Question

Two years ago, the age of a father was three and a half times the age of his
daughter then. Six years hence, the age of father will be ten years more than twice
the age of his daughter then. Find their present ages.
(1) Let the present age of the father be x years and that of his daughter be y years.
(2) Form two equations from the given conditions.
(3) Solve the equations and find the answer.​

Answer 1

Answer:

The present age of the father is 44 years and present age of daughter is 14 years.

Step-by-step explanation:

Let the present age of father be “x” years and the present age of daughter be “y” years.

For equation 1: Ages of father and daughter 2 years ago

According to the question,

[x-2] = 3 ½ * [y-2]

⇒ [x-2] = (7/2)[y-2]

⇒ 2x – 4 = 7y -14

⇒ 2x – 7y + 10 = 0 ……. (i)

For equation 2: Ages of father and daughter 6 years hence

According to the question,

[x+6] = 10 + 2[y+6]

⇒ x + 6 = 10 + 2y + 12

⇒ x – 2y – 16 = 0 ……. (ii)

On multiplying 2 throughout the eq. (ii) and then subtracting the eq. (i) from (ii), we get

2x – 4y – 32 = 0

2x – 7y + 10 = 0

-      +       -

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        3y = 42

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∴ y = 14 years

Now, substituting the value of y = 14 in eq. (ii), we get

x – (2*14) – 16 = 0

⇒ x = 16 + 28

⇒ x = 44 years

Thus, their present age of father is 44 years and of daughter is 14 years.

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