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Prathu1221
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 21:44
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D=4/7, P=1/5,V=E=O=S=2/35, now Press became 2/5 and E became zero.
So total number present will be sum of all ie 39/35 which seems wierd but after we take ratio of donors ie. 20/35 now we get 20/39 ie 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


 


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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 21:58
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Lets take sweet number; lcm of the fractions (5, 7) = 35

inv_donors (4/7) = 20/35 = 20
inv_p (1/5) = 07/35 = 7
remaining = 8 = equally divided among rest => 2 in each of the four category
---
showed up
show_p (x2) = 7x2 = 14
show_v (/2) = 2/1 = 1
show_e = 0
..rest showed up as expected
show_d = 20
show_(remaining 2 categories = 2, 2

total showed = 14 + 1 + 0 + 20 + 2 + 2 = 39
faction_of_donors_showed = show_d / total_showed = 20/39
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Let d, p, v, e, o & s represent donors, press, volunteers, entertainers, organizers, & sponsors.

Given,
d = 4/7 = 20/35
p = 1/5 = 7/35
v + e + o + s = 8/35
v = e = o = s = 2/35
Let the total number of guests be 35.

Attendees:
v(only half attended) -> 1 guest (-1)
e(none attended) -> 0 guests(-2)
p(twice as many attended) -> 14 guests(+7)
All remaining attended as planned.
Total = 35 - 1 - 2 + 7 = 39
Fraction of the actual attendees who were donors = d/39 = 20/39

Therefore the correct option is Option B
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Correct Answer: Option B 20/39
Given information:
Donors 4/7
Press 1/5
If we take Donors only than remaining other will 3/7
Deduct press invitation from remaining we get 8/35
Now we will divide equally among volunteers, entertainers, organizers, and sponsors.
So invited groups is below mentioned:
Donors 4/7
Press 1/5
Volunteers 2/35
Entertainers 2/35
Organizers 2/35
Sponsors 2/35

Here we found attended groups using given information,

GroupsInvitedAttended
Donors4/74/7
Press1/52/5
Volunteers2/351/35
Entertainers2/35Not Attended
Organizers2/352/35
Sponsors2/352/35

Total Attendance= 4/7+2/5+1/35+2/35+2/35
= 39/35
To find percentage of donors attended= (4/7)/(39/35)
= 20/39
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Find a common total Since we have fractions 4/7 and 1/5, let's use 35 invited guests (LCM of 7 and 5).

Invited guests breakdown
  • Donors: (4/7) × 35 = 20
  • Press: (1/5) × 35 = 7
  • Remaining: 35 - 20 - 7 = 8
  • Each of the 4 remaining groups gets 8 ÷ 4 = 2 people each
So: Volunteers = 2, Entertainers = 2, Organizers = 2, Sponsors = 2


Actual attendees
  • Donors: 20 (as expected)
  • Press: 7 × 2 = 14 (twice as many)
  • Volunteers: 2 × 0.5 = 1 (half attended)
  • Entertainers: 0 (boycotted)
  • Organizers: 2 (as expected)
  • Sponsors: 2 (as expected)
Total attendees = 20 + 14 + 1 + 0 + 2 + 2 = 39

Fraction of attendees who were donors = 20/39
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 23:37
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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


 


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Here, we can assume the total number of guests invited to be a multiple of 7, 5, and 4 to make our calculations easier.
Let total guests invited be 1400

Invited guests:
Donors: 4/7*1400
= 800
Press: 1/5*1400
= 280
Remaining: 1-(4/7+1/5)
= 8/35
Volunteers, Entertainers, Organizers, Sponsors = 8/35*1400
= 80

Guests who attended:
Press: 2*280 =560
Volunteers: 80/2= 40
Entertainers: 0
Donors: 800
Organizers: 80
Sponsors: 80
Total: 1560

Fraction of attendees who were donors: 800/1560= 20/39

Answer: B 20/39
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Let's assume total attendees as x, and represent the six groups as D,P,V,E,O,S

