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12 Days of Christmas GMAT Competition 2025 - Day 6: At a professional

1 2 3 4

AviNFC

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22 Dec 2025, 11:01

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only NT=x
only DA=6x
both NT & DA but not PM =p
only PM=12 p

given: x= 0.25*12p =3p
12p+k=220

so, 7x+p+220=308; x=3p
p=4

reqd only one= 7x+12 p =33p = 33*4 =132

Ans D

linnet

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Only P.M- X
NT- Y
DA- Z
DA+NT= A
only PM= 12(A) Total PM IS 220 & Total Participants= 308
None PM are, only DA= Z
Only NT = 3A
only DA+NT= A
Z+3A+A=Z+4A since Total PM is 220 , Z+4A=308-220=88
Z= 88-4A So only PM+ only NT +only DA= 12A+3A+(88-4A)= 11A+88
we find valid integer values of A- lets test small valid values
try A= 4 so only PM=48, Only NT=12, Only DA= 88-16= 72
Exactly one program will e 48+12+72= 132
Final answer is 132

Xdsa

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only data analysis =18x
only project management =12x
only negotiation=3x
data analysis and negotiation only=x
project management= 220
remaining is 88
18x + 3x + x=88
x=4
exactly 1 = (12+3+18)x =33.4=132 is the answer

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22 Dec 2025, 11:19

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Let x be the number of participants attending only Negotiation training. 6x will be the number of participants attending only Data Analysis. Further let y be the number of participants attending both Data Analysis and Negotiation but not project management. 12y will be the number of participants attending only Project Management. Further , the number who attended only Negotiation Training was one-quarter of the number who attended only Project Management.

So , x = 0.25*12y = 3y

if 220 participants attended the project management workshop , then we need to find the number of participants attended exactly of the three workshops (z):

z= 6x+x+12(x/3) = 7x+4x = 11x

220 = 308-6x-x/3-x

x=12

z=12*11 = 132

So , answer is 132.

Bunuel

At a professional conference, 308 participants signed up for at least one of three evening workshops: Data Analysis, Project Management, and Negotiation Training. Exactly six times as many participants attended only Data Analysis as attended only Negotiation Training. The number of participants who attended only Project Management was twelve times the number who attended both Data Analysis and Negotiation Training but not Project Management. The number who attended only Negotiation Training was one-quarter of the number who attended only Project Management. If 220 participants attended the Project Management workshop, how many participants attended exactly one of the three workshops?

A. 48
B. 88
C. 120
D. 132
E. 136

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gemministorm

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22 Dec 2025, 11:36

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Acc. to question - 308 participants signed for at-least one DA/PM/NT -> none = 0.
let us assume only NT = x, DA & NT but not PM = y (this is for one of the statements mentioned in the question)
Given: DA only = 6x, PM only = 12y and x = 12y/4 = 3y -- (i)
also apart from 220 remaining from 308 are 6x (DA only) + x (NT only) + y (DA & NT but not PM) = 7x + y = 308 ---(ii)
from (i) and (ii) we get y = 4.
we need to find -> exactly 1 of 3 -> x (NT only) + 6x (DA only) + 12y (PM only) = 7x + 12y = 132

flippedeclipse

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22 Dec 2025, 11:49

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Bunuel

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Image attached/at this link for reference (image uploader is not cooperating so apologies if it's double-posted).https://ibb.co/Kp4wCQLm

x, y and z represent those who only attended that session.

From the lengthy stem we know that:
$x=\frac{1}{4}y=3a$
$z=6x=18a$
$y=12a$

We're looking for $x+y+z$. From the above equations, it becomes clear that unlocking a value for A will help unlock this sum.

If 220 people attended the project management session, it follows that $308-220=88$ did not attend the project management session.

Using the diagram, that means $88=x+a+z=3a+a+18a$, thus $a=4$.

