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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:00
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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. 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 those invited, 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. 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 Official Explanation:



Let total invited = 35 (LCM of 7 and 5, since fractions are 4/7 and 1/5).

  • Donors = 4/7 * 35 = 20
  • Press = 1/5 * 35 = 7
  • Remaining = 35 - 20 - 7 = 8, split equally: 2 each to volunteers, entertainers, organizers, sponsors

Actual attendees:

  • Donors = 20 (unchanged)
  • Press = 2 * 7 = 14
  • Volunteers = 1 (half of 2)
  • Entertainers = 0 (boycott)
  • Organizers = 2
  • Sponsors = 2

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

Donors = 20, so the fraction of actual attendees who were donors is 20/39.

Answer: B.
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:08
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Information given:
- Total invited guests: 100
- Guests belong to exactly one of six groups: donors, press, volunteers, entertainers, organizers, sponsors
- Fractions of invited guests: Donors 4/7, Press 1/5, Remainder: Split equally among volunteers, entertainers, organizers, sponsors
- On the day: twice as many press showed up, half the volunteers attended, entertainers did not attend, donors/organizers/sponsors showed up as invited

Question:
- What fraction of the actual attendees were donors?

Solution
- First, find the invited breakdown
- Donors 4/7 so 20/35
- Press 1/5 so 7/35
- Remaining fraction: 1 - 27/35 is 8/35
- This 8/35 is split equally among four groups, so each gets 2/35

- Then, find actual attendees
- Donors attended 20/35
- Press attended 14/35
- Volunteers attended 1/35
- Entertainers attended 0/35
- Organizers attended 2/35
- Sponsors attended 2/35
- So, in total attendance was 39/35 (more people showed up than were invited originally)

- Donors made up 20/39, so B

Answer: B, 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, 08:13
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GIven,
Fraction of D = 4/7
Fraction of P = 1/5
Fraction of (V+E+O+S) =1-(4/7 + 1/5)=8/35
also, V=E=O=S=2/35

Total attendance
D=4/7
P=2/5
V=1/35
E=0
O=2/35
S=2/35

Total attendees = sum of above = 39/35

Hence fraction of the actual attendees of donors= (4/7)/(39/35)=20/39

Option B
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:13
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Let's say there were 35 invited guests initially ('m choosing 35 because it's divisible by both 7 and 5, which will make our calculations cleaner).

Step 1: Calculate the number of invited guests in each category:
- Donors: 4/7 of 35 = 20 people

- Press: 1/5 of 35 = 7 people


- Remaining: 35 - 20 - 7 = 8 people

- These 8 people are divided equally among 4 groups (volunteers, entertainers, organizers, sponsors), so each group has 2 people

Step 2: Calculate the actual attendance based on the given conditions:
- Donors: All 20 showed up


- Press: Twice as many as invited, so 2 × 7 = 14 people

- Volunteers: Half showed up, so 1/2 × 2 = 1 person

- Entertainers: None showed up (0 people)
- Organizers: All 2 showed up

- Sponsors: All 2 showed up


Step 3: Calculate the total actual attendance:

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

Step 4: Calculate the fraction of donors among actual attendees:
Fraction of donors = 20/39
Therefore, the answer is B 20/39
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:14
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Let D, P, V, E, O, and S be donors, press, volunteers, entertainers, organizers, or sponsors

D+P = 4/7+1/5 = 27/35 remaining would be 2/35 each

Let's say total people were 35 comprising D=20, P=7, V,E,O,S=2 each

Actual people arrived:
D=20, P=14, V=1, E=0, O=2, S=2

D/Actual = 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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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 total number of guests to be 700.
Therefore according to question,
donor = 4/7 * 700 = 400
press = 1/5 * 700 = 140

Remaning = 700 - (400+140) = 160

And then volunteers, entertainers, organizers, or sponsors(equally divided) = 40

On the event day => Donors = 400,
Press= 2 * 140 = 280
Volunteers. = 40/2 = 20
And entertainers = 0,
organizers= 40
sponsors = 40
Then the total = 780
Hence the percentage of donors = 400/780 = 20/39 Option B Answer.
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:15
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Donors = 4/7 Press = 1/5
Let the total no of invited people be 35 (LCM of 7 & 5)
Then
Donors (D) = 20
Press (P) = 7
Remaining = 8
Volunteers (V) = Entertainers (E) = Organizer (O) = Sponsors (S) = 2 each

Day of the event, actual count of people
P = 2*7 = 14
V = 1/2 *2 = 1
E = 0
D = 20
O = 2
S = 2

Total = 39
Hence donors are 20/39 of the total
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Initially, we have the following invited :
Donors D = 4/7
Press P = 1/5
Volunteers V , Entertainers E, Organizers O, Sponsors S = (1-(D+P))/4 = 8/35/4 = 2/35 each.

