Three years hence a man's age will be three times his son's age,and 7 years ago he was seven times as old as his son. How old are they now.
Answers
Step-by-step explanation:
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Present age of son is 12 years and present age of father is 42.
Given :
Three years hence a man's age will be = three times his son's age
7 years ago he was = seven times as old as his son
To find :
Their present ages
Solution :
Let the present age of son be x years
Hence,
3 years later
Age of son = (x + 3) years
Age of man = 3 (x + 3) years
= 3x + 9 years
7 years ago
Age of son = x - 7 years
Age of man = 7(x-7) years
= 7x - 49 years
Difference in 3years later and 7years ago = 7+3 = 10 years
According to the problem,
3x + 9 = 7x - 49 + 10
=> 3x + 9 = 7x - 39
=> 3x - 7x = -39 - 9
=> -4x = -48
=> x = -48/(-4)
=> x = 12
Son is 12 years old now
According to the first problem, 3 years later , age of father is three times of that of son
Hence, age of father 3 years later = (12+3)×3
= 15 × 3
= 45
Difference in ages = 45 - 15
= 30
Hence, father's present age = 12 + 30
= 42
Hence, present age of son is 12 years and present age of father is 42.
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