 1/6 times of his father's age.

=> Father age = 6*R  years

The present sum of the ages of Ramu, father and mother is 70

=> Mother age = 70 - (6R + R) = 70 - 7R

When Ramu's age gets doubled their sum is twice the present sum.

=> Ramu age = 2R  that means after 2R - R = R years

Fathers age = 6R + R = 7R

Mother age = 70 - 7R + R  = 70 - 6R

2R + 7R + 70 - 6R  = 2 * 70

=> 3R = 70

=> Ramu age = 70/3  Years

Father Age = 6 * 70/3 = 140 Years

There is some mistake in data

as sum of their ages = 70 years hence father age can not be 140 years

Found a similar question

The sum of their (father, mother and son) ages is 70. The father is 6 times as old as the son.When the sum of their ages is twice 70, the father will be twice as old as the son.How old is the Father

Let say Father Present age  = 6R   years

Then Ramu Present Age = (1/6) 6R  = R   Years

The present sum of the ages of Ramu, father and mother is 70.

=> Mother Age  = 70 - 6R - R

= 70 - 7R  years

Let say after X years the sum of their ages is twice 70

=> 3X = 70   => X = 70/3

Father Age = 6R + 70/3

Son age = R + 70/3

6R + 70/3 = 2(R + 70/3)

=> 4R = 70/3

=> R = 35/6

Father Age = 6R = 35 Years

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