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Farah got married 8 years ago, Today her age is $$1\frac{2}{7}$$ times her age at the time of her marriage. At present her daughter's age is one-sixth of her age. What was her daughter's age 3 years ago ?

Correct Answer: 3 Years
Let Farah's age 8 years ago be x years.
Then , her present age = (x + 8)
$$\eqalign{ & \therefore x + 8 = \frac{9}{7}x \cr & \Rightarrow 7x + 56 = 9x \cr & \Rightarrow 2x = 56 \cr & \Rightarrow x = 28 \cr} $$
∴ Farah's age now
= (x + 8) years
= (28 + 8) years
= 36
Her daughter's age now
=$$\left( {\frac{1}{6} \times 36} \right)$$   years
= 6 years
Her daughter's age 3 years ago
= (6 - 3) years
= 3 years

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