At a certain auto dealership, 500 cars were sold last week.

why did you assume the other days would have 74,73 and 72?

We can have something like 77 + 76 + 75 + 75 + 75 + 75 = 453. Hence, Friday could have 47 cars sold as well.

Please correct me if I am missing anything.

Question states that different number of cars are sold on each of the seven days.. So you can't assume same number of cars to be sold on 4 days.

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For these types of questions, the statement: a different number of cars were sold on each of the seven days is very important.


(1) On Monday, the dealership sold 75 cars, which was the third-highest number of cars sold on any day that week.

The majority of cars could be sold on the top 3 days, leaving Friday with 0 cars sold. Alternately, Friday could have sold more than 50 if it were the 4th highest.

Insufficient

(2) On Saturday, the dealership sold 77 cars, which was the highest number of cars sold on any day that week.

We want Friday to be at the lowest possible, so have all other days at the maximum possible. To obtain the maximum sales:
Take the sum of the 6 consecutive integers counting down from 77:
\(77 + 76 + \ldots +72 = 6 \times \frac{1}{2} \times (77 + 72) = 447\)

Therefore the minimum number of cars sold on Friday is 53

Sufficient

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.

A) M = 75, from the other 5 days
let 2nd highest be 76 and highest be 77

take the 6 days other than fri as (in increasing order) : 72, 73, 74, 75, 76, 77
Total is less than 250, hence Fri can be more than 50, but need not be. If the max number of car sold exceeds say its 100, and the rest remain the same, Fri will be less than 50 ----> not sufficient

B) as proved above, if highest is 77 then least is more than 50. -----> sufficient (B)

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.

For these types of max/min questions, is the approach to number picking choosing those numbers that cluster around the max? e.g. in this case

77 is surrounded by 75,76, etc.?

i.e. what is an ideal starting point?

CEdward, when you're trying to minimise an entity (in this case - the number of cars sold on Friday), it generally helps to maximise the other entities.

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