Two years ago, a man's age was three times the square of his daughter's age . Three years hence , his age will be four times his daughter's age . Find their present ages.

Let present age of daughter be x years Two years before daughter's age = (x - 2) years Man's age two years before = 3(x - 2) 2 So, present age of man = 3(x - 2) 2 + 2 Three year's later Daughter's age = (x + 3) years Man's age = (3(x - 2) 2 + 2 + 3) years According to question, Since, age cannot be in fraction so x ≠ ∴ x = 5, 3(x - 2) 2 + 2 = 3(5 - 2) 2 + 2 = 3(3) 2 + 2 = 29 The present age of daughter is 5 years and of man is 29 years.

If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14cm². However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm². Determine the sides of two squares.

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The length (in cm) of the hypotenuse of a right angled triangle exceeds the length of one side by 2 cm and exceeds twice the length of other side by 1 cm. Find the length of each side. Also find the perimeter and the area of the triangle.

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The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than 3 times the age of his son. Find their present ages.

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Paul is x years old and his father's age is twice the square of Paul's age. Ten years hence, father's age will be four times Paul's age. Find their present ages.

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Mathematics

Two years ago, a man's age was three times the square of his daughter's age . Three years hence , his age will be four times his daughter's age . Find their present ages.

Quadratic Equations

Related Questions

If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14cm². However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm². Determine the sides of two squares.

The length (in cm) of the hypotenuse of a right angled triangle exceeds the length of one side by 2 cm and exceeds twice the length of other side by 1 cm. Find the length of each side. Also find the perimeter and the area of the triangle.

The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than 3 times the age of his son. Find their present ages.

Paul is x years old and his father's age is twice the square of Paul's age. Ten years hence, father's age will be four times Paul's age. Find their present ages.

Mathematics

Two years ago, a man's age was three times the square of his daughter's age . Three years hence , his age will be four times his daughter's age . Find their present ages.

Quadratic Equations

Related Questions

If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14cm2. However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm2. Determine the sides of two squares.

The length (in cm) of the hypotenuse of a right angled triangle exceeds the length of one side by 2 cm and exceeds twice the length of other side by 1 cm. Find the length of each side. Also find the perimeter and the area of the triangle.

The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than 3 times the age of his son. Find their present ages.

Paul is x years old and his father's age is twice the square of Paul's age. Ten years hence, father's age will be four times Paul's age. Find their present ages.

If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14cm². However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm². Determine the sides of two squares.