D = 4x/7
P = x/5

rest of them are equally divided among 4 groups, let's assume y, so

4y = x - (P+D) = 8x/35
y = 2x/35

so, we get,
D = 4x/7
P = x/5
V = 2x/35
E = 2x/35
O = 2x/35
S = 2x/35

On the day of the event

P = 2x/5
V = x/35
E = 0
O = 2x/35
S = 2x/35
D = 4x/7

Now the total number of actual attendees = D+P+V+E+O+S = 39x/35

Fraction of donors to actual attendees = (4x/7)/(39x/35) = 20/39
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other invitations = 1-(4/7+1/5)=8/35
equally divided among 4 groups= 2/35 each
donorpressvolunteerentertainerorganisersponser
invitations4/71/52/352/352/352/35
attendees4/72/51/3502/352/35

total attendees = 39/35

fraction of the actual attendees that were donors = (4/7)/(39/35) = 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


 


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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


 


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Let total guest invited be x, donors = 4x/7, press=x/5, Remaining = x- (4x/7+x/5) = x- 27x/35= 8x/35
volunteers = entertainers = organizers = sponsors = 2x/35

Guest Arrived ---> Press = 2x/5 , entertainers = 0, volunteers = x/35, organizers = sponsors = 2x/35, donors = 4x/7.
Total Attendees = 2x/5+4x/7+4x/35 +x/35= (14+20+5)x/35 = 39x/35.

Requisite fraction = (4x/7)/ (39x/35) = 4x * 35/7*39x =20/39 B.
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20/35=D
7/35= P
remaining are 2/35 each

Arrived:
D=20/35, P=14/35, V=1/35, E=0, S=O=2/35 each

So Donors/ total= (20/35) / (39/35)=20/29
Ans B
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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


 


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donors - 4/7
press - 1/5
Remaining = 1 - 4/7 - 1/5 = 1 - 27/35 = 8/35

Dividing equally we get

volunteers - 2/35
entertainers - 2/35
organizers - 2/35
sponsors - 2/35

donors - 4/7
press - 2/5
volunteers - 1/35
entertainers - 0
organizers - 2/35
sponsors - 2/35

New total = 39/35

We get donor/total = 4/7 * 35/39 = 20/39

Alternatively

Lets say the total guests = 35
Then
donors - 20
press - 7
volunteers - 2
entertainers - 2
organizers - 2
sponsors - 2

On the day of the event below people arrived
donors - 20
press - 14
volunteers - 1
entertainers - 0
organizers - 2
sponsors - 2

=> donor/total = 20/39

Option B
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I used Smart-Number for this ratio problem: Intuitively, i will use pick 35 as the total number of attendees, then:
- 20 as donors, 7 as press, and 8 remains splitting equally among volunteers, entertainers, organizers, or sponsors.
After the unexpected results change we have: still 20 as donors, 14 now as press, 1 for volunteers, 4 for the rest
- Therefore, we have 20/39 as the fraction of the actual attendees were donors
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Let the total attendees =35

Initially,

Donors: 4/7* 35= 20
Press: 1/5* 35= 7

The remainder= 35-20+7
= 8

This was split equally among the 4 teams

Hence 2 attendees for volunteers, entertainers, organisers, sponsors.

Actual attendees
Press: 2*7= 14
Donors: 20
Volunteers: 1/2 * 2= 1
Entertainers= 0
Organisers= 2
Sponsors= 2

New total= 39

New fraction for donors= 20/39
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Let the total guest be T
Invitee:
Donor=4T/7
Press=T/5
Remaining member {V,E,O,S}=T-(4T/7+T/5)
=> T-(27/35) T = 8T/35
donors, press, volunteers, entertainers, organizers, or sponsors.

