Last visit was: 21 Sep 2026, 00:56 It is currently 21 Sep 2026, 00:56
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2   3   4   5   6   
User avatar
Rishika2805
User avatar
Joined: 13 Dec 2023
Last visit: 19 Sep 2026
Posts: 31
Own Kudos:
24
 [1]
Given Kudos: 31
Products:
Forum Quiz
Posts: 31
Kudos: 24
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:29
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let the total number of an invited guests be x

So, number os invited guest as per the question stem,

Donors: 4x/7 , Press: 1x/5, The remainder were divided equally among the other four groups(volunteers, entertainers, organizers, or sponsors)= x-4x/7-1x/5=8x/35 and since it is equally divided so each group(volunteers, entertainers, organizers, or sponsors) have =8x/35*1/4 =2x/35 members

Now, actual attendees as per the question stem

Press= 2*1x/5
volunteers= 2x/35*1/2= x/35
entertainers=2x/35*0=0 (since they did not attend)
Donors= 4x/7
sponsors=2x/35
organizers=2x/35

Sum all for total attendees = 2x/5+x/35+0+4x/7+2x/35+2x/35=39x/35

fraction of the actual attendees were donors= 4x/7/39x/35= 4x*35/39x*7= 20/39. Option B


Bunuel
At a charity gala, each invited guest belonged to exactly one of six groups: donors, press, volunteers, entertainers, organizers, or sponsors. Among the invited guests, 4/7 were donors, 1/5 were press, and the remainder were divided equally among the other four groups. On the day of the event, however, twice as many press guests arrived as were invited, only half of the invited volunteers attended, and the entertainers did not attend due to a last-minute boycott. The donors, organizers, and sponsors showed up as expected. What fraction of the actual attendees were donors?

A. 4/7
B. 20/39
C. 4/9
D. 2/5
E. 10/39


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more
User avatar
chasing725
User avatar
Joined: 22 Jun 2025
Last visit: 16 Aug 2026
Posts: 240
Own Kudos:
235
 [1]
Given Kudos: 6
Location: United States (OR)
Schools: Stanford
Schools: Stanford
Posts: 240
Kudos: 235
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:30
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
At a charity gala, each invited guest belonged to exactly one of six groups: donors, press, volunteers, entertainers, organizers, or sponsors. Among the invited guests, 4/7 were donors, 1/5 were press, and the remainder were divided equally among the other four groups. On the day of the event, however, twice as many press guests arrived as were invited, only half of the invited volunteers attended, and the entertainers did not attend due to a last-minute boycott. The donors, organizers, and sponsors showed up as expected. What fraction of the actual attendees were donors?

A. 4/7
B. 20/39
C. 4/9
D. 2/5
E. 10/39


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more

We can assume that the number of guests is the LCM of (4,7,5) = 35 * 4

Planned Number of Donors = 4/7 * (35*4) = 80
Planned Number of Press = 2/5 * (35*4) = 28
Planned Number of Volunteers, entertainers, organizers, and sponsors = 8

Actual Number of Donors = 80
Actual Number of Press = 56
Actual Number of Volunteers = 4
Actual Number of Entertainers = 0
Actual Number of Organizers = 8
Actual Number of Sponsors = 8

Therefore ratio = 80/(80+56+16+4) = 80/156

Solving we get 20/39

Option B
avatar
ManifestDreamMBA
avatar
Joined: 17 Sep 2024
Last visit: 01 Jul 2026
Posts: 1,381
Own Kudos:
934
 [1]
Given Kudos: 243
Posts: 1,381
Kudos: 934
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:32
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
DPVEOS
Invited4/7=20/351/5=7/35=(1-27/35)*1/4 = 2/352/352/352/35
Attended20x14x1x0x2x2x

What fraction of the actual attendees were donors?
=20x/(20x+14x+1x+2x+2x) = 20/39

Bunuel
At a charity gala, each invited guest belonged to exactly one of six groups: donors, press, volunteers, entertainers, organizers, or sponsors. Among the invited guests, 4/7 were donors, 1/5 were press, and the remainder were divided equally among the other four groups. On the day of the event, however, twice as many press guests arrived as were invited, only half of the invited volunteers attended, and the entertainers did not attend due to a last-minute boycott. The donors, organizers, and sponsors showed up as expected. What fraction of the actual attendees were donors?

