This problem requires calculating the average age of the family members at a specific point in the past, just before the youngest member was born.
The family has 6 members with an average age of 25 years.
Current Total Age = Number of Members \(\times\) Average Age
Current Total Age = \(6 \times 25 = 150\) years.
The youngest member is 8 years old. This means 8 years ago, this member was just born (age 0). At that time, the family consisted of the other 5 members.
First, find the current total age of the 5 older members:
Total Age of 5 Older Members (Now) = Current Total Family Age - Age of Youngest Member
Total Age of 5 Older Members (Now) = \(150 - 8 = 142\) years.
Now, calculate their total age 8 years ago. Each of the 5 members was 8 years younger:
Total Age of 5 Older Members (8 years ago) = Total Age of 5 Older Members (Now) - (Number of Older Members \(\times\) 8 years)
Total Age of 5 Older Members (8 years ago) = \(142 - (5 \times 8) = 142 - 40 = 102\) years.
Average Age (8 years ago) = Total Age of 5 Older Members (8 years ago) / Number of Older Members
Average Age (8 years ago) = \(\frac{102}{5} = 20.4\) years.
Therefore, the average age of the family just before the birth of the youngest member was 20.4 years.
The ages of two persons P and Q are in the ratio 5 : 7. Eight years ago, the ratio of P and Q was 7 : 13. The present ages of P and Q, respectively, are:
One year ago, the ratio of the ages of A and B was 4 : 3. The ratio of their ages, after 7 years from now, will be 9 : 7. What is the present age (in years) of B?
In 8 years, Subhash will be 3 times as old as he is now. After how many years will Subhash be 5 times as old as he is now?
The average ages of parents and two children are 30 years and 8 years respectively. The average age of the family is
A. 16 years
B. 19 years
C. 18 years
D. 17 years
A father is presently 3 times his daughter’s age. After 10 years he will be twice as old as her. Find the daughter’s present age.