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 then twice the age of his daughter then. find their present ages.
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Let the present age of father be x years and the present age of daughter be y years.
For equation 1: According to the question,Ages of father and daughter 2 years ago
[x-2] = 3 ½ * [y-2]
⇒ [x-2] = (7/2)[y-2]
⇒ 2x – 4 = 7y -14
⇒ 2x – 7y + 10 = 0 (1)
For equation 2: According to the question, Ages of father and daughter 6 years hence
[x+6] = 10 + 2[y+6]
⇒ x + 6 = 10 + 2y + 12
⇒ x – 2y – 16 = 0 (2)
On multiplying 2 throughout the eq. (1) and then subtracting the eq. (1) from (2), we get
2x – 4y – 32 = 0
2x – 7y + 10 = 0
3y = 42
∴ y = 14 years
Now, substituting the value of y = 14 in eq. (2), 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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