Discussion
If no car in 1993 was involved in more than four accidents, what is the minimum number of cars that could have been in accidents in 1993?
*This question is included in Nova Math - Problem Set W: Graphs, question #20
| (A) | 50 thousand |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 12/19/2012 13:07
Please I need more explanation than the one given
Posted: 12/21/2012 16:29
In 1993, there are 360,000 car accidents, and 12 million cars. Not all cars are involved in accidents. For the cars involved in accidents, the maximum number of accidents per car is 4. If all accident-involved cars have 4 accidents, we have the minimum number of cars that have been in accidents, i.e., 360,000 divided by 4 = 90,000. If some cars have less than 4 accidents, then we will have more than 90k accident-involved cars. I hope this helps.