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If ninety percent of the expected number of managers

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If ninety percent of the expected number of managers  [#permalink]

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New post 20 Jun 2011, 23:25
1
7
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

47% (01:22) correct 53% (01:32) wrong based on 176 sessions

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If ninety percent of the expected number of managers attended a conference, how many managers attended the conference?

(1) No more than ninety percent of all managers were expected to attend the conference

(2) 7 managers did not attend the conference

Source: Master Gmat 18 Q in sample test

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Re: managers at a party  [#permalink]

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New post 20 Jun 2011, 23:35
BlueRobin wrote:
If ninety percent of the expected number of managers attended a conference, how many managers attended the conference?

(1) No more than ninety percent of all managers were expected to attend the conference

(2) 7 managers did not attend the conference

Source: Master Gmat 18 Q in sample test


I think the wording are more important here...
90% of expected number of manager

st1: no more than 90% .. can be 50% -- can be 89%......... insufficient
st2: 7 manager dint come ... dont get mistaken by assuming this is 10%( 100- 90%). 90% is of the expected managers... insufficient

1&2 : we still dont know how many managers were expected..
suppose there are 100 managers.. 7 dint come = 93.
we expected 80 to come for conference. the question stem says 90% of these 80 manager came .. = 72.
so this number(72) can change dependending upon other parameters.

hence E
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Re: managers at a party  [#permalink]

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New post 20 Jun 2011, 23:39
1) Max 90% of all manageres expected. If 90% of that maximum showed up it states that max 81% of the expected showed up. But it's a maximum so it could easily also be just 50% we don't know - Insufficient
2) 7 did not attend but that does not give any information on the rest

1+2) Since we don't know the actual number of managers that were really expected to be present, you can't really tell how much managers showed up even though you know 7 managers were not present. For example if it would be exactly 90% that was expected and thus 90% of them showed up, you would know that 81% showed up leaving 19% that didnt and you could relate those 7 managers to that 19% and come up with a total of around 37 managers. Right now, you're not sure so E is the best choice.
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Re: managers at a party  [#permalink]

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New post 05 Sep 2011, 09:48
**
Quote:
If ninety percent of the expected number of managers attended a conference, how many managers attended the conference?

(1) No more than ninety percent of all managers were expected to attend the conference

(2) 7 managers did not attend the conference

Source: Master Gmat 18 Q in sample test


From statement 1
Let total no. of managers be x.
a maximum of 0.9x managers were expected to attend.
at max, 0.9(0.9x) managers attended.
Insufficient

From statement 2
We do not know the number of managers expected to attend. Unable to derive any concrete values with given info.
Insufficient.

Statement 1 + 2
We are not sure how many percent of total managers attended the conference. It can be anywhere between 0% to 90%.
As such, knowing the number of managers who did not attend is insufficient for us to derive answer.
Insufficient.

Answer: E
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Re: managers at a party  [#permalink]

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New post 05 Sep 2011, 09:51
**
Quote:
If ninety percent of the expected number of managers attended a conference, how many managers attended the conference?

(1) No more than ninety percent of all managers were expected to attend the conference

(2) 7 managers did not attend the conference

Source: Master Gmat 18 Q in sample test


From statement 1
no actual values to work with. Unable to derive number of managers who attended.
Insufficient

From statement 2
We do not know the number of managers expected to attend. Unable to derive any concrete values with given info.
Insufficient.

Statement 1 + 2
We are not sure what is the actual percentage/number of managers expected to attend the conference. It can be anywhere between 0% to 90%.
As such, knowing the number of managers who did not attend is insufficient for us to derive answer.
Insufficient.

Answer: E
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Re: managers at a party  [#permalink]

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New post 12 Oct 2011, 19:45
I chose C mistaking that Stat1 is absolute. If Stat1 did not say "No more than", then would I be able to say that 1+2 is sufficient?

Thx
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Re: If ninety percent of the expected number of managers  [#permalink]

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New post 14 Aug 2014, 15:30
1
arcanis2000 wrote:
I chose C mistaking that Stat1 is absolute. If Stat1 did not say "No more than", then would I be able to say that 1+2 is sufficient?

Thx

Dear arcanis2000,
I'm happy to respond. :-)

Because we are dealing with people, it would be impossible for both percents to equal 90% and for the number of managers not coming to equal only 7. You see, when the prompt says, "ninety percent of the expected number of managers attended a conference," that's not 90%, rounded to the nearest whole number. It actually means exactly, mathematically, 90%. In other words, the number of expected managers must be divisible by 10, and 9/10 of them came to the conference, and 1/10 didn't. It also must be true that
(1/10)* (expected number) + (those not expected) = 7
For example, consistent with this statement:
16 managers, ten expected to go, 9 of the expected went, and 7 didn't attend. Expected/whole = 10/16 = 5/8 = .625, lower than 90%
25 managers, 20 expected to go, 18 of the expected went, and 7 didn't attend. Expected/whole = 20/25 = 0.8, lower than 90%
34 managers, 30 expected to go, 27 of the expected went, and 7 didn't attend. Expected/whole = 30/34 ---
Is 30/34 less than 90%? Well, 1/10 of 34 is 3.4, and 30 is less than (34 - 3.4), so it's below 90%.
43 managers, 40 expected to go, 36 of the expected went, and 7 didn't attend. Expected/whole = 40/43 ---
Is 40/43 less than 90%? Well, 1/10 of 43 is 4.3, and 40 is more than (43 - 4.3), so it's more than 90%.

You see, none of these had the percent of expected/whole equal precisely 90%. In order for both percents to equal precisely 90%, the total number of managers would have to be a multiple of 100. Say:
Total managers = 100
Percent expected to do = 90
Percent of the expected who went = 81
Number who didn't go = 100 - 81 = 19
Any multiple of this scenario would require that the number who didn't go be a multiple of 19. That's inconsistent with this scenario, which is why both percents can equal precisely 90 at the same time.

Does this make sense?
Mike :-)
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Re: If ninety percent of the expected number of managers  [#permalink]

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New post 25 Aug 2018, 11:59
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Re: If ninety percent of the expected number of managers &nbs [#permalink] 25 Aug 2018, 11:59
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