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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 12:26
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First of all, we need to calculate the ratio of the 4 groups mentioned. For simple calculations, we multiply the ratio by 35, so we have whole numbers for donors and press.
For the groups besides donors and pres,s we have: \(4a+20+7=35\) so, \(a=2\)

For the event, though, twice as many press showed up, which is \(2*7=14\)
half of the volunteers, which is \(2/2=1\)
And no entertainer which is 0.
The rest attended as the same as invited.
So the new total is \(20+14+1+0+2+2=39\)
And the ratio of donors to total attendance is : \(20/39\)

donors press volunteers entertainers organizers sponsors Total
Ratio invited 4/71/5a/35a/35a/35a/351
Ratio invited 207222235
Ratio attended 2014102239

Option B is correct.


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, 12:27
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donors, press, volunteers, entertainers, organizers, sponsors
Invited 4/7 1/5 2/35 2/35 2/35 2/35

Actual 4/7 2/5 1/35 0 2/35 2/35


donors/Actual = 4/7 / (4/7 + 2/5 + 1/35 + 0 + 2/35 + 2/35) = 20/39

Therefore, option 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


 


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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 12:28
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Topic(s)- Non-Overlapping Sets, Fractions, Least Common Denominator
Strategy- Algebra
Variable(s)- donors = "D"; press = "P"; volunteers = "V"; entertainers = "E"; organizers = "O"; sponsors = "S";
guests invited = subscript "i"; guests arrived = subscript "f"; total # invited = "Ti"; total # arrived = "Tf"
Rephrase the Question: What is Df/Tf?

1. Ti = Sum(invited)/Ti = Ti/Ti = 1/1 (Given each guest belongs to one group)
Ti = Di + Pi + Vi + Ei + Oi + Si
i) Di = (4/7)(Ti) = (20/35)(Ti)
ii) Pi = (1/5)(Ti) = (7/35)(Ti)
iii) Vi = Ei = Oi = Si = (1/4)*(Remaining guests invited)
= (1/4)*(Ti - (Di + Pi))
= (1/4)*(Ti - (4/7)(Ti) + (1/5)(Ti))
= (1/4)*(Ti)*(35/35 - ((20/35)+ (7/35))
= (2/35)(Ti)

2. Find Tf
Tf = Df + Pf + Vf + Ef + Of + Sf
i) Df = Di = (20/35)(Ti)
ii) Pf = 2*Pi = 2*(7/35)(Ti)
=(14/35)(Ti)
iii) Vf = (1/2)*(Vi) = (1/2)*(2/35)(Ti)
= (1/35)(Ti)
iv) Ef = 0
v) Of = Oi = Sf = Si = (2/35)(Ti)
vi) Tf = (Ti/35)*( (20) + (14) + (1) + (0) + (2) + (2) )
= (39/35)(Ti)

3. Df/Tf
(20/35)(Ti) / (39/35)(Ti)

Answer: B
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 12:34
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let T = total number of attendees and let T = 100

100 x4/7 = about 57 people; round to 57 since you can't have partial a person. = 57 Donors

100 x 1/5 = 20 = 20 press =

57+20 = 77; 100 -77= 23 people remaining divided by 4 categories; let's round up to 24 to make the math easier, leaving 6 people divided evenly in the remaining 4 categories.

we're told on the day of:
twice as many press guests showed up: 20 x 2= 40 press

half the volunteers attended: i.e. 6/2 = 3 volunteers

NO entertainers: 3-3= 0

Donors, Organizers, and sponsors were as expected: 57, 6, and 6 Respectively = 69 total+3+40+= 112 total attendees

of which donors are 57/112

57 x2 = 114, so 57/122 = a little more than double: 20/39 looks like a good guestimate for this: Answer choice 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, 13:04
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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’s take the total guest to be 35. And the donors, press, volunteers, entertainers, Organizers and sponsors be denoted
as D, P,V ,E,O and S respectively.

Total guest 35:

D = (4/7)*35 = 20

P = (1/5)*35 = 7

remaining = 35-27 = 8 and it’s divided equally amongst V,E,O and S = 2 each.

V = 2

E =2

O =2

S =2

New information about guests based on the new constraints : 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.

