Last visit was: 19 Nov 2025, 05:34 It is currently 19 Nov 2025, 05:34
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.
Close
Request Expert Reply
Confirm Cancel
User avatar
ODST117
Joined: 15 Aug 2024
Last visit: 29 Oct 2025
Posts: 173
Own Kudos:
85
 [1]
Given Kudos: 149
Posts: 173
Kudos: 85
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Jarvis07
Joined: 06 Sep 2017
Last visit: 19 Nov 2025
Posts: 295
Own Kudos:
236
 [1]
Given Kudos: 160
GMAT 1: 750 Q50 V41
GMAT 1: 750 Q50 V41
Posts: 295
Kudos: 236
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
asingh22
Joined: 31 Jul 2024
Last visit: 18 Nov 2025
Posts: 68
Own Kudos:
57
 [1]
Given Kudos: 8
Location: India
GMAT Focus 1: 635 Q84 V78 DI82
GMAT Focus 2: 655 Q89 V80 DI78
GPA: 2.5
Products:
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
D3N0
Joined: 21 Jan 2015
Last visit: 19 Nov 2025
Posts: 587
Own Kudos:
572
 [1]
Given Kudos: 132
Location: India
Concentration: Operations, Technology
GMAT 1: 620 Q48 V28
GMAT 2: 690 Q49 V35
WE:Operations (Retail: E-commerce)
Products:
GMAT 2: 690 Q49 V35
Posts: 587
Kudos: 572
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Ans: B
As per the question:
expected attendance
D = 4/7, P = 1/5, and ( V+E+O+S ) = x (lets say)
then
V+E+O+S+D+P =1,
x + 4/7+ 1/5 = 1,
x = 8/35,
then V = E = O = S = 2/35

Now, as per the question for actual attendance
D = 4/7, as expected
P = 2/5, twice as many as expected
V = 1/35, only half of the expected
E = 0, none attended
O = S = 2/35, as expected

AA = Actual total attendance = V+E+O+S+D+P = 39/35
We need to find = What fraction of the actual attendees were donors = D/ AA

= (4/7) / (39/35)
= 20/39 Answer
User avatar
Raome
Joined: 21 Apr 2025
Last visit: 18 Nov 2025
Posts: 109
Own Kudos:
30
 [1]
Given Kudos: 84
Location: India
Posts: 109
Kudos: 30
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Invited:-

Total = 35

Donors= 20
Press=7
Volunteers=2
Entertainers=2
Organizers=2
sponsors=2

actual attendees:-

Donors= 20
Press=7*2=14
Volunteers=2/2=1
Entertainers=0
Organizers=2
sponsors=2

New total= 39

20/39. Correct 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

 

User avatar
iCheetaah
Joined: 13 Nov 2021
Last visit: 17 Nov 2025
Posts: 81
Own Kudos:
72
 [1]
Given Kudos: 1
Location: India
Posts: 81
Kudos: 72
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
We start off with the following,

D = 4/7
P = 1/5
V = 2/35
E = 2/35
O = 2/35
S = 2/35

Values of V to S each, are calculated by: [1-(4/7+1/5)]*1/4

Let's bring all values to a denominator of 35 for easier calculations:

D = 20/35
P = 7/35
V = 2/35
E = 2/35
O = 2/35
S = 2/35

Now, these are the values that we have at the day of the event:

D = 20/35
P = 14/35
V = 1/35
E = 0
O = 2/35
S = 2/35


Total number of attendees = sum of all of the above = 20/35 + 14/35 + 1/35 + 0 + 2/35 + 2/35 = 39/35
Number of donors = D = 20/35

Dividing the latter by the former, we get:

Answer: B. 20/39
User avatar
findingmyself
Joined: 06 Apr 2025
Last visit: 18 Nov 2025
Posts: 230
Own Kudos:
157
 [1]
Given Kudos: 57
Posts: 230
Kudos: 157
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
DPVEOSTotal
invited4/71/52/352/352/352/35X
attended4/72/51/3502/352/3539X/35

