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eduardofp
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In a certain company, the ratio of the number of managers to the numbe 25 Jul 2016, 17:10
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C
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63% (01:47) correct 37% (01:57) wrong based on 270 sessions

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In a certain company, the ratio of the number of managers to the number of non-managers in any department must always be greater than 5 : 24. In the company, what is the maximum number of non-managers in a department that has 8 managers:

(A) 36
(B) 37
(C) 38
(D) 39
(E) 40
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C
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In a certain company, the ratio of the number of managers to the numbe 25 Jul 2016, 20:20
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M/NM > 5x/24x

M = 8
8/NM > 5x/24x
8/NM - 5x/24x > 0
(192x - 5x * NM)/(24x * NM) > 0
192x - 5x*NM > 0
5x*NM < 192x
NM < 192x/5x
NM < 38.4
Maximum number of non managers = 38

Answer: C
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In a certain company, the ratio of the number of managers to the numbe 25 Jul 2016, 23:29
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eduardofp
In a certain company, the ratio of the number of managers to the number of non-managers in any department must always be greater than 5 : 24. In the company, what is the maximum number of non-managers in a department that has 8 managers:

(A) 36
(B) 37
(C) 38
(D) 39
(E) 40

This is a copy of the following GMAT Prep question: https://gmatclub.com/forum/at-a-certain- ... 03543.html

Hope it helps.
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In a certain company, the ratio of the number of managers to the numbe 29 Aug 2017, 13:53
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eduardofp
In a certain company, the ratio of the number of managers to the number of non-managers in any department must always be greater than 5 : 24. In the company, what is the maximum number of non-managers in a department that has 8 managers:

(A) 36
(B) 37
(C) 38
(D) 39
(E) 40

Let M = managers
Let x = non-managers

"[T]he ratio of the number of managers to the number of non-managers [] must always be greater than 5 : 24" -->

\(\frac{M}{x} >\\
\frac{5}{24}\)

Maximum number of non-managers if there are 8 managers?

\(\frac{8}{x} >\\
\frac{5}{24}\)

Cross multiply (the numbers are positive):

\((24)(8) > 5x\)

\(192 > 5x\)

\(38.4 > x\), or \(x < 38.4\)

The number of non-managers must be fewer than 38.4
For the maximum possible number of non-managers,
we need the integer just below 38.4, which is 38.

Answer C
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In a certain company, the ratio of the number of managers to the numbe 05 Sep 2017, 18:21
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eduardofp
In a certain company, the ratio of the number of managers to the number of non-managers in any department must always be greater than 5 : 24. In the company, what is the maximum number of non-managers in a department that has 8 managers:

(A) 36
(B) 37
(C) 38
(D) 39
(E) 40

We can create the following inequality:

8/n > 5/24

192 > 5n

38.4 > n

So, the maximum number of non-managers is 38.

Answer: C
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In a certain company, the ratio of the number of managers to the numbe 09 Jan 2023, 23:08
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eduardofp
In a certain company, the ratio of the number of managers to the number of non-managers in any department must always be greater than 5 : 24. In the company, what is the maximum number of non-managers in a department that has 8 managers:

(A) 36
(B) 37
(C) 38
(D) 39
(E) 40

Ratio of managers:non managers >= 5:24

If there are 8 managers, multiplier is 8/5 so number of non managers must not exceed 24 * (8/5) = 38.4
So maximum number of non managers must be 38

Answer (C)
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In a certain company, the ratio of the number of managers to the numbe 11 Jan 2023, 06:38
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eduardofp
In a certain company, the ratio of the number of managers to the number of non-managers in any department must always be greater than 5 : 24. In the company, what is the maximum number of non-managers in a department that has 8 managers:

(A) 36
(B) 37
(C) 38
(D) 39
(E) 40

Solution:

  • It is given that ratio of \(\frac{Manager}{NonManager}>\frac{5}{24}\)
    \(⇒\frac{8}{NonManager}>\frac{5}{24}\)
    \(⇒\frac{8\times 24}{5}>NonManager\)
    \(⇒NonManager<\frac{8\times 24}{5}\)
    \(⇒NonManager<38.4\)
  • The maximum value of Non Manager \(= 38\)

Hence the right answer is Option C
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In a certain company, the ratio of the number of managers to the numbe 23 Jun 2025, 03:36
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