$x+y+z=33a=132$

Thus, D is the answer.

batman10bigman

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22 Dec 2025, 12:14

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Bunuel

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Let the categories be:
D only = d
P only = p
N only = n
D & P only = x
D & N only = y
P & N only = z
All three = t

For exactly six times as many participants attended DA as attended only NT:
d = n

Only PM was twelve times the number who attended both DA and NT but not PM:
p = 12y

Only NT was one quarter of only PM:
n = (1/4)p

from these, we get
n = 3y
d = 6n = 18y
p = 12y

PM total is 220:
p + x + z + t = 220

Total participants is 308:
d + p + n + x + y + z + t = 308

substitute d, p ,n in terms of y:
18y + 12y + 3y + x + y + z + t = 308
34y + x + z + t = 308

Subtract the P equation:
(34y + x + z + t) - (12y + x + z + t) = 308 - 220
22y = 88 => y = 4

n = 3y = 12
d = 18y = 72
p = 12y = 48

Participants who attended exactly one workshop:
d + p + n = 72 + 48 + 12 = 132

Option D

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22 Dec 2025, 12:28

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Let

numbers of participants in only NT be x
then number of participants in only DA = 6x

number of participants in both NT and DA and not PM be y
then number of participants for only PM = 12y
total PM = 220

x = 1/4 of 12y = 3y (eq 2)

Equation:

x+6x+y+220 = 308
Solving using eq 2, y =4

We have to find, x+6x+12y = 33y = 132

Hence (D) is the answer

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22 Dec 2025, 12:47

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since given n(none)=0
so E1+E2+E3=308 (union of the set)-----i
let P, Po, Do, No, (D&N)o represents n(project), n(project only), n(data only), n(negotiation only), n(data & negotiation only) respectively.
from conditions given P=220
Do=6*No=6x let-------ii
Po=12*(D&N)o=12y let-------iii
and No=1/4*12y=3y--------iv
from ii & iv , x=3y
so Do=18y, Po=12y & No=3y
i.e. exactly one E1 =18y+12y+3y=33y----v
and at least two=E2+E3=P - Po + (D&N)o=220-12y+y=220-11y----------vi
from i , v and vi
308 - 33y= 220-11y
22y=88 or y=4
so E1=33*4=136
so E (can be solved fast when represented on a venn diagram)

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I solved it using venn diagram, attaching it's picture for reference (why is it so difficult to attach a picture). We have total = 308, Neither = 0

a+b+c+f = 220
g = 6e
f = 12b
e = 1/4 f

we need g+e+f?

a+b+c+d+e+f+g = 308
220 +b+e+g = 308

putting in b,e & g in terms of f, using above

f/12 + f/4 + 6/4 f = 88
22f = 1056
f = 48

Hence g + e + f = 48 +12 + 72 = 132

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22 Dec 2025, 13:50

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Bunuel

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We can draw an overalpping set

Only DA = a
Only PM = b
Only NT = c

Both DA and PM = d + g

Both DA and NT = d + g + e

Both NT and PM = e + g + f

All three = g

Given a = 6c
b = 12e
c = 1/4 b

a + b + c + d + e + f + g = 308

b + d + g + f = 220

a + c + e = 88

6c + c + c/3 = 88

c = 12

a = 72

b = 48

c + b + a= 132

Option D

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I filled out my Venn diagram with the following:

Data Analysis only = 6X
Project Management Only = 12Y
Negotiation Training = X = 1/4(12Y)

Data Analysis and Negotiation Training = Y

Now I just solved to make every space in the Venn diagram the same variable.

Only Data Analysis is 18Y
Only Project Management is 12Y
Only Negotiation Training is 3Y.