Actual attendees :
P = 2/5
V = 1/35
E = 0
D = 4/7
O = S = 2/35

So now, fraction = D/(P+V+E+D+O+S)
On substituting above values and solving the equation, we get ,
Fraction = 20/39
Answer : Option B
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:17
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Let, total number of attendees = 70
D= (4/7)*70=40 ......all attended =40
P= (1/5)*70=14 ........twice attended =2*14=28
Remaining =70-40-14=16 were divided equally among V,E,O, S
V=4 ..........only half attended =4/2=2
E=4........ all cancelled or no one attended =0
O=4....... all attended =4
S=4 ....... all attended =4

Total people who attended = 40+28+2+4+4=78

Fraction of actual attendees who were donors= 49/78=20/39

B
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:19
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To solve this complex fraction problem, the most efficient method is to assume a convenient integer for the total number of invited guests. A "smart number" that is a multiple of the denominators in the problem (7 and 5) and also of 4 (for the four groups sharing the remainder) will avoid cumbersome fraction arithmetic. The least common multiple of 7, 5, and 4 is 140. Assuming a total of 140 invited guests, the number of invitees in each category can be calculated. Invited donors are (4/7) * 140 = 80. Invited press are (1/5) * 140 = 28. The remaining 140 - 80 - 28 = 32 guests are divided equally among the four other groups, resulting in 8 invitations each for volunteers, entertainers, organizers, and sponsors.

Next, the number of actual attendees for each group is determined based on the changes described. The number of attending donors, organizers, and sponsors remained the same at 80, 8, and 8, respectively. The number of attending press doubled to 2 * 28 = 56. The number of attending volunteers was halved to 8 / 2 = 4. The entertainers did not attend, so their count is 0. The total number of actual attendees is the sum of these new figures: 80 + 56 + 4 + 0 + 8 + 8 = 156.

The final step is to find the fraction of actual attendees who were donors. This is the ratio of actual donors to total actual attendees, which is 80/156. This fraction simplifies by dividing both the numerator and the denominator by their greatest common divisor, 4. The resulting fraction is 20/39.

The correct answer is (B).
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:21
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Read the question, understood that since the fraction have 7 and 5 in denominations, Assumed 35 as total invited guest (LCM of 7,5)
Check If the fraction works or we could take multiple of 35.

Invitees: 35,
Donor: \(\frac{4}{7}* 35 = 20\),
Press: \(\frac{1}{5} * 35 = 7\)
remaining= 8, equally divided.
V=2, E = 2, O =2, S= 2.

Now attended.
Donor = 20, Press = 14, V =1, E =0, O =2, S =2 => Total = 39.

\(\frac{Donor}{ Actual Attendees} = \frac{20}{39}\)

B is the 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


 


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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:22
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Initially, we assume the total number of individuals to be N.
The problem stipulates that there is no overlap.
Thus, the expected distribution of invitations is as follows:

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

Upon actual attendance, it was observed that:
D = (20/35)N
P = (7/35)N *2 = (14/35)N
V = (2/35)N *0.5 = (1/35)N
E = (2/35)N * 0 = 0
O = (2/35)N = (2/35)N
S = (2/35)N = (2/35)N

The proportion of D among the total attendance is calculated as:
(20/35)N / ((20/35)N + (14/35)N + (1/35)N + (2/35)N + (2/35)N) = 20/39
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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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Solution:

Fraction of donor guests invited = 4/7

Fraction of press guests invited = 1/5

Remaining guests were equally divided among volunteers, entertainers, organizers and sponsors.

So, lets assume the total number of guests invited as 35 (LCM of 7 and 5)

Therefore,

Number of donor guests invited = 20
Number of press guests invited = 7
Number of volunteers = Number of entertainers = Number of sponsors = Number of organizers = (35-20-7)/4 = 2

Further,
Only half the volunteers attended the chartity gala.
Therefore, the number of volunteers who attended the charity gala = 1.

Also, twice as many press guests arrived as were invited
Therefore, the number of press guests who attended the charity gala = 14

Number of entertainers who attended the charity gala = 0.

Therefore, total number of guests who attended the charity gala = 20+14+1+2+2 = 39.

Therefore the required fraction of donors is 20/39.
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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. 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?

