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# M19 Q16

Author Message
Manager
Joined: 16 Feb 2011
Posts: 193
Schools: ABCD
Followers: 1

Kudos [?]: 143 [0], given: 78

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20 Jul 2012, 07:35
In a certain company, employees receive their salary on a monthly basis and are paid on either 10th or 20th of each month. If 65 percent of all managers are paid on 10th of each month and 79% of employees who are paid on 20th of each month are not managers, what percent of the company employees are managers?

12%
15%
20%
21%
the answer cannot be derived from the data provided

Here's what I did:

on the 20th, 0.35M = .21 E (because 35% of managers will get paid on the 20th; 21% of employees are managers)

therefore, M/E = 60%

Thoughts?
Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)
Followers: 90

Kudos [?]: 819 [0], given: 43

### Show Tags

20 Jul 2012, 09:13
voodoochild wrote:
In a certain company, employees receive their salary on a monthly basis and are paid on either 10th or 20th of each month. If 65 percent of all managers are paid on 10th of each month and 79% of employees who are paid on 20th of each month are not managers, what percent of the company employees are managers?

12%
15%
20%
21%
the answer cannot be derived from the data provided

Here's what I did:

on the 20th, 0.35M = .21 E (because 35% of managers will get paid on the 20th; 21% of employees are managers)

therefore, M/E = 60%

Thoughts?

21% of those paid on the 20th are managers, it doesn't mean that 21% of all employees are managers.
There are employees paid on the 10th, those paid on the 20th are not all the non-manager employees.
Also, managers can be divided into two groups: those paid on the 10th, and those paid on the 20th.
And you don't know anything about the employees (non-managers) paid on the 10th.
Try to use the 2 X 2 table method and you will see that you cannot fill it out. Or the Venn diagram.

_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Manager
Joined: 16 Feb 2011
Posts: 193
Schools: ABCD
Followers: 1

Kudos [?]: 143 [0], given: 78

### Show Tags

22 Jul 2012, 18:29
EvaJager wrote:
voodoochild wrote:
In a certain company, employees receive their salary on a monthly basis and are paid on either 10th or 20th of each month. If 65 percent of all managers are paid on 10th of each month and 79% of employees who are paid on 20th of each month are not managers, what percent of the company employees are managers?

12%
15%
20%
21%
the answer cannot be derived from the data provided

Here's what I did:

on the 20th, 0.35M = .21 E (because 35% of managers will get paid on the 20th; 21% of employees are managers)

therefore, M/E = 60%

Thoughts?

21% of those paid on the 20th are managers, it doesn't mean that 21% of all employees are managers.
There are employees paid on the 10th, those paid on the 20th are not all the non-manager employees.
Also, managers can be divided into two groups: those paid on the 10th, and those paid on the 20th.
And you don't know anything about the employees (non-managers) paid on the 10th.
Try to use the 2 X 2 table method and you will see that you cannot fill it out. Or the Venn diagram.

It's definitely solvable. Please see the attached image. I am still waiting for Bunuel to reply...
Attachments

sawal.png [ 19.09 KiB | Viewed 966 times ]

Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)
Followers: 90

Kudos [?]: 819 [1] , given: 43

### Show Tags

23 Jul 2012, 01:40
1
KUDOS
voodoochild wrote:
EvaJager wrote:
voodoochild wrote:
In a certain company, employees receive their salary on a monthly basis and are paid on either 10th or 20th of each month. If 65 percent of all managers are paid on 10th of each month and 79% of employees who are paid on 20th of each month are not managers, what percent of the company employees are managers?

12%
15%
20%
21%
the answer cannot be derived from the data provided

Here's what I did:

on the 20th, 0.35M = .21 E (because 35% of managers will get paid on the 20th; 21% of employees are managers)

therefore, M/E = 60%

Thoughts?

21% of those paid on the 20th are managers, it doesn't mean that 21% of all employees are managers.
There are employees paid on the 10th, those paid on the 20th are not all the non-manager employees.
Also, managers can be divided into two groups: those paid on the 10th, and those paid on the 20th.
And you don't know anything about the employees (non-managers) paid on the 10th.
Try to use the 2 X 2 table method and you will see that you cannot fill it out. Or the Venn diagram.

It's definitely solvable. Please see the attached image. I am still waiting for Bunuel to reply...

Non-managers paid on the 10th is not $$0.21T_1$$ and therefore, $$0.79T_1\neq0.65x$$.
Nowhere is stated that also 21% of those paid on the 10th are non-managers.
Nobody is paid both on the 10th and on the 20th, and what is valid for those paid on the 20th, is not necessarily
valid for those paid on the 10th regarding their division into the two types (managers and non-managers).
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Manager
Joined: 14 Jun 2011
Posts: 85
Followers: 2

Kudos [?]: 30 [0], given: 15

### Show Tags

05 Sep 2013, 11:34
In a certain company, employees receive their salary on a monthly basis and are paid on either 10th or 20th of each month. If 65 percent of all managers are paid on 10th of each month and 79% of employees who are paid on 20th of each month are not managers, what percent of the company employees are managers?

A) 12%
B) 15%
C) 20%
D) 21%
E) the answer cannot be derived from the data provided
_________________

Kudos always encourages me

Re: M19 Q16   [#permalink] 05 Sep 2013, 11:34
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# M19 Q16

Moderator: Bunuel

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