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During a 6·day local trade show, the least number of people 13 Dec 2012, 07:42
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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.
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During a 6·day local trade show, the least number of people 13 Dec 2012, 07:49
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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 ?

The question basically asks whether more than 6*90=540 people registered during the show.

(1) For the 4 days with the greatest number of people registered, the average (arithmetic mean) number registered per day was 100 --> in these 4 days total of 4*100=400 people registered. Since the minimum number of people registered on the other two days is 80*2=160, then the total number of people registered is at least 400+160=560. Sufficient.

(2) For the 3 days with the smallest number of people registered, the average (arithmetic mean) number registered per day was 85. Clearly insufficient. Consider: {85, 85, 85, 86, 86, 86} and {85, 85, 85, 1,000, 1,000, 1,000}.

Answer: A.
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During a 6·day local trade show, the least number of people 13 Jan 2014, 07:59
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Walkabout
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.
Show more


Basically, they're both telling us that there's AT LEAST 480 people in total, for all days, and they're asking us if there were more than 540 people registered, in total.

1) 4* 100 = 400, + the smallest value which is 80. And since we know that the 6:th value has to be greater than 80, then we know that more than 540 people were registered. So 1) is sufficient

2) This only tells us that more than 255 people were registered, but this gives us no further insight in if total people registered is > 540.. So insufficient

Answer is A
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During a 6·day local trade show, the least number of people 19 Sep 2016, 11:02
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The question says that the least number of people registered in a single day was 80. So do we have to consider 80 to be the least value on only one day or all 6 days ?
The solution given in OG takes it for one day, 80+a+b+c+d+e/6.

Please guide me. I think I am a little lost in the wording of the question.
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During a 6·day local trade show, the least number of people 19 Sep 2016, 14:51
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Krishna2016
The question says that the least number of people registered in a single day was 80. So do we have to consider 80 to be the least value on only one day or all 6 days ?
The solution given in OG takes it for one day, 80+a+b+c+d+e/6.

Please guide me. I think I am a little lost in the wording of the question.
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It actually doesn't matter! All you know is that the lowest number of people to ever register was 80. So, there was at least one day where exactly 80 people registered, and there were no days where fewer than 80 people registered. The other days could have had 80 registrants, or any number of registrants greater than 80. That's why the other five days are represented as variables - you can't really say much about them, until you get to the statements.

You could argue about whether 'least' implies that there was exactly one day with 80 registrants, but it isn't important to the solution.

And if they wanted to specifically say that exactly 80 people registered on every day, they'd have to state that clearly. Since they don't, you can't make that assumption.
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During a 6·day local trade show, the least number of people 19 Jul 2023, 20:42
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Bunuel
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 ?

The question basically asks whether more than 6*90=540 people registered during the show.

(1) For the 4 days with the greatest number of people registered, the average (arithmetic mean) number registered per day was 100 --> in these 4 days total of 4*100=400 people registered. Since the minimum number of people registered on the other two days is 80*2=160, then the total number of people registered is at least 400+160=560. Sufficient.

(2) For the 3 days with the smallest number of people registered, the average (arithmetic mean) number registered per day was 85. Clearly insufficient. Consider: {85, 85, 85, 86, 86, 86} and {85, 85, 85, 1,000, 1,000, 1,000}.

Answer: A.
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Expanding on Statement 2:

The average of number of people registered per day = 85, but the question stems says the minimum number in any single day was 80. So the least value for one out of the 3 days can be 80.

In that case : Average of 3 days = 3x85 = 255, so the sum of number of people in the 2 remaining days = 255-80 = 175.

This remaining number of 175 people for Day 2 and 3 can be split in 2 ways:
Min possible number of people registered for Day 2 = 80, so Day 3= 95
Max possible number of people registered for Day 2 = (175/2) rounded down to 87, so Day 3 = 88

So the range of possible cases for the 3 days with the smallest number of people registered becomes:
{80, 80, 95} and {80, 87, 88}

Remaining Day 4 to Day 6 will have at least as many people as Day 3. So the minimum possible number of people registered for Day 4, 5 and 6 will be at least 88.
(80, 87, 88, 88, 88, 88}

This gives a total of = 519. So the minimum possible number of people registered for all 6 days >= 519

As Bunuel rightly pointed out, the question basically asks whether more than 540 people registered during the show.

Statement 2 is not sufficient!
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During a 6·day local trade show, the least number of people 20 Jul 2023, 01:04
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Krishna2016
The question says that the least number of people registered in a single day was 80. So do we have to consider 80 to be the least value on only one day or all 6 days ?
The solution given in OG takes it for one day, 80+a+b+c+d+e/6.

Please guide me. I think I am a little lost in the wording of the question.
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I chose C at first and then when I redid the question after knowing the answer now I understand.
The least day = 80
Four greatest days = 400 with a mean of 100

Therefore, the one day that left (let's call it x) must vary between 80 and 100 (if it's more than 100 than it must be one of those 4 greatest days, while we are now looking for the second least day).

In scenario when x = 80, you see the result in other comments in this topic
In scenario when x = 100, the mean of 6 days is (400+100+80)/6 = 580/6 = 8x (I stopped calculating when I see the tenth number being 8).

Then in all scenario, the mean is always less than 90.

Hope this helps
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During a 6·day local trade show, the least number of people 16 Mar 2025, 07:24
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using weighted averages -
correct A - if we take 4 max days with avg. of 100 and then remaining 2 days with minimum of 80 then also if the weights (days here) were equal the weighted avg. will be 90 but here the weight of 100 is more than 80 (4>2) so weighted avg. will shift more towards 100 so it will be more than 90 hence sufficient.
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