Let the number of total guests be x

Invited guests : -
The number of donors = 4x/7
The number of press = x/5
Remaining number of guest = x - 4x/7 - x/5 = (35 - 20 - 7)x/35 = 8x/35
The number of volunteers = The number of entertainers = The number of organizers = The number of sponsors = 2x/35

Guests attended:
The number of donors = 4x/7
The number of press = 2x/5
The number of volunteers = x/35
The number of entertainers = 0
The number of organizers = 2x/35
The number of sponsors = 2x/35

Total number of guests attended = 4x/7 + 2x/5 + x/35 + 2x/35 + 2x/35 = 39x/35

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

IMO B
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Taking numbers, let us assume total guests are 35
donors = 4/7(35) = 20
press = (1/5)35 = 7
rest all 4 groups 2 each

---> after change press doubled --> (14), leading to +7
volunteers is 1/2(2) --> 1, leading to -1
entertainers --> 0, leading to -2
total new count is 35+7-3 = 39

Answer is 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, 08:25
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let total guests be x
D = 4x/7
P= x/5
v=e=o=s=x/2
total x be 70
so D = 40
P = 14
v=e=o=s = 4
day of event attendance is
p = 28 v=2 e=0 d=40 o=s=4
total 78

actual donors atttendes are 40/78 ; 20/39

OPTION B 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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nissamis00
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:26
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To determine the fraction of actual attendees who were donors, I began by analyzing the distribution of invited guests:
  • Donors: 4/7 of the total
  • Press: 1/5 of the total
To work with common denominators, I converted both to have a denominator of 35:
  • Donors: 4/7 = 20/35
  • Press: 1/5 = 7/35
So donors and press together accounted for 20/35 + 7/35 = 27/35 of the guests, leaving 8/35 for the remaining four groups (volunteers, entertainers, organizers, sponsors). That means each of these four groups had 2/35 of the invited guests.
Now, adjusting for actual attendance:
  • Donors: attended as expected → 20/35
  • Press: twice the invited → 2 × 7/35 = 14/35
  • Volunteers: half attended → 1/2 × 2/35 = 1/35
  • Entertainers: none attended → 0
  • Organizers: full attendance → 2/35
  • Sponsors: full attendance → 2/35
Now summing the fractions for actual attendees:
20/35 + 14/35 + 1/35 + 0 + 2/35 + 2/35 = 39/35
So, the fraction of actual attendees who were donors is:
(20/35) ÷ (39/35) = 20/39
IMO Answer: (B) 20/39
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Harika2024
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:27
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Based on given conditions, let's check the initial fraction of each group invited:

Donors: 4/7

Press: 1/5

Remaining: 1−(4/7+1/5)

Find a common denominator for 7 and 5, which is 35.

4/7=20/35

1/5=7/35

Remaining: 1−(20/35+7/35)=1−27/35=8/35

The remaining 8/35 is divided equally among Volunteers, Entertainers, Organizers, and Sponsors.

Each of these four groups: (8/35)/4=2/35

So, the initial invited fractions are:

Donors: 20/35

Press: 7/35

Volunteers: 2/35

Entertainers: 2/35

Organizers: 2/35

Sponsors: 2/35


Now let's calculate the fraction of each group that actually attended:

We can assume a total number of invited guests to make calculations easier. Let's assume there were 35 invited guests

Donors: 4/7 of 35=20 donors invited. They showed up as expected, so 20 donors attended.

Press: 1/5 of 35=7 press invited. Twice as many arrived, so 7∗2=14 press attended.

Volunteers: 2/35 of 35=2 volunteers invited. Only half attended, so 2/2=1 volunteer attended.

Entertainers: 2/35 of 35=2 entertainers invited. They did not attend, so 0 entertainers attended.

Organizers: 2/35 of 35=2 organizers invited. They showed up as expected, so 2 organizers attended.

Sponsors: 2/35 of 35=2 sponsors invited. They showed up as expected, so 2 sponsors attended.

Now let's calculate the total number of actual attendees:

Total attendees = Donors + Press + Volunteers + Entertainers + Organizers + Sponsors
Total attendees = 20+14+1+0+2+2=39

4. Calculate the fraction of actual attendees who were donors:

Fraction of donors = (Number of donors who attended) / (Total number of actual attendees)
Fraction of donors = 20/39
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 08:27
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Let there are 140 guests:
Donors=4*140/7=80
press=0.2*140=28
vol=ent=org=spon=8

new pres=56
ent=0
volun=4

frac=80/156=20/39

Ans B
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