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116. During a 6-day local trade show, the least number of people registered in a single day was 80. Was the average (arithmetic mean) number of people registered per day for the 6 days greater than 90 ?
(1) For the 4 days with the greatest number of people registered, the average (arithmetic mean) number registered per day was 100.
(2) For the 3 days with the smallest number of people registered, the average (arithmetic mean) number registered per day was 85.
附OG的解释：Let a, b, c, d, and e be the numbers of people registered for the other 5 days, listed in increasing order. Determining if （80+ a + b + c + d + e ）/6> 90
is equivalent to determining if (80 + a + b + c + d + e) > (6)(90) = 540, or if a + b + c + d + e > 460.
(1) Given that( b + c + d + e)/4 = 100, then b + c + d + e = 400. Th erefore, since a ≥ 80 (because 80 is the least of the 6 daily
registration numbers), it follows that a + b + c + d + e ≥ 80 + 400 = 480, and hence a + b + c + d + e > 460; SUFFICIENT.
(2) Given that (80+ a + b）/3 = 85, then 80 + a + b =(3)(85), or a + b = 175. Note that this is possible with each of a and b being an
integer that is at least 80, such as a = 87 and b = 88. From a + b = 175, the condition a + b + c + d + e > 460 is equivalent to
175 + c + d + e > 460, or c + d + e > 285. However, using 3 integers that are each at least 88 (recall that the values of c, d, and
e must be at least the value of b), it is possible for c + d + e > 285 to hold (for example, c = d = e = 100) and it is possible
for c + d + e > 285 not to hold (for example, c = d = e = 90); NOT suffi cient.
Th e correct answer is A; statement 1 alone is suffi cient.
如题，关于第二个条件的解释我看不太懂，麻烦各位大侠帮个忙。。。谢谢！ |
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发表于 2010-11-27 12:54:36
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