D/attended=(4X/7)/(39X/35)
Answer: 20/39

The above question can be answered best when made as a table since it reduces chances of error given so many segments
User avatar
sahayrahul
Joined: 07 Jan 2019
Last visit: 01 Oct 2025
Posts: 31
Own Kudos:
8
 [1]
Given Kudos: 29
Location: United States (NJ)
Concentration: Technology, General Management
GPA: 3.5
WE:Project Management (Technology)
Posts: 31
Kudos: 8
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
We know two ratio from given 6 groups which is 1/5 for P and 4/7 for D..
We can solve very easily by assuming total number and to correctly assume number it is always better to take ration Denom. LCM (5,7) = 35.
Assuming total guest be 35 , we have 1/5 of 35 = 7 for P || 4/7 of 35 = 20 for D || Remaining 8 which is 2 each.
Now on event day P = 2 time = 14 || V = 1⁄2 time = 1 ||E = 0 ;
So now total becomes P(14) + D (20) + V (1) + E(0) + O (2) + S(2) = 39
Ration of D = 20/39 B
User avatar
gioz
Joined: 16 Nov 2024
Last visit: 21 Sep 2025
Posts: 10
Own Kudos:
7
 [1]
Given Kudos: 9
Posts: 10
Kudos: 7
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Quote:
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
We are dealing with fractions, so let’s pick a smart number.
The LCM of 5 and 7 = 35 total invited
Now calculate:
Donors = 4/7*35= 20
Press= 1/5*35= 7
Remainder = 35 - 20 - 7 = 8 (divided equally among the other four groups as the text says)
volunteers= 2
entertainers=2
organizers=2
sponsors=2

Now apply the attendance conditions:
twice as many press guests arrived as were invited = 7*2=14
only half of the invited volunteers attended =2/2=1
entertainers did not attend due to a last-minute boycott=0

The donors (20) , organizers (2), and sponsors (2) showed up as expected.

Total attendees= 14 (press) + 1 (volunteers) + 20 (donors) + 2 (organizers) + 2 ( sponsors) = 39

\(\frac{Donors}{Total}\)=\(\frac{20}{39}\)


ANSWER B
User avatar
sabareeshmc1405
Joined: 09 Jun 2025
Last visit: 17 Nov 2025
Posts: 18
Own Kudos:
Given Kudos: 60
Posts: 18
Kudos: 14
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

 


Assume the total number of invited guests = 70



Derive the table based on give Stats:

GroupInvitedAttended
Donors4040 (as expected)
Press142×14=28
Volunteers42 (half invited)
Entertainers40 (boycott)
Organizers44 (as expected)
Sponsors44 (as expected)
Total7040+28+2+0+4+4=78

Now Answer is :

Fraction=40/78=20/39
Attachment:
GMAT-Club-Forum-og357a4g.png
GMAT-Club-Forum-og357a4g.png [ 9.5 KiB | Viewed 175 times ]
User avatar
kvaishvik24
Joined: 31 Mar 2025
Last visit: 15 Oct 2025
Posts: 81
Own Kudos:
65
 [1]
Given Kudos: 16
Posts: 81
Kudos: 65
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let assume the total invitations be 35. (LCM of 5,7)
And donors, press, volunteers, entertainers, organizers, or sponsors as D,P,V,E,O and S.

As per question,
4/7 of 35 => 20 were D
1/5 of 35 => 7 were P
And rest were equally divided, so 35- (27) = 8 which means rest 4( V,E,O and S) have 2 each.