A. 4/7
B. 20/39
C. 4/9
D. 2/5
E. 10/39


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more
User avatar
Cana1766
User avatar
Joined: 26 May 2024
Last visit: 17 May 2026
Posts: 85
Own Kudos:
80
 [1]
Given Kudos: 12
Posts: 85
Kudos: 80
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:35
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
First find the fraction for donors(d), press(p), volunteers(v), entertainers(e), organizers(o), sponsors(s).

d= 4/7=20/35
p = 1/5 =7/35

Fraction left=1-(20/35)-(7/35)=8/35 [this needs to be distributed equally among v,e,o,s]
v=e=o=s=[(8/35)/4]=2/35

Now according to question for p it became=14/35 and v became =1/35

Now lets add it up =d+p+v+o+s [we are excluding e because of the question]
=20/35 + 7/35 + 1/35 + 2/35 + 2/35 = 29/35

Now for the final fraction=d/total=(20/35)/(29/35)=20/29

Hence the answer is B



Bunuel
At a charity gala, each invited guest belonged to exactly one of six groups: donors, press, volunteers, entertainers, organizers, or sponsors. Among the invited guests, 4/7 were donors, 1/5 were press, and the remainder were divided equally among the other four groups. On the day of the event, however, twice as many press guests arrived as were invited, only half of the invited volunteers attended, and the entertainers did not attend due to a last-minute boycott. The donors, organizers, and sponsors showed up as expected. What fraction of the actual attendees were donors?

A. 4/7
B. 20/39
C. 4/9
D. 2/5
E. 10/39


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more
User avatar
Emkicheru
User avatar
Joined: 12 Sep 2023
Last visit: 12 Sep 2025
Posts: 119
Own Kudos:
22
 [1]
Given Kudos: 11
Location: Kenya
Schools: Anderson (UCLA) Sloan (MIT) Columbia '27
GMAT 1: 780 Q50 V48
GRE 1: Q167 V164
GPA: 3.7
Schools: Anderson (UCLA) Sloan (MIT) Columbia '27
GMAT 1: 780 Q50 V48
GRE 1: Q167 V164
Posts: 119
Kudos: 22
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:40
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
At a charity gala, each invited guest belonged to exactly one of six groups: donors, press, volunteers, entertainers, organizers, or sponsors. Among the invited guests, 4/7 were donors, 1/5 were press, and the remainder were divided equally among the other four groups. On the day of the event, however, twice as many press guests arrived as were invited, only half of the invited volunteers attended, and the entertainers did not attend due to a last-minute boycott. The donors, organizers, and sponsors showed up as expected. What fraction of the actual attendees were donors?

A. 4/7
B. 20/39
C. 4/9
D. 2/5
E. 10/39


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more
Press were 1/5,Donors were 4/7 implies 1/5+4/7=7+20=27/35 the remainder divided equally is 8/35*1/4=1/35 organisers did not show up ,press were twice ie 7/35*2=14/35 ,donors show up as expected 20/35, half of volunteers show up 2/35 sponsors show up as expected 2/35 ,entertainers show up as expected 2/35 total attendees were 20/35+14/35+1/35+2/35+2/35=20+14+1+2+2=39/35 implies fraction of donors in the attendees is 20/39
User avatar
iamchinu97
User avatar
Joined: 14 Dec 2020
Last visit: 26 Jun 2026
Posts: 136
Own Kudos:
154
 [1]
Given Kudos: 35
Products:
My Reviews
Posts: 136
Kudos: 154
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:41
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
let total be invited be x
then we know that
donors=4x/7, press=x/5,
Then remaining invited will be = x-(4x/7+x/5) = 8x/35, so for each
volunteers=2x/35, entertainers=2x/35, organizers=2x/35, sponsors = 2x/35

Now Actual Attended
donors=4x/7, press=2x/5, volunteers=x/35, entertainers=NA, organizers=2x/35, sponsors=2x/35
Total of above = 39x/35

