Question

four year ago marina was three times old as her daughter.Six years from now the mother will be twice as old as her daughter.Find their present ages.



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Answer 1

Let the present age of mother and daughter be X and Y year's.

Before four years age of mother = (X-4) years.

Before four years age of daughter = (Y-4) years.

According to question,

(X-4) = 3(Y-4)

X - 4 = 3Y - 12

X - 3Y = -12 + 4

X - 3Y = -8 ---------(1)

After 6 years age of mother = (X+6) years.

After 6 years age of daughter = (Y+6) years.

According to question,

(X+6) = 2(Y+6)

X + 6 = 2Y + 12

X - 2Y = 12-6

X - 2Y = 6 ---------(2)

From equation (1) we get,

X - 3Y = -8

X = -8 + 3Y -----------(3)

Putting the value of X in equation (2)

X - 2Y = 6

-8 + 3Y - 2Y = 6

Y = 6+8

Y = 14 years

Putting the value of Y in equation (3)

X = -8 + 3Y = -8 + 3 × 14 = -8 + 42

X = 34 years

Age of mother = X = 34 years

And,

Age of daughter = Y = 14 years.

Answer 2

Step-by-step explanation:

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