A man is 37 years old and his two sons are 8 years and 3 years old respectively. After how many years will he be twice as old as the sum of the ages of his sons?
Answers
Answer:
5 years
Step-by-step explanation
Given man’s age 37 and his elder son’s age is 8, So age difference is ED =37–8 = 29 between the man and his elder son. And his younger son’s age is 3, age difference is YD= 37–3 = 34 between the man and his younger son.
Let x be the required man’s which is such that x equals twice as old as the sum of the age of his sons and note that differences remain same always
x = 2 * [(x - ED) + (x - YD)] = 2 * [ 2x - ED - YD ]
x = 2 * [ 2x -29 -34] = 2 * [ 2x - 63]
x = 4x - 126
3x = 126 => x = 42
Thus after (42–37 = ) 5 years he becomes twice as old as the sum of the age of his sons.
Answer:
Age of man: 37
Age of son 1: 8
Age of son 2: 3
Age of man at given condition: 37+x
Sum of age of sons at given condition: 8+x + 3+x
It is to be noted that the age of father is twice the sum of age of sons.
Therefore
37+x = 2( (8+x) + (3+x) )
37+x = 2( 11+2x )
37+x = 22+4x
3x = 15
x=5
Hence after 5 years your condition will be the answer.