So fraction of the actual attendees were donors = (4x/7)/(39x/35)
=> 20/39

hence Ans B
User avatar
ektalakra
User avatar
Joined: 29 Jul 2024
Last visit: 26 Mar 2026
Posts: 38
Own Kudos:
16
 [1]
Given Kudos: 84
Posts: 38
Kudos: 16
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:42
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
let total invited guest be 350
so d=200
p=70
v=20,e=20,o=20,s=20
now, actual attended
d=200
p=140
v=10
e=0
o=20
s=20
required ratio=200/390=20/39 .....ans b
Bunuel
At a charity gala, each invited guest belonged to exactly one of six groups: donors, press, volunteers, entertainers, organizers, or sponsors. Among the invited guests, 4/7 were donors, 1/5 were press, and the remainder were divided equally among the other four groups. On the day of the event, however, twice as many press guests arrived as were invited, only half of the invited volunteers attended, and the entertainers did not attend due to a last-minute boycott. The donors, organizers, and sponsors showed up as expected. What fraction of the actual attendees were donors?

A. 4/7
B. 20/39
C. 4/9
D. 2/5
E. 10/39


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more
User avatar
Decimal
User avatar
Joined: 20 Jul 2022
Last visit: 03 Oct 2025
Posts: 9
Own Kudos:
9
 [1]
Given Kudos: 16
Posts: 9
Kudos: 9
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:47
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
At a charity gala, each invited guest belonged to exactly one of six groups: donors, press, volunteers, entertainers, organizers, or sponsors. Among the invited guests, 4/7 were donors, 1/5 were press, and the remainder were divided equally among the other four groups. On the day of the event, however, twice as many press guests arrived as were invited, only half of the invited volunteers attended, and the entertainers did not attend due to a last-minute boycott. The donors, organizers, and sponsors showed up as expected. What fraction of the actual attendees were donors?

A. 4/7
B. 20/39
C. 4/9
D. 2/5
E. 10/39


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more
So, the Givens :
1. Six groups were to attend the charity gala: Donors, Press, Volunteers, Organisers, Entertainment, and Sponsors
2. Their distribution as per invitation => Donors = 4/7; Press = 1/5, and the remaining four categories were equally distributed.
3. Final Attendees - Press = double the invitation; no one from entertainment; and half the volunteers.

Asked :
What fraction of the final attendees were Donors

Solution :
Simplify the distribution as per the invitation first
=> Equating 4/7 of Donors and 1/5 of Press
=> 4/7 (20/35) + 1/5 (7/35) = 27/35
=> The remaining four categories will be 8/35
=> Each of the remaining four categories will be 2/35.
Now, taking only the numerator of the fractions to represent the invitation distribution:
=> Donors = 20; Press = 7; Volunteers = 2; Organisers = 2; Entertainment =2; and Sponsors = 2.

Now, actual attendees were - Press = double the invitation; no one from entertainment; half the volunteers and remaining group same.
=> Actual attendees after applying the given condition = Donors = 20; Press = 14; Volunteers = 1; Organisers = 2; Entertainment =0; and Sponsors = 2.
=> Total = 39 attendees
=> Donors = 20 (same as invitation)
Hence, fraction of Donors = 20/39

B is the answer.
User avatar
BeachStudy
User avatar
Joined: 30 Jun 2025
Last visit: 18 Aug 2025
Posts: 61
Own Kudos:
37
 [1]
Given Kudos: 4
Posts: 61
Kudos: 37
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:48
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Notes:
Six Groups:
Donor, Press, Volunteer, Entertainers, Organizers, Sponsors

Donors - 4/7
Press - 1/5

bring these to a common denominator:

Donors - 20/35
Press - 7/35
Remainder - 8/35 - (2/35 each)

Changes for day of:
Day of event 2x as many press came - 14/35.
Volunteers 1/2 as many- 1/35
Entertainers 0 - 0/35

Total count of people in attendance out of 35 original planned (estimate)
Donors - 20
Press- 14
Volunteers - 1
Entertianers -0
Sponsors - 2
Organizers - 2

Total count- 39
Donors 20/39

ANSWER: B 20/39
User avatar
amansoni5
User avatar
Joined: 16 Nov 2021
Last visit: 21 Sep 2026
Posts: 94
Own Kudos:
64
 [1]
Given Kudos: 219
Products:
GMAT Club Tests
Posts: 94
Kudos: 64
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:58
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let T be the total number of invited guests
Donors - 4/7 of T
Press - 1/5 of T
Therefore,
Volunteers - 2/35T
Entertainers - 2/35T
Organisers - 2/35T
Sponsors - 2/35T