Now, Attendees twice of P came i.e. 14 and Half of V came i.e. 1. No E came means E=0
Rest as per invitation, So we have:

DPVEOS
20141022

Total attendees= 20+14+1+0+2+2 => 39

Fraction of donors in actual attendees = 20/39
User avatar
Sherlock94
Joined: 04 Sep 2022
Last visit: 13 Nov 2025
Posts: 6
Own Kudos:
2
 [1]
Given Kudos: 45
Location: India
Concentration: Sustainability, Entrepreneurship
GPA: 7.5/10
WE:Business Development (Energy)
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let, Total number of guests invited = T

D=4/7 T
P=1/5 T

Remainder = T- 27/35 T= 8/35 T

V=E=O=S= 2/35 T

Actual Attendance:
D'=D = 4/7 T = 20/35 T
P'=2P = 2/5 T= 14/35 T
V'=V/2 = 1/35 T
E'=0
O'=O = 2/35 T
S'=S = 2/35 T
T'= 30/35 T

D'/T' = 20/39 T
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

 

User avatar
AnujSingh0147
Joined: 17 Jun 2024
Last visit: 31 Oct 2025
Posts: 38
Own Kudos:
16
 [1]
Given Kudos: 126
Location: India
GPA: 6.5
Posts: 38
Kudos: 16
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Total = 1
D = 4/7
P = 1/5
Remaining = (1-(4/7+1/5)) = 8/35
Therefore,
V,E,O,S = 2/35 each

On the event day
2P = 2/5
V/2 = 1/35
E = 0

Adjusting the changed value and add:
Changed total = D+P+V+E+O+S = 4/7+2/5+1/35+0+2/35+2/35 = 39/35

Actual attendees of Donors = (4/7) / (39/35) = (20/35)/(39/35) = 20/39
User avatar
MrBombastic
Joined: 07 Aug 2023
Last visit: 15 Nov 2025
Posts: 20
Own Kudos:
11
 [1]
Given Kudos: 49
Location: India
GMAT 1: 700 Q45 V40
GMAT 1: 700 Q45 V40
Posts: 20
Kudos: 11
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Since there are many categories involved, I would like to solve by assuming numbers. I take the LCM of of 4/7 and 1/5 as 35 but since it is not evenly divisible for the other 4 categories, I will assume the total to be 70. Based on the criteria given the split looks as follows:

DPVEOS
Invites40144444

Accounting for the mentioned events, the number of guests arrived as follows -
DPVEOS
Arrived40282044

Therefore the number of donors / total = 40/ 78 = 20 / 39. B is the correct answer.


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

 

User avatar
radhikamehta
Joined: 31 Jan 2023
Last visit: 12 Nov 2025
Posts: 9
Own Kudos:
10
 [1]
Given Kudos: 8
Location: India
GMAT Focus 1: 565 Q80 V80 DI74
GMAT Focus 2: 615 Q86 V79 DI76
GMAT Focus 3: 655 Q85 V81 DI81
GMAT Focus 3: 655 Q85 V81 DI81
Posts: 9
Kudos: 10
 [1]
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

 

Let the total invited guests be 35 (LCM of 5 & 7).

Donors = 4/7*35 = 20, Press = 1/5*35 = 7
Remaining = 35-27 = 8
Equally divided among the other 4 groups; Volunteers = 2, Entertainers = 2, Organizers = 2, Sponsors = 2.

Guests Arrived:
Press = 2*7 = 14 , Volunteers = 1/2*2 = 1 , Entertainers = 0, Donors = 20, Organizers = 2, Sponsors = 2

Total = 14+1+0+20+2+2 = 39

Donors as a fraction of actual attendees = Donors/Actual Attendees = 20/39

Answer: B. 20/39
User avatar
BongBideshini
Joined: 25 May 2021
Last visit: 16 Nov 2025
Posts: 46
Own Kudos:
39
 [1]
Given Kudos: 58
Location: India
Concentration: Entrepreneurship, Leadership
GPA: 3.8
WE:Engineering (Aerospace and Defense)
Posts: 46
Kudos: 39
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
D=\(\frac{4}{7}\), P=\(\frac{1}{5}\),
So, V+E+O+S=1- \(( \frac{4}{7} + \frac{1}{5} )\) =\(\frac{8}{35}\)
As, V=E=O=S,
So, V=E=O=S=\(\frac{2}{35}\)

