(d) 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:
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. Thus after (42–37 = ) 5 years he becomes twice as old as the sum of the age of his sons.
Step-by-step explanation:
Answer:
Here, we have to form a linear equation in order to find the value of x or his son's age. So, suitable equation according to the given information is :
Math
5 points
\sf { 5x + 2 = 37 }
➳
\sf { 5x = 37-2 }
➳
Transposing 2 from LHS to RHS. So, its sign changed.
\sf { 5x = 35}
➳
\sf { x = \dfrac{35}{5} }
➳
\sf { x = 7 }
➳
\sf \red { x \: years = 7 \: years }
➳
Therefore, age of his son is 7 years.
Verification:
As the question states that his age is two years more than 5 time his son's age and his present age is 37 years. So,
\sf { 5x + 2 = 37 }
➳
LHS :
→ 5x + 2