To make calculations easy, consider T to be 35
Therfore,
Donors - 20
Press - 7
Volunteers - 2
Entertainers - 2
Organisers - 2
Sponsors - 2

Now, guests who actually attended
Donors - 20
Press - 14
Volunteers - 1
Entertainers - 0
Org - 2
Sponsors - 2
Total 39
Therefore, Donors / Total actually attended = 20/39
User avatar
Gmat232323
User avatar
Joined: 22 Sep 2022
Last visit: 02 Aug 2026
Posts: 96
Own Kudos:
26
 [1]
Given Kudos: 162
Location: India
Schools: ISB '27
GPA: 3.04
Schools: ISB '27
Posts: 96
Kudos: 26
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:00
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Invited Guests
Let the total number of invited guests be \(x = 35\) (LCM of 7 and 5)

Invited donors \(= \frac{4}{7}*35 = 20\)
Invited press \(= \frac{1}{5}*35 = 7\)
Remaining guests \(= 35 - (20+7) = 8\)
Remaining guests were divided equally among the volunteers, entertainers, organizers, and sponsors => \(8/4 = 2\)

Actual Guests
Press \(= 2*7 = 14\)
Volunteers \(= 2*\frac{1}{2}\)
Entertainers \(= 0\)
Donors \(= 20\)
Organizers \(= 2\)
Sponsors \(= 2\)

Total actual guests \(= 14+1+0+20+2+2 = 39\)

Donors/Total = 20/39 => Option B.
Bunuel
At a charity gala, each invited guest belonged to exactly one of six groups: donors, press, volunteers, entertainers, organizers, or sponsors. Among the invited guests, 4/7 were donors, 1/5 were press, and the remainder were divided equally among the other four groups. On the day of the event, however, twice as many press guests arrived as were invited, only half of the invited volunteers attended, and the entertainers did not attend due to a last-minute boycott. The donors, organizers, and sponsors showed up as expected. What fraction of the actual attendees were donors?

A. 4/7
B. 20/39
C. 4/9
D. 2/5
E. 10/39


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more
User avatar
AVMachine
User avatar
Joined: 03 May 2024
Last visit: 26 Mar 2026
Posts: 190
Own Kudos:
158
 [1]
Given Kudos: 40
Posts: 190
Kudos: 158
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:11
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
At a charity gala, each invited guest belonged to exactly one of six groups: donors, press, volunteers, entertainers, organizers, or sponsors.

Assume Total Invited Guests are X;

Among the invited guests, 4/7 were donors(4x/7), 1/5 were press(1x/5), and the remainder were divided equally among the other four groups.

Remainder -> x - (4x/7) - (x/5) = 8x/35

The remainder were divided into four groups, so each group will have 2x/35 guests.

On the day of the event, however, twice as many press guests arrived as were invited (2x/5), only half of the invited volunteers attended (x/35), and the entertainers did not attend due to a last-minute boycott (0). The donors (4x/7), organizers (2x/35), and sponsors (2x/35) showed up as expected. What fraction of the actual attendees were donors?

Donors (d) = (4x/7)

Total Actual Attendees (t) : (2x/5) + (x/35) + (4x/7) + (2x/35) + (2x/35)

fraction of the actual attendees who were donors: d/t = 20/39;
User avatar
APram
User avatar
Joined: 23 Jun 2024
Last visit: 12 Sep 2026
Posts: 705
Own Kudos:
283
 [1]
Given Kudos: 248
Location: India
Schools: ISB '26 HEC INSEAD Jan '26
GMAT Focus 1: 605 Q86 V78 DI76
GPA: 3.608
Schools: ISB '26 HEC INSEAD Jan '26
GMAT Focus 1: 605 Q86 V78 DI76
Posts: 705
Kudos: 283
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:29
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let total number of guests invited LCM(7,5) = 35
Number of donors invited = 20
Number of Press invited = 7
Volunteers, entertainers, organizers, or sponsors be 2 each

