One-tenth of the employees at a company dinner are vice-presidents

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One-tenth of the employees at a company dinner are vice-presidents, and one-third of the remaining employees are assistant vice-presidents. If there are e employees in attendance, how many employees in attendance are neither vice-presidents nor assistant vice-presidents?

A. e/30
B. 2e/5
C. 13e/30
D. 17e/30
E. 3e/5

Answer: E
Source: Official guide
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BTGModeratorVI wrote:
Thu Aug 27, 2020 12:09 pm
One-tenth of the employees at a company dinner are vice-presidents, and one-third of the remaining employees are assistant vice-presidents. If there are e employees in attendance, how many employees in attendance are neither vice-presidents nor assistant vice-presidents?

A. e/30
B. 2e/5
C. 13e/30
D. 17e/30
E. 3e/5

Answer: E
Source: Official guide
One approach is to use the INPUT-OUTPUT approach (described in our free video - https://www.gmatprepnow.com/module/gmat- ... /video/933)

We have two fractions (1/10 and 1/3), so let's let e = the least common denominator of these two fraction.
We'll let e = 30

If there are 30 employees, then the number of vice-presidents = (1/10)(30) = 3

one-third of the remaining employees are assistant vice-presidents
There are 27 employees remaining.
1/3 of 27 = 9, so there are 9 assistant vice-presidents

So, the total number of vice-presidents AND assistant vice-presidents = 3 + 9 = 12
So, the number of employees who are NEITHER vice-presidents nor assistant vice-presidents = 30 - 12 = 18

So, when the INPUT e = 30, the OUTPUT is 18 people who are NEITHER vice-presidents nor assistant vice-president

Now check the answer choices to see which one yields an OUTPUT of 18, when we INPUT e = 30

A. e/30 = 30/30 = 1 - NO - ELIMINATE
B. 2e/5 = 2(30)/5 = 12 - NO - ELIMINATE
C. 13e/30 = 13(30)/30 = 13 - NO - ELIMINATE
D. 17e/30 = 17(30)/30 = 17 - NO - ELIMINATE
E. 3e/5 = 3(30)/5 = 18 - YES - KEEP

Answer: E

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Brent
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BTGModeratorVI wrote:
Thu Aug 27, 2020 12:09 pm
One-tenth of the employees at a company dinner are vice-presidents, and one-third of the remaining employees are assistant vice-presidents. If there are e employees in attendance, how many employees in attendance are neither vice-presidents nor assistant vice-presidents?

A. e/30
B. 2e/5
C. 13e/30
D. 17e/30
E. 3e/5

Answer: E
Source: Official guide
Here's the algebraic approach:
There are e employees

One-tenth of the employees at a company dinner are vice-presidents
So, (1/10) of e = number of employees who are vice-presidents
So, e/10 = number of employees who are vice-presidents

one-third of the remaining employees are assistant vice-presidents
So, number of remaining employees = e - e/10 = 9e/10
1/3 of 9e/10 = (1/3)(9e/10) = 3e/10
So, 3e/10 = number of employees who are assistant vice-presidents

TOTAL # of vice-presidents AND assistant vice-presidents = e/10 + 3e/10 = 4e/10 = 2e/5

How many employees in attendance are neither vice-presidents nor assistant vice-presidents?
We get e - 2e/5 = 3e/5

Answer: E

Cheers,
Brent
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BTGModeratorVI wrote:
Thu Aug 27, 2020 12:09 pm
One-tenth of the employees at a company dinner are vice-presidents, and one-third of the remaining employees are assistant vice-presidents. If there are e employees in attendance, how many employees in attendance are neither vice-presidents nor assistant vice-presidents?

A. e/30
B. 2e/5
C. 13e/30
D. 17e/30
E. 3e/5

Answer: E
Source: Official guide
Solution:

The number of employees in attendance who are neither vice-presidents nor assistant vice-presidents is:

e - e/10 - ⅓(e - e/10) = e - e/10 - e/3 + e/30 = 30e/30 - 3e/10 - 10e/30 + e/30 = 18e/30 = 3e/5

Answer: E

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