Vincen wrote: ↑

Sat Nov 27, 2021 3:59 am

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. \(\dfrac{e}{30}\)

B. \(\dfrac{2e}5\)

C. \(\dfrac{13e}{30}\)

D. \(\dfrac{17e}{30}\)

E. \(\dfrac{3e}5\)

Answer: E

Source: GMAT Paper Tests

One approach is to use the INPUT-OUTPUT approach

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

Cheers,
Brent