Number of press arrived = 14
Number of volunteers attended = 1
Donors, Organizers and sponsors attended as invited
Total attendees = 20+14+1+2+2 = 39
required fraction = 20/39

B iscorrect

Bunuel
At a charity gala, each invited guest belonged to exactly one of six groups: donors, press, volunteers, entertainers, organizers, or sponsors. Among the invited guests, 4/7 were donors, 1/5 were press, and the remainder were divided equally among the other four groups. On the day of the event, however, twice as many press guests arrived as were invited, only half of the invited volunteers attended, and the entertainers did not attend due to a last-minute boycott. The donors, organizers, and sponsors showed up as expected. What fraction of the actual attendees were donors?

A. 4/7
B. 20/39
C. 4/9
D. 2/5
E. 10/39


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more
User avatar
hakzarif
User avatar
Joined: 31 May 2025
Last visit: 20 Sep 2026
Posts: 65
Own Kudos:
30
 [1]
Given Kudos: 10
Products:
My Reviews
Posts: 65
Kudos: 30
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:29
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Assume 35 (LCM of 7 and 5) guests were invited. Donors were 20, press 7, and the remaining 8 guests were split evenly: 2 each for volunteers, entertainers, organizers, and sponsors. On the day of the event, 20 donors came, press doubled to 14, only 1 volunteer came, entertainers didn’t show (0), and organizers and sponsors came as expected (2 each). Total attendees: 20 + 14 + 1 + 0 + 2 + 2 = 39. Donors were 20 out of 39 attendees. Final answer: 20/39. Option B.
User avatar
Ryga
User avatar
Joined: 12 Aug 2023
Last visit: 19 Aug 2025
Posts: 68
Own Kudos:
51
 [1]
Given Kudos: 4
Location: India
Concentration: General Management, Leadership
Schools: INSEAD Jan '26 NUS '27 Stern '26
GMAT Focus 1: 695 Q90 V80 DI83
Schools: INSEAD Jan '26 NUS '27 Stern '26
GMAT Focus 1: 695 Q90 V80 DI83
Posts: 68
Kudos: 51
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:33
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let total invited guests be N
Find invited numbers for each group
  • Donors: (4/7)N
  • Press: (1/5)N
  • Remaining: N - (4/7)N - (1/5)N
    = N × [35/35 - 20/35 - 7/35]
    = N × (8/35)
  • This remainder is split equally among 4 groups (volunteers, entertainers, organizers, sponsors):
    Each group: (1/4) × (8/35)N = (2/35)N

Who actually attended
  • Press: Twice as many as invited = 2 × (1/5)N = (2/5)N
  • Volunteers: Half attended = (1/2) × (2/35)N = (1/35)N
  • Entertainers: None attended = 0
  • Donors = (4/7)N
  • Organizers: (2/35)N
  • Sponsors: (2/35)N

Total attendees = (4/7)N + (2/5)N + (1/35)N + (2/35)N + (2/35)N
= (20/35)N + (14/35)N + (5/35)N
= (39/35)N

Fraction of attendees who were donors
= (Donors) / (Total attendees)
= [(4/7)N] / [(39/35)N]
= 20/39
Answer: B
User avatar
WrickR
User avatar
Joined: 22 Dec 2024
Last visit: 11 Dec 2025
Posts: 37
Own Kudos:
26
 [1]
Given Kudos: 49
Location: India
GPA: 3.7
Posts: 37
Kudos: 26
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:35
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Assume the initial number of attendees to be 35 for ease of calculation(LCM of 7 and 5).

So, Donors = \(\frac{4}{5}\times35 = 20\) and Press = \(\frac{1}{5}\times35=7\)
No. of attendees left = \(35-(20+7) = 8\)

So, 8 attendees are divided equally among the remaining groups.
Volunteers = Entertainers = Organizers = Sponsors = 2

Actual attendees:
Donors = 20
Press = 2 times original = 14
Volunteers = 1/2 of original = 1
Entertainers = 0
Organizers = Sponsors = 2
Total attendees = 39

