Nevernevergiveup wrote:
At a certain auto dealership, 500 cars were sold last week. If a different number of cars were sold on each of the seven days, did the dealership sell at least 50 cars on Friday of that week?
(1) On Monday, the dealership sold 75 cars, which was the third-highest number of cars sold on any day that week.
(2) On Saturday, the dealership sold 77 cars, which was the highest number of cars sold on any day that week.
Veritas Prep Official Explanation
To solve these Minimum/Maximum problems, you must organize the information and consider different scenarios. From statement (1) it is given that the 75 cars sold on Monday represents the third-highest number sold that week. To use this information, consider the following scenario that lists possible number of cars sold on each day starting, highest to lowest, with 75 as the third-highest: 200, 100, 75, 60, 35, 20, 10. Since the other numbers of cars could represent any of the remaining days, it is clear that Friday could be a day with less than 50 or more than 50, so this information is not sufficient.
In statement (2) it is given that on Saturday the dealership sold 77 cars, the most number of cars sold on any one day. This leaves 423 cars that must be accounted for on the other days. To see how this number of cars could be distributed, maximize the number of cars sold on five of the remaining six days and see what is left on the day with the least number sold. Since 77 was the most and no two days had the same number sold, that would be 76 + 75 + 74 + 73 + 72, or 370 cars. Since the remaining six days must account for 423 cars, this would leave 53 cars left on the lowest day. In other words, if 77 is the most number of cars sold on one day, then it is impossible to have fewer than 53 cars sold on any one day, because the sum of 500 could not be achieved. The correct answer choice is B; statement (2) ALONE is sufficient.