guerrero wrote:A new electric car company holds a limited-time sales event. On the first day, 3 cars are sold. On each subsequent day of the event, 3 more cars are sold than on the previous day. For how many days does the event last?
(1) If the event had lasted 2 more days, the average number of cars sold per day would have increased by 3.
(2) On exactly 9 days during the event, the number of cars for the day is a multiple of 9.
kindly help me tackle this problem.thanks!
The number of cars sold increases by 3 each day:
3, 6, 9, 12, 15...
The result is an EVENLY SPACED SET.
Statement 1: If the event had lasted 2 more days, the average number of cars sold per day would have increased by 3.
3
3,
6, 9
3, 6,
9, 12, 15
3, 6, 9,
12, 15, 18, 21
With an evenly spaced set, average = median.
As shown here, EVERY TIME two more days are included, the median -- and thus the average -- increases by 3.
Thus, statement 1 is IRRELEVANT.
ANY NUMBER OF DAYS will satisfy statement 1.
If the event lasts 1 day, then adding 2 more days will increase the average number of cars sold per day by 3.
If the event lasts 2 days, then adding 2 more days will increase the average number of cars sold per day by 3.
If the event lasts 3 days, then adding 2 more days will increase the average number of cars sold per day by 3.
And so on.
Since statement 1 does not constrain the number of days in any way, eliminate A, D, and C.
Statement 2: On exactly 9 days during the event, the number of cars for the day is a multiple of 9.
The number of cars sold per day could look like this:
3, 6, 9...75, 78, 81.
The list above will include exactly 9 multiples of 9:
9, 18, 27, 36, 45, 54, 63, 72, 81.
But if the event lasts one more day, it will still be true that on exactly 9 days the number of cars sold per day will be a multiple of 9:
3, 6, 9...75, 78, 81...84.
Since the number of days can be different values, INSUFFICIENT.
Eliminate B.
The correct answer is
E.
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