So, the fraction of attendees that are donors = \(\frac{20}{39}\) (B)
User avatar
DataGuyX
User avatar
Joined: 23 Apr 2023
Last visit: 02 Apr 2026
Posts: 111
Own Kudos:
77
 [1]
Given Kudos: 161
Location: Brazil
Concentration: Entrepreneurship, Technology
Schools: Booth '26 (A$) Kellogg '26 (A) Sloan '26 (D) Stanford '26 (D) Haas '26 (D) HBS '28 (D)
GMAT Focus 1: 675 Q79 V85 DI86
GPA: 7.3
Schools: Booth '26 (A$) Kellogg '26 (A) Sloan '26 (D) Stanford '26 (D) Haas '26 (D) HBS '28 (D)
GMAT Focus 1: 675 Q79 V85 DI86
Posts: 111
Kudos: 77
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:38
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let D= Donors, P = Press, V = Volunteers, E = Entertainers, O = Organizers, and S = Sponsors.
Di = Donors invited, Pi = Press invited, Vi = Volunteers invited, Ei = Entertainers invited, Oi = Organizers invited, and Si = Sponsors invited.

Invites
Di = 4/7 total
Pi = 1/5 total
LCM ( 5 , 7 ) = 35
Di + Pi = (20/35) + (7/35) = 27/35

Remainders = 35/35 - 27/35 = 8/35. Equally divided between the other 4 categories, so: Vi = Ei = Oi = Si = (8/35)/4 = 2/35

Since our denominator is equal to 35, I'll consider that our list size is 35 (but you can consider any multipe of 35) to simplify our calculations. This way, the number of invited per category is simply the numerators.

Attendance:
P = 2*(Pi) = 2*(7) = 14
V = 1/2*(Vi) = (1/2)*(2) = 1
E = 0
D = Di = 20
O = Oi = 2
S = Si = 2

Total attendees = 14 + 1 + 0 + 20 + 2 + 2 = 15 + 24 = 39

Donors fraction = 20/39

Answer
B. 20/39
User avatar
ckavya2691
User avatar
Joined: 23 Jun 2023
Last visit: 12 Feb 2026
Posts: 15
Own Kudos:
18
 [1]
Given Kudos: 31
Location: India
GMAT Focus 1: 615 Q84 V78 DI79
GMAT Focus 1: 615 Q84 V78 DI79
Posts: 15
Kudos: 18
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:45
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let total members invited = T
Donors = 4/7 T
Press = 1/5 T
And volunteers, entertainers, organisers and sponsors = x
Adding all = 4/7 T + 1/5 T + 4X = T, this gives X=2T/35
Now, on the actual day, donors = same = 4/7 T, press = T/5, volunteers = X/2, entertainers = 0, organisers = X and sponsors = X
Donors/Sum of actual attendees = 4/7 T % (4/7 T + 2T/5 + X/2 + 2X). Substitute X=2T/35 and solving the equation gives the complete ratio as 20/39.
User avatar
cabii
User avatar
Joined: 30 Jul 2024
Last visit: 21 Sep 2025
Posts: 29
Own Kudos:
18
 [1]
Given Kudos: 2
Posts: 29
Kudos: 18
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:48
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
let say total invited guest= 35 (5*7)
donors= 4/7*35=20
press= 1/5*35=7
others: 35-27=8 let say 2 each

attended:
donors: 20
press: 2*7=14
volunteers: 1/2*2=1
orani.& spon:2 each
enter...=0
total+ 20+14+1+2+2=39
fraction= 20/39
User avatar
muuss
User avatar
Joined: 10 Aug 2024
Last visit: 06 May 2026
Posts: 108
Own Kudos:
83
 [1]
Given Kudos: 38
GMAT Focus 1: 615 Q84 V81 DI76
GMAT Focus 1: 615 Q84 V81 DI76
Posts: 108
Kudos: 83
 [1]
GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:48
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let the total attendees be x
Initial Attendees: D=4x/7 , P=x/5 , V=2x/35, E=2x/35,O=2x/35,S=2x/35
Actual Attendees: D=4x/7,P=2x/5, V=x/35, E=0, O=2x/35 , S=2x/35
This makes the total=(14x+x+2x+2x+20x)/35=39x/35
Ratio of D/new total= (4x/7)/39x/35= 20/39
Ans : B
   1   2   3   4   5   6   
Moderator:
Bunuel
Math Expert
113703 posts