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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:29
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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


 


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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:30
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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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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
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:32
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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


 


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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:35
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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


 


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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:40
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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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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
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:41
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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
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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


 


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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:47
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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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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.
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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
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:58
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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
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:00
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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


 


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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;
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:29
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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


 


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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:29
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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.
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:33
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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
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:35
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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)
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 09:38
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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
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ckavya2691
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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.
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cabii
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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
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muuss
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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
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