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DS-OG 12 sum no 116

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DS-OG 12 sum no 116

by mitaliisrani » Mon Mar 22, 2010 5:00 am
During a 6-day local trade show, the least number of people registered in a single day was 80. Was the average 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 number registered per day was 100.
2) For the 3 days with smallest number of people registered, the average number registered per day was 85

OA: A

This is a DS OG12 sum no 116...please help to solve

Thanxxx!!!
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Source: — Data Sufficiency |
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by scoobydooby » Mon Mar 22, 2010 5:22 am
1) the number registered for the 4 days with the greatest number: (4*100)=400. the minimum registrations for the other 2 days : (80*2=160) given that the minimum on any day is 80
the min average registration works out to (400+160)/6>90
sufficient


2) the number of registrations for the 3 days with the smallest numbers : 85*3=255.
the other 3 days can have any number of registrations. if they have 86 each, the overall average: (255+258)/6=<90
if the other 3 days have 100 each, overall average =(255+300)/6 >90
not sufficient

hence, A
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by kstv » Mon Mar 22, 2010 5:27 am
The least registration in any day = 80
1) Avg. 4 highest Registration = 100 Total of these 4 days = 400
The second lowest Registration is > 80. cos lowest is 80.
Total of 6 days will be > 560 (400+80+ a value > 80 )
So avg of 6 days > 90 cos 540/6 is 90
Here the total is > 560 Sufficient
2.) Lowest was 80
avg of three lowest is 85 , total of three days is 85*3
we need the total of 6 days
Insuff.

IMO A
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by sanju09 » Mon Mar 22, 2010 5:31 am
mitaliisrani wrote:During a 6-day local trade show, the least number of people registered in a single day was 80. Was the average 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 number registered per day was 100.
2) For the 3 days with smallest number of people registered, the average number registered per day was 85

OA: A

This is a DS OG12 sum no 116...please help to solve

Thanxxx!!!
(1) If sum of registrations for these best 4 days is 400, then the other two less fortunate days would total to greater than 160, thus making the 6 days' total to be greater than 560, whose average would no doubt be greater than 560/6 or greater than 90. Sufficient

(2) If sum of registrations for these best 3 days is 255, then the other three less fortunate days would total to greater than 240, thus making the 6 days' total to be greater than 495, whose average would be greater than 495/6, which is insecure to be greater than 90. Insufficient

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by mitaliisrani » Mon Mar 22, 2010 5:36 am
mitaliisrani wrote:During a 6-day local trade show, the least number of people registered in a single day was 80. Was the average 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 number registered per day was 100.
2) For the 3 days with smallest number of people registered, the average number registered per day was 85

OA: A

This is a DS OG12 sum no 116...please help to solve

Thanxxx!!!
The problem was solved in some of the previous post too..please check url https://www.beatthegmat.com/average-arit ... 15640.html
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