D: P:V:E:O:S=20:7:2:2:2:2
Or,
D=20x, V=7x, V=E=O=S=2x

Actual attended guest on the day of the event,
D'=20x, P'=14x, V'=x, E'=0, O'=2x, S'=2x
Total attended guest=20x+14x+x+2x+2x=39x

Therefore, the fraction of the actual attendees that were donors =\(\frac{20x}{39x}\)=\(\frac{20}{39}\)

IMO, answer 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

 

User avatar
LucasH20
Joined: 13 Apr 2023
Last visit: 31 Aug 2025
Posts: 52
Own Kudos:
35
 [1]
Given Kudos: 384
Posts: 52
Kudos: 35
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
To solve this problem we can first determine the fraction of the total for each group: donors are given to be [4][/7], the press is [1][/5], using these two values we can determine the remainder which will be divided equally among the other four groups --> Remaining value which will be divided equally among the four other groups = 1 - [4][/7]+[1][/5]=1 - [20+7][/35]=1 - [27][/35]=[8][/35]. Because we have four groups we can divide the remainder equally to [2][/35] for each group. Now that we know all the fraction values we can apply the changes --> press doubles to [14][/35] donors stays [20][/35], volunteers will half to [1][/35], entertainers wont show up and the organizers and sponsers remain [2][/35] each. Thus [donors][/total]=[20][/35]/[39][/35]=[20][/39]. Answer B.

Regards,
Lucas
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

 

User avatar
Rahul_Sharma23
Joined: 05 Aug 2023
Last visit: 12 Nov 2025
Posts: 114
Own Kudos:
82
 [1]
Given Kudos: 17
Location: India
GMAT Focus 1: 695 Q87 V83 DI83
GPA: 2.5
Products:
GMAT Focus 1: 695 Q87 V83 DI83
Posts: 114
Kudos: 82
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
let suppose total 350 people were invited

D = 200
P = 70

remaining = 80 divided in 4 equal parts

V = E = O = S = 20

Actual people attended

D = 200
O = 20
S = 20

E = nil
P = 140
V = 10

Total = 200+20+20+140+10 = 390

Donors fractions = 200/390 = 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

 

User avatar
Manu1995
Joined: 30 Aug 2021
Last visit: 11 Nov 2025
Posts: 81
Own Kudos:
55
 [1]
Given Kudos: 18
Posts: 81
Kudos: 55
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let's assume total number of invited guests = 140

Donors(D) = (4/7)*140 = 80
Press(P) = (1/5)*140 = 28
Remaining = 140-80-28 = 32
These 32 are equally divided among 4 groups,
Volunteers(V)=8, Entertainers(E)=8, Organizers(O)=8, Sponsors(S)=8

Actual attendance:
D= 80,
P= 2*28= 56
V=(1/2)*8 = 4
E= 0, O=S= 8

Total= 156

Fraction of actual attendees who were Donor = 80/156 = 20/39
User avatar
missionmba2025
Joined: 07 May 2023
Last visit: 07 Sep 2025
Posts: 341
Own Kudos:
427
 [1]
Given Kudos: 52
Location: India
Posts: 341
Kudos: 427
 [1]
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

 





We can assume the number of candidates to be a multiple of 35 * 4 as that's the minimum number required.

we can then work the ratios as shown in the above calculation.

Ratio = 80 / 156 = 40 / 78 = 20 / 39

Option B
Attachment:
GMAT-Club-Forum-r5dyw8bu.jpeg
GMAT-Club-Forum-r5dyw8bu.jpeg [ 93.86 KiB | Viewed 157 times ]
   1   2   3   4   5   6   
Moderators:
Math Expert
105385 posts
Tuck School Moderator
805 posts