Question

10 years ago, a man had 2 children, and the sum of their ages was was exactly half of his age. 7 years ago, he had another child. Today, the sum of the ages of all the children is exactly equal to his age. What is his age today?

Answer 1

Answer:

50 years old after

Answer 2

Final Answer:

The present age (as of today) of the man with three children is 50 years.

Given:

10 years ago, a man had 2 children, and the sum of their ages was exactly half of his age.

7 years ago, he had another child.

Today, the sum of the ages of all the children is exactly equal to his age.

To Find:

The present age of the man with three children, today, is to be found out.

Explanation:

Note that the present age is calculated based on the present date considering the date of birth of the concerned person.

Step 1 of 3

Assume that the present ages (as of today) of the man, his two children born 10 years ago, and the child born 7 years ago, are $m,x,y,z$ respectively.

As of today, the sum of the ages of all the children is exactly equal to his age, so write the following equation.

$x+y+z=m\\2x+2y+2z=2m$

Step 2 of 3

Now, 10 years ago, the man had 2 children, and the sum of their ages was exactly half of his age, so write this equation.

$(x-10)+(y-10)+(z-10)=\frac{(m-10)}{2} \\2(x-10)+2(y-10)+2(z-10)=m-10\\2x+2y+2z-60=m-10$

Step 3 of 3

Deduct the second equation from the first equation.

$(2x+y+2z)-(2x+2y+2z-60)=2m-(m-10)\\2x+y+2z-2x-2y-2z+60=2m-m+10\\m=50$

Therefore, the required present age (as of today) of the referred man with three children, is 50 years.

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https://brainly.in/question/50540762

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