D = 20
P = 2*7 = 14

V = 1

E = 0

O = 2

S =2

Total new guest count = 20+14+1+0+2+2 = 39

Fraction of actual attendees as donors = 20/ 35

= 4*5 / 5*7

= 4/7

Option A
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 13:58
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It is given that invitations were distributed as follows:
D: 4/7
P: 1/5
V: Remaining/4
E: Remaining/4
O: Remaining/4
S: Remaining/4

Remaining = 1-(4/7+1/5) = 8/35
Hence, V = E = O = S = 2/35
Let the total invitations = 35. Then D = 20, P = 7, V = E =O = S = 2

Attendees: D = 20; P = 14; V = 1; E=0; O=2; S=2

Required fraction: D/ T = 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, 14:40
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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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Kinda tricky - i got caught up by the denominator.
Here, its easy to work through absolute numbers. Do decide which to use, we just use the given relations (4/7 and 1/5)
With 7*5 = 35 we get a nice multiple to use.

35 is our total, 20/35 d and 7/35 p are planned. besides that, 2/35 are the reminder for each of the other groups.
That said, we know can go on with the conditions given.
Therese are:
14 instead of 7 p. 1+0+2+2 of the remainders, and 20d.
If we add those together, we get as a result: 39.
Therefore, we have 39 attendees and 20 of them are donors.

That results in B)
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 16:59
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Donors= 4/7 = 20/35
Press= 1/5 = 7/35
The other 4 ratio= (1-4/7 - 1/5)* 1/4 = 2/35
Therefore before the event anticipated ratios are as follow( initials used):
D:P:V:O:S:E= 20:7:2:2:2:2

Making adjustments based on attendance- multiply press guestx2, volunteers by 1/2 and remove entertainers the ratios will now be as follows:
20:14:1:2:2
Therefore actual attendees who were donors =20/ (20+14+1+2+2)= 20/39
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 17:29
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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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donorpressvolunteerentertainerorganizersponcer
Before Event(Invited Number)4/71/52/352/352/352/35
During Event (Actual Number)4/72/51/3502/352/35

So, we just need to find, (d)/(Sum of all fraction for actual number). When we solve it, we get 20/39. So, (B) is right answer
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 17:54
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B. 20/39

Get all to common denominator:
Donors = 20/35
Press = 7/35
Volunteers = ((35-20-7)/4)/35 = 2/35
Entertainers = 2/35
Organizers = 2/35
Sponsors = 2/35

Adjustments:
Press 7 -> 14
Volunteers 2 -> 1
Entertainers 2 -> 0

Total: +4

Donors = 20/(35+4) = 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, 18:38
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Let total no of attendees = 35

No of donors=4/7*35= 20
No of press= 1/5*35= 7
Remaining attendees= 35-7= 8
They are equally divided to volunteers, sponsors, entertainers, organizers

So each will have 2 attendees as 8/4 =2

Now day of event

Press twice attended= 2*7= 14
Volunteers half attended= 2/2=1
Entertainers cancelled = 0
Donors, organizers and sponsors as invited attended

So Donors= 20, organizers= 2, Sponsors= 2

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

Fraction of actual attendees are donors= 20/39
Answer should be B.
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 18:43
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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's assume that 35 guests were invited. Now we can divide them among the groups.

D : P : V :E :O : S = 20 : 7 : 2 : 2 : 2 : 2
Now we need to look at who attended the gala.

D : P : V : E : O : S = 20 : 14 : 1 : 0 : 2 : 2 => 39 TOTAL.

Donors to Attendees = 20/39.

IMO B
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 18:55
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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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Answer: B (20/39)

Fraction of total guests invited = 1
Donors = 4/7
Press = 1/5
Remaining guests = 1 - 4/7 - 1/5 = 8/35
Fraction of each remaining group = [(8/35)/4] = 2/35

Using the net change(s):
-Twice as many invited press came [(2*1/5) - 1/5], so net increase of: +1/5
-Only half of the volunteers came[(1/2*2/35)-2/35], so net decrease of: -1/35
-Entertainers did not show, so automatic net decrease: -2/35
Total net change = (1/5)-(1/35)-(2/35) = 4/35

Fraction of total actual attendees = 1+(4/35) = 39/35
Fraction of actual donors who attended = [(4/7) / (39/35)] = 4/7*35/39 = 20/39
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 18:57
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Let us assume total invited attendees were 35
Donors = 4/7 of 35 = 20
Press = 1/5 of 35 = 7
Rest = 35-27 =8, equally divided into Volunteers, Sponsors, Organizers, and Entertainers, so each will have 2


