Question

7 years ago, a woman had 3 daughters, and the sum of their ages was was exactly half of the woman's age. 2 years ago, she had a 4th daughter. Today, the sum of the ages of her 4 daughters is exactly equal to her age. What is her age today?

Answer 1

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for answer

Answer 2

Answer:

The present age of women is 39 yrs.

Step-by-step explanation:

Given 7 years ago, a woman had 3 daughters, and the sum of their ages was was exactly half of the woman's age. 2 years ago, she had a 4th daughter. Today, the sum of the ages of her 4 daughters is exactly equal to her age. we have to find the present age of women.

Let the present age of women is x yrs and of her 3 daughters are a, b, c

Since,2 years ago, she had a 4th daughter ∴ 4th daughter present age is 2 yrs.

Also, 7 years ago, a woman had 3 daughters, and the sum of their ages was was exactly half of the woman's age.

∴ $(a-7)+(b-7)+(c-7)=\frac{1}{2}(x-7)$

⇒ $2a+2b+2c-42=x-7$

⇒ $2a+2b+2c=x+35$  →  (1)

Now, given at present the sum of the ages of her 4 daughters is exactly equal to her age.

∴ $a+b+c+4th daughter=x$

⇒  $a+b+c+2=x$

⇒ $2a+2b+2c=2x-4$    → (2)

Subtracting (1) from (2), we get

x-90=0 ⇒ x=39

Hence, the present age of women is 39 yrs.