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During a 6·day local trade show, the least number of people [#permalink]

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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.

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}.

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.

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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Re: During a 6·day local trade show, the least number of people [#permalink]

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13 Jan 2014, 07:59

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Walkabout wrote:

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.

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

Re: During a 6·day local trade show, the least number of people [#permalink]

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27 Dec 2012, 08:50

(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.[/quote]

Bunuel,

From my understanding. It says the first 3 days the avg is 85. IT does not talk about the other 3 days. It can vary from 80 to anything. So is there any specific reason you wrote {85, 85, 85, 86, 86, 86}??
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Re: During a 6·day local trade show, the least number of people [#permalink]

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20 Feb 2013, 19:40

rajathpanta wrote:

(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.

Bunuel,

From my understanding. It says the first 3 days the avg is 85. IT does not talk about the other 3 days. It can vary from 80 to anything. So is there any specific reason you wrote {85, 85, 85, 86, 86, 86}??[/quote]

Hi,

I might be able to help you there. You were right in saying that the values can range from 85 to anything. However, it cannot be 85 since three days with the least value have already been mentioned. Let's take the immediate bigger value=86. Similarly since it can range to anything you can take 100,000 as the upper limit. This might be the reason that Bunuel took those examples.

In such questions, it might help you to test extreme cases or cases that might result in the option yielding both yes and no answers.

During a 6-day local trade show, the least number of people registered [#permalink]

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10 Dec 2014, 18:02

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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Re: During a 6-day local trade show, the least number of people registered [#permalink]

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10 Dec 2014, 18:15

it has been given that least number of people registered per day was 80 Option 1: The total number of people registered for 4 days will be 400. So the average will be 400+80(least no of people)+81(>80)/6=561(>)/6=90+(>) .Sufficient

Option 2: The total number of people registered for 3 days with least people is 255. total number of people for 2 days is 255-80=175 leaves us the option of 87 and 88 people attending for other 2 days.(keeping the maximum of 2 days minimum ). Assume other 3 days attendance was 89 per each day or 95 per each day. Since different value yield a different result this option is not sufficient

Re: During a 6-day local trade show, the least number of people registered [#permalink]

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10 Dec 2014, 19:52

kukr007 wrote:

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.

The question "Was the average (arithmetic mean) number of people registered per day for the 6 days greater than 90?" can be rewritten as "Was the total number of people registered for the 6 days greater than 540?".

Statement 1: the total number of people during the 4 days was 400. The least number of people in a single day was 80. The total number of people during the 5 days was 480. The 6th day has to be between 80 and 100. Therefore, total sum ranges from 560 to 580, both of which greater than 540. Statement 1 is sufficient.

Statement 2: the total number of people during the 3 days was 255 and this figure includes the day with the least number of people. The sum of the number of people during the remaining 3 days must be greater than 285, meaning, on average, there should be more than 95 people attending the the local trade show per day. Note that the total number of people during the 3 days was 255. If day with the least number of people was excluded, the sum of the number of people during the 2 days would be 175, implying an average of 87.5 people per day or more. This number could be either greater or lower than 95. Therefore, statement 2 is insufficient.

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.

Merging topics. Please search before posting. Thank you.
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Re: During a 6·day local trade show, the least number of people [#permalink]

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20 Aug 2015, 14:05

Bunuel wrote:

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.

Hi Bunuel, in your answers I have marked 85 because it cannot be 85.. the minimum value is 80, in this case we're left with 175 (3*85-80) and we have 2 scenarios for statement 2: 1. (min) [80,87,88 then the values for each of the remaining 3 days =>88 --> {80,81,88,88,88,88} Average = 85,5 2. (max) {80,81,94,1000,1000,1000 --> Average >90
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Re: During a 6·day local trade show, the least number of people [#permalink]

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19 Sep 2016, 11:02

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.

Re: During a 6·day local trade show, the least number of people [#permalink]

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20 Sep 2016, 23:03

Thanks ccooley. Now I understand the rationale behind taking 80 as the least number of registrations for all 6 days. As 80 is the least number to ever register, the sum of registrations on other five days would be greater than or equal to 400.

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Re: During a 6·day local trade show, the least number of people [#permalink]

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30 Jun 2017, 04:54

Bunuel wrote:

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.

Bunuel For the 2nd statement , We can't consider the values as {85, 85, 85, 86, 86, 86} Since the least value is given as 80.

And average of least 3 is 85. So sum of other 2 numbers = 85 x 3 - 80 = 175..

So we can take the number {80, 87, 88, 89, 89, 89}. Also {80, 87, 88, 1000, 1000,1000} . So insufficient.

We need to take care of the minimum number of people as 80 in a single day and then solve the problem, otherwise we might end up with a wrong answer...
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Though it does not impact the final answer in any way,I too agree with BrainLab and shashankism here.

IMHO too the Highlighted portion below cannot be a possible set when taking S2. This is due to the limitation set in the stem on the LEAST possible value in the set (80).

I reasoned it this way. Please correct me if I am wrong. To get the minimum possible values for the remaining 3 days lets consider the set to be {80, 87, 88, x, y, z}. Here x,y,z cannot be 80 or 87. Else, the average of the 3 days with least attendance will no longer be 85. The lowest possible value for x,y,z is x = y = z = 88. The smallest possible set then becomes {80, 87, 88, 88, 88, 88} in this case the average is less then 90.

On the other extreme of course the set can be something like {80, 81, 94, 1,000, 10000, 10000} and the average could be greater than 90. Two answer for S2 and hence insuff.

Please correct me if I am wrong in my reasoning. If not, maybe its better to modify the highlighted statement. Some people like me may get confused.

Bunuel wrote:

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}.