Type of guest Invited Attended
Donors2020
Press714
Volunteers21
Sponsors22
Organizers22
Entertainers20
Total3539

From above table, fraction of actual attendees, that were donors = 20/39
Answer is B
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 19:49
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Invitees:
Donors \(=\frac{4}{7}= \frac{20}{35}\)
Press \(=\frac{1}{5}=\frac{7}{35}\)

Remaining \(\frac{8}{35}\) were divided equally among the other four groups

\(Volunteers=Entertainers=Organizers=Sponsors =\frac{2}{35}\)

Lets say that 35 guests were invited

Attendees:

\(Press=2*7=14\)

\(Volunteers=\frac{2}{2}=1\)

\(Entertainers=0\)

\(Donors=20\)

\(Organizers=2\)

\(Sponsors =2\)

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

\( \text{Fraction of Donors }= \frac{20}{39}\)

Answer: B
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Invited:
donors + press + volunteers + entertainers + organizers + sponsors = 4/7 + 1/5 + x = 27/35 + x = 1

x = 1 - 27/35 = 8/35

volunteers, entertainers, organizers and sponsors have 2/35 each one

On the day of the event:
donors + press + volunteers + entertainers + organizers + sponsors = 4/7 + 2/5 + 1/35 + 0 + 2/35 + 2/35 = 39/35

donors/actual attendees = (4/7)/(39/35) = (20/35)/(39/35) = 20/39

IMO B
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 20:17
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Data: Invited "I" = D + P + V + E + O + S | D = 4I/7; P = I/5, rest equally divided (D+P) = 27 I/35 -> rest = 8I/35 (2I/35 each)
Question: From attendees "A", what fraction were D?

Analysis: A = Da + Pa + Va + Ea + Oa + Sa -> Pa = 2P = 2I/5; Va = V/2 = I/35; Ea = 0; Da = D = 4I/7; Oa = O = 2I/35; Sa = S = 2I/35 -> A = 39I / 35

Answer: Da / A = 20I/35 / 39I/35 = 20/39 -> B

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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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SaanjK26
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 20:32
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donors = 4/7, press = 1/5 total = 27/35.

Remaining 8/35 split equally among 4 groups →
each = 2/35 (volunteers, entertainers, organizers, sponsors).

On event day: donors (4/7), press (2/5), volunteers (1/35), entertainers (0), organizers (2/35), sponsors (2/35).

Total attendees = 20/35 + 14/35 + 1/35 + 0 + 2/35 + 2/35 = 39/35.

5. Donor fraction = (20/35) ÷ (39/35) = 20/39 →

Answer: (B)
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 20:55
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Invited People:
donors=4/7
press=1/5
volunteers, entertainers, organizers, sponsors = 2/35

But attended people:
donors=4/7
press=2/5
organizers, sponsors = 2/35
Volunteers= 1/35
entertainers=0

fraction of donors= 4/7/(4/7+2/5+1/35+4/35)= 20/39.
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GMAT Club Olympics 2025 (Day 6): At a charity gala, each invited guest 07 Jul 2025, 21:03
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Given,
Invitation of Charity Gala sent in below manner,
1. The fraction of Donor = 4/7
2. The fraction of Press = 1/5
3. The fraction of Volunteer = (1 – (4/7 + 1/5))/4 = 2/35
4. The fraction of Entertainer = 2/35
5. The fraction of Organizer = 2/35
6. The fraction of Sponsor = 2/35
Assume the value of total invited people = q, & total people who came = r
Actual attendees,
1. The fraction of Donor, D = 4q/7
2. The fraction of Press, P = 2*1/5 = 2q/5
3. The fraction of Volunteer, V = (2/35)/2 = 1q/35
4. The fraction of Entertainer, E = 0
5. The fraction of Organizer, O = 2q/35
6. The fraction of Sponsor, S = 2q/35

Now, solve,
D + P + V + E + O + S = r
4q/7 + 2q/5 + q/35 + 0 + 2q/35 + 2q/35 = r
(20 + 14 + 1 + 0 + 2 + 2) *q/35 = r
39q/35 = r
q = 35r/39

Now, we need to find the fraction of the actual attendees who were donors, means (fraction of Donor in terms of r)

= 4q/7
= 4/7 *(35r/39)
= 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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