If ninety percent of the expected number of managers

**

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

**

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

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

I dont get this. Its an obvious C for me.

x = all of the managers

0,9x is expected to come

0,9*0,9x actually came = 0,81x

7 of the total numbers o managers did not come. x-7 did not come.

x-7 = x-0,81x = 0,19x

0,19x = 7

x = 36,8

0,81x = 29,8 managers attended the conference

Edit: Ok, I get it. "No more than" doesnt mean exactly 90 %. But its weird then how they could even say that "If ninety percent of the expected number of managers attended a conference" - which implies there is an exact number, and then as of a sudden they say that "no more than 90 percent of all managers". Thats a contradiction in what they even say. First they know the exact number, and then in B) they are not sure anymore. Idiots.

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