Discussion
If Ajay drives at 4/5th of his usual speed to his office, he is 6 minutes late. What is his usual time to reach his office?
Answer: B
Let t be his usual time to reach his office and v be his usual speed.
v = d/t ……….(d is the distance Ajay travels while going to his office)
vt = d
At v1 = 4v/5 ; t1 = t + 6
4v/5 = d/(t + 6)
4v/5* (t + 6) = d
4v/5* (t + 6) = vt
On solving we get,
t = 24 minutes
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