Now this will be divided into 4 parts so, 2T/35 each

On the actual day:
Press =2T/5
Volunteers= T/35
Entertainers =0
sponsors and Organizers =2T/35 * 2=4T/35
Donors=4T/7

FRACTION OF DONORS in this =

(4/7)T/(2T/5+4/7T+T/35+4T/35)

(4/7)T / (14+20+1+4)/35T

4/7*35/39
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


 


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Assume the number of all guests invited is 35. Then there is:
  • 4/7 * 35 = 20 donors
  • 1/5 * 35 = 7 press
  • and the remaining 35 - 20 - 7 = 8 people is split equally among 4 other groups, 2 people for each.
On the day of event, we have:
  • no. of donors as expected: 20
  • no. of press doubled: 14; no. of volunteers half: 1
  • no. of entertainers: 0
  • Other 2 groups have 2 people each.
Totally, we would have 20 + 14 + 1+ 2*2 = 39 guests, in which 20 of them is donors.
=> The fraction shall be 20/39, which 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


 


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Let's assume that 35 (LCM of 7 and 5) people were invited
4/7 of 35 is 20 donors and all of them come.
1/5 of 35 is 7 press, but double of that or 14 actually show up.
The remaining 8 guests are split evenly into 4 groups.
So each of volunteers, entertainers, organizers, and sponsors has invited 2.
-> Only half the volunteers come, so 1 volunteer shows up.
-> The entertainers did not attend, so 0 show up.
-> All 2 organizers and all 2 sponsors come.

We can add up everyone who arrived :
20 donors + 14 press + 1 volunteer + 0 entertainers + 2 organizers + 2 sponsors = 39 guests in total
Out of these, 20 are donors. So the fraction of attendees who are donors is 20/39
Answer : B
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Simplest way i found for this question was taking the LCM of the two denominators we are given. that is, 7 and 5 , which comes to 35. Let's assume total to be 35.

So, we can arrive at the individual count for all the categories -

2 (Remaining dividing equally)2 (Remaining dividing equally)
TypeInvitedChange?Actuals
Donors 4/7 of 35 = 20 No Change20
Press1/5 of 35 = 7Double Attended14
Volunteers 2 (Remaining dividing equally)Half Attended1
Entertainers2 (Remaining dividing equally)None attended 0
Organizers2 (Remaining dividing equally)No Change2
Sponsors2 (Remaining dividing equally)No Change2
TOTAL39


Hence donors as a fraction of the actual attendees - 20 / 39. (B)
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Choose invited guests = 35
As donors = 20 and press = 7, there are 35-27 = 8 guests in the other four groups, 2 each one:
volunteers = 2
entertainers = 2
organizers = 2
sponsors = 2

Actual attendes:
donors = 20
press = 14
volunteers = 1
entertainers = 0
organizers = 2
sponsors = 2

total: 39

fraction = donors/total = 20/39

The right answer is B
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Ans : B
Let Donors : D, Press : P, Volunteers : V, Entertainers : E , Organisers : O & Sponsors : S & "n" be the total number of invited guests
Below are the numbers for invited and attended guests for each guest type

D : Invited = 4n/7 Attended = 4n/7
P : Invited = n/5 Attended = 2n/5 (given that press numbers is twice the invited number )
since the remaining invitees were equally divided => n - (4n/7 + n/5) = 8n/35 was equally divided among the 4 remaining guest types => no of each guest type = 2n/35
V : Invited = 2n/35 Attended = n/35
E : Invited = 4n/35 Attended = 0
O : Invited = 4n/35 Attended = 2n/35
S : Invited = 4n/35 Attended = 2n/35

Required fraction = Donor attended / actual attendees = (4n/7) / (39n/35) = 20/39
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 08 Jul 2025, 05:30
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This can be solved relatively easily if taken some numbers, because we have 7 and 5 in denominators, lets take 35 for the cal to be easier, convert the fractions to integers by multiplying by 35, Donors would come to be 20 and the total when added up according to the conditions given would come up as 39, so B, it is.
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


 


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