Discussion
Father is aged three times more than his son. After 8 years, father would be two and a half times of son's age. After further 8 years, how many times would father be of son's age?
Answer: A
Assume that son's present age \(= x \)
Then, father's present age = \(3x+x=4x\)
After \(8\) years, father's age \(=2\tfrac{1}{2}\) times of son's age.
\(\Rightarrow\) \((4x+8)=2\tfrac{1}{2}(x+8)\)
\(\Rightarrow\) \(4x+8=\tfrac{5}{2}(x+8)\)
\(\Rightarrow\) \(8x+16=5x+40\)
\(\Rightarrow\) \(3x=40-16=24\)
\(\Rightarrow\) \(x=\frac{24}{3}=8\)
After further \(8\) years,
Son's age \( =x+8+8=8+8+8=24\)
Father's age \(=4x+8+8=4×8+8+8=48\)
Father's age/Son's age \(=\frac{48}{24}=2\)
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development