A man has 2 daughters and one son. The sum of the ages of his 2 daughters and one son is equal to the age of the father. In 15 years, the sum of the ages of his children will be one and half times their father’s age. What is the father’s age now?

# A man has 2 daughters and one son. The sum of the ages of his 2 daughters and one son is equal to the age of the father. In 15 years, the sum of the ages of his children will be one and half times their father's age. What is the father's age now?

1. A
45 years
2. B
44 years
3. C
46 years
4. D
47 years

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### Solution:

Given that the sum of age of 2 daughters and one son is equal to the age of father and after 15 years sum of ages of children will be $1\frac{1}{2}$ times the age of father.
Let the present age of the father be x years and that of sum of all his children be y years.
According to the first condition,
$⇒x=y$……..(1)
Since the man has three children, so according to the second condition,
$⇒1\frac{1}{2}\left(x+15\right)=y+15×\left(3\right)$   ……(2)
Substituting y for x into equation 2,
$⇒1.5\left(y+15\right)=y+15×\left(3\right)$ $⇒1.5y+22.5=y+45$ $⇒0.5y=22.5$
$⇒y=45$
Substituting 45 for y into equation 1,
$⇒x=45$
Therefore, the present age of the father is 45 years.
Hence, option (1) is correct.

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