Together it sums up to 33Y. I looked at the answer choices to see which is a multiple of that. Answer 132.

jefferyillman

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22 Dec 2025, 16:43

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Answer D

data analysis only = D
project management only = P
Negotiation Training only = N
data and negotiation = x

D = 6N
P = 12x
4N = P

Project management total = P only + (P+D) + (P+N) = 220

308 (attend at least one) = D + P+ N + DN(x) + PN + PD - PDN
308 = 220 + D + N + DN - PDN
308 = 220 + 6N + N + x - PDN

Only D = 72
Only P = 48
only N =12

total 132

Bunuel

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GmatTest2026

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22 Dec 2025, 19:19

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Solution:

Signed up at least one of the three workshop = 308
Project Managemnet = 12(Data Analysis+Negotiation Training)
Negotiation Training = 1/4PM
Total PM = 220

NT = 1/4*220=55
We have now PM as 220 and NT as 55. So, if we subtract 275 from 308 we get 88 as our answer.

Bunuel

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aka007

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22 Dec 2025, 20:29

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Will need to make a pie diagram with all three intersecting each other.
Let x,y,z be only DA, only PM and only NT respectively.

The intersection between
both DA and PM: c
both DA and NT: a
both NT and PM: b

The intersection between all three is d.

Given.

x=6z
y=4z
z=z
y=12a=> 4z=12a=> a=z/3
c+b+d+y( which is 4z)=220=> c+b+d+4z=220 => c+b+d=220-4z

all are having some relation with z

also given
x+y+z+a+b+c+d=308
which can be substituted and written as
11z+z/3+220-4z=308

hence z=12

question is asking exactly one i.e. x+y+z= 11z= 11x12=132

Ans is D

Bunuel

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Aarushi100

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Bunuel

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Let the number of attendees:
Only Data Analytics (DA): a
Only Project Management(PM): b
Only Negotiation Training (NT): c
DA and NT but not PM: x

According to the ratios given:

1. a=6c
2. b=12x
3. b=4c

We have to find a+b+c=?
Total of PM = 220.
Total = 308
Now, if we remove the total of PM from the total we will get the value of a+c+x:
=> a+c+x=308-220
a+c+x=88
6c+c+c/3=88
22c/3=88

c=12
=> b=48
=> a= 72
a+b+c= 72+48+12= 132

Answer D. 132

jai169

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total participant=T.p=308
ONLY negotiation training=ONA= x
only data anaylsis=O.DA= 6x
number of participants who only attended both data analysis and negotiation training but not project managment=y
Number of parti who attended only project managment=OPA=12 y
the number who attended only negotiation training= 1/4(12y)=3y
total no of people attending the project managment workshop is =220
remaining people= who attended only DA(6x)+who attended only negotiation training(ONA(x)+who attended both negotition and DA but not project managment(Y)=308-220=88
6x+x+y=88
7x+y=88
21y+y=22y=88
y=4
The number who attended only negotiation training was one quarter of the number who attended only project management =x=3y
so question asks how many attended exactly one of three workshop which is= ONA+ODA+OPA=12y+7x=12y+7(3y)=12y+21y=33y=33(4)=132
ans is option D 132

Bunuel

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truedelulu

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22 Dec 2025, 22:47

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Let 6x, x be the number of participants in only Data Analysis (D) and only Negotiation (N) respectively.
Let 12a, x be the number of participants in only Project Management (P) and only D+N respectively.

From the given information:
x = 1/4 * 12a = 3a => 6x = 18a

It's better to draw a Venn chart to visualize so i attach the image.

Total participants = 308 = 18a + 3a + a + 220 => a = 4

Exactly one = 18a + 3a + 12a = 33a = 33*4 = 132

Answer: D

Bunuel

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22 Dec 2025, 23:27

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Data analysis only = d
Negotiation Training only = n
Project management only = p

Given, d = 6n
Then, if people who attended both data analysis and negotiation training but not project management = a, then given p = 12a.
Also given n = 12a / 4, so n = 3a, then d = 18a.

For the number of participants who attended exactly one = d + n + p = 18a + 3a + 12a = 33a. Note here that a is an integer, so the needed answer is a multiple of 33. Only one option satisfies this: 132.

Option D.

Bunuel

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