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The organizers of a conference offered a certain number of simultaneous seminars with the intention that each seminar

Expert replies
by BTGmoderatorDC » Mon Apr 19, 2021 3:13 pm

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Answers

A

B

C

D

E

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Difficulty

The organizers of a conference offered a certain number of simultaneous seminars with the intention that each seminar would be attended by 18 conference attendees. However, space limitations allowed only up to 15 conference attendees to participate in each of a number of the seminars, leaving 4 remaining seminars that together would be attended by at least 93 conference attendees. How many seminars were there?

(A) 10
(B) 11
(C) 15
(D) 20
(E) 26


OA B

Source: GMAT Prep
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Source: — Problem Solving |

BTGmoderatorDC wrote:
Mon Apr 19, 2021 3:13 pm
The organizers of a conference offered a certain number of simultaneous seminars with the intention that each seminar would be attended by 18 conference attendees. However, space limitations allowed only up to 15 conference attendees to participate in each of a number of the seminars, leaving 4 remaining seminars that together would be attended by at least 93 conference attendees. How many seminars were there?

(A) 10
(B) 11
(C) 15
(D) 20
(E) 26


OA B

Source: GMAT Prep
Algebraic approach,

# of seminar\(= s\)
# of seminar with space limitations\(= x\)

Now remaining seminar without limitation:
\(s-x = 4 \quad (1)\)

Total number of INTENDED attendees\(= 18\cdot s\)
Total number of attendees in seminar with limitations\(= 15\cdot x\)

Number of remaining attendees \(=\) intended attendees \(-\) attendees with limitations \(\geq 93\)

\(18s-15x \geq 93 \quad (2)\)

Solving \((1)\) and \((2)\)
\(x=7\)
\(s=4+7\)
\(s=11\quad\Longrightarrow\quad\)B
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BTGmoderatorDC wrote:
Mon Apr 19, 2021 3:13 pm
The organizers of a conference offered a certain number of simultaneous seminars with the intention that each seminar would be attended by 18 conference attendees. However, space limitations allowed only up to 15 conference attendees to participate in each of a number of the seminars, leaving 4 remaining seminars that together would be attended by at least 93 conference attendees. How many seminars were there?

(A) 10
(B) 11
(C) 15
(D) 20
(E) 26


OA B

Solution:

We can let x = the number of seminars in the conference and create the equation:

18x = 15(x - 4) + 93

18x = 15x - 60 + 93

3x = 33

x = 11

Answer: B

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