During a 6-day local trade show, the least number of people

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

14 May 2011, 12:08

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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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Re: Local Trade Show [#permalink]

14 May 2011, 12:44

St - 1==> rough estimation, 4 days avg is 100 i.e. 4 * 100 = 400 and 1 day least is given as 80. so considering the least at the last day as 80 also, 400+80+80 = 560
avg of 6 days 560/6 >90; sufficient

St - 2==> given 3 days with smallest # of people registered- avg as 85
rest 3 days could be 86,87,88 people registered or 91,92,93 - no info is provided.
if you consider 86,87,88 than mean of 6 days <90
and if you consider 91,92,93 than mean of 6 days > 90
not sufficient

Answer A.

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Re: Local Trade Show [#permalink]

15 May 2011, 00:25

a for 4 days with greatest number of people - total = 400
5th day it is 80 then total = 480.
for average to be >90 the total for 6 days should be > 540.

540-480 = 60 which is not possible. Thus 6th day must be greater than 80. Hence sufficient.

b total for 3 days with least number of people registered = 3*85 = 255.

remaining = 540-255 which can be distributed in any fashion for the remaining 3 days. Hence not sufficient.

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Re: Local Trade Show [#permalink]

15 May 2011, 00:30

Suppose 6 days are represented by {x1, x2, x3, x4, x5, x6} each >=80

Stmt1: For the 4 days with the greatest number of people registered, the average (arithmetic mean) number registered per day was 100.
Let x3+x4+x5+x6/4 = 100.
x3+x4+x5+x6=400
Lets suppose x1 and x2 to be minimum i.e 80.
So, x1+x2+x3+x4+x5+x6/6 = 80+80+400/6 = 560/6 = 93.3 > 90
Now, Lets suppose x1 and x2 to be more than minimum, 90
So, x1+x2+x3+x4+x5+x6/6 = 90+90+400/6 = 580/6 = 96.6 > 90 Sufficient.

Stmt2: For the 3 days with the smallest number of people registered, the average (arithmetic mean) number registered per day was 85.

Let x1+x2+x3/3 = 85
x1+x2+x3=255
Lets suppose x4,x5 and x6 to be minimum. i.e 80
x1+x2+x3+x4+x5+x6/6= 255+80+80+80/6=495/6=82.5 < 90
Lets suppose x4,x5 and x6 to be more than minimum, 100
x1+x2+x3+x4+x5+x6/6= 255+100+100+100/6=/6=92.5 < 90
Hence two answers. Insufficient.

OA A.
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Re: Local Trade Show [#permalink]

14 Aug 2011, 08:49

Easy way:

Statement (1):

(80+4*100+x)/6 > 90 ?
480+x > 540
x>60

There would be more than 90 people per day if on the 5th day there were at least 60 people. We know the least number of people registered at any day was 80. Sufficient.

Easy to check:
(80+400+60)/6 = 90 --> 540 = 540
(80+400+70)/6 > 90 --> 550 > 540

Statement (2):

(3*85+x+y)/6 > 90?
255+x+y > 540?
x+y > 285?

Do you know? Me neither. Insufficient.

A

Re: Local Trade Show [#permalink] 14 Aug 2011, 08:49

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During a 6-day local trade show, the least number of people

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During a 6-day local trade show, the least number of people

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