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605-655 (Medium)|   Overlapping Sets|         
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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:15
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First we need to set up the initial equation. We know that the total participants a should be equal to b + c + d while subtracting the number of people who attended both. We can call this variable X

Therefore, a = b + c + d - x and the fraction of the participants who attended both sessions would be x / a

In this case we solve for x in the first equation and divide it by a, x then equals b + c + d - a ... we then divide that by a

The answer is C
Bunuel
At a tech seminar with a total participants, b attended a session on cybersecurity, and c attended a session on cloud computing. If exactly d participants attended neither session, then in terms of a, b, c, and d, what fraction of the participants attended both sessions?

A. \(\frac{a - b - c + d}{a}\)

B. \(\frac{a - b - c - d}{a}\)

C. \(\frac{b + c + d - a}{a}\)

D. \(\frac{a - b - c + 2d}{a}\)

E. \(\frac{b + c - a + d}{a-d}\)


 


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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:15
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cyber no cyber total
cloud b+c+d-aa-b-dc
no cloud a-c-dda-c
totalba-ba

Fraction of participants who attended both = b+c+d-a/a (C)

Bunuel
At a tech seminar with a total participants, b attended a session on cybersecurity, and c attended a session on cloud computing. If exactly d participants attended neither session, then in terms of a, b, c, and d, what fraction of the participants attended both sessions?

A. \(\frac{a - b - c + d}{a}\)

B. \(\frac{a - b - c - d}{a}\)

C. \(\frac{b + c + d - a}{a}\)

D. \(\frac{a - b - c + 2d}{a}\)

E. \(\frac{b + c - a + d}{a-d}\)


 


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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:15
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Simple question.

We know the formula AunionB=A+B-(AinteresctionB)
Since we have participants not attending the sessions,that should also be added(let that be d)
So
Universal set=A+B-(A intersection B)+d
Substitute the values from the question

Universal set=total participants=a
A=b (people who attended session on cybersecurity)
B= c (people who attended session on cloud computing)

a=b+c-(b interesction c)+d
(b intersection c)=b+c+d-a[Note that b intersection c means participants attending both the sessions]

The fraction then becomes=(b intersection c)/total participants=b+c+d-a/a [Answer is C]

Bunuel
At a tech seminar with a total participants, b attended a session on cybersecurity, and c attended a session on cloud computing. If exactly d participants attended neither session, then in terms of a, b, c, and d, what fraction of the participants attended both sessions?

A. \(\frac{a - b - c + d}{a}\)

B. \(\frac{a - b - c - d}{a}\)

C. \(\frac{b + c + d - a}{a}\)

D. \(\frac{a - b - c + 2d}{a}\)

E. \(\frac{b + c - a + d}{a-d}\)


 


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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:17
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C. (b+c+d−a)/a

We have to find out b ∩ c

We know b ∪ c = b + c - (b ∩ c) -- 1
and we know reading the question i.e. b ∪ c = a - d

Substituting in 1 we get
a - d = b + c - (b ∩ c)
(b ∩ c) = b + c + d - a

Final answer = participants who attended both sessions/total participants = (b + c + d - a)/a
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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:18
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Information given:
- Total participants: a
- Cybersecurity session: b
- Cloud computing session: c
- Neither session: d

Question:
- What fraction of participants attended both sessions?

Solution:
- Using the formula for overlapping sets, we find: Total = Cybersecurity + Cloud - Both + Neither
- Or: a = b + c - Both + d
- So: both = b + c + d - a
- Since we want the fraction of participants that attended both sessions, we want to find both / a
- Both / a = (b + c + d - a) / a

Answer: C, (b + c + d - a) / a
Bunuel
At a tech seminar with a total participants, b attended a session on cybersecurity, and c attended a session on cloud computing. If exactly d participants attended neither session, then in terms of a, b, c, and d, what fraction of the participants attended both sessions?

A. \(\frac{a - b - c + d}{a}\)

B. \(\frac{a - b - c - d}{a}\)

C. \(\frac{b + c + d - a}{a}\)

D. \(\frac{a - b - c + 2d}{a}\)

E. \(\frac{b + c - a + d}{a-d}\)


 


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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:19
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Given 'a' total participants, b attended cybersecurity, c attended cloud computing, and d neither

Total 'a' = b + c -both + d

hence both = b + c + d - a

Hence fraction of both = (b + c + d - a)/a Option C
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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:19
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Bunuel
At a tech seminar with a total participants, b attended a session on cybersecurity, and c attended a session on cloud computing. If exactly d participants attended neither session, then in terms of a, b, c, and d, what fraction of the participants attended both sessions?

A. \(\frac{a - b - c + d}{a}\)

B. \(\frac{a - b - c - d}{a}\)

C. \(\frac{b + c + d - a}{a}\)

D. \(\frac{a - b - c + 2d}{a}\)

E. \(\frac{b + c - a + d}{a-d}\)


 


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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:20
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Total no. of participants = a
Total no. of participants who did not attend any session = d
Total no. of participants who attended session on cybersecurity = b
Total no. of participants who attended session on cloud computing = c
Let total no. of participants who attended both session be x
From above,
total no. of participants who attended either of the sessions = a-d or b+c-x
equating both a-d = b+c-x
or, x = a-d-b-c
Fraction of participants attending both sessions = \(\frac{a - b - c - d}{a}\)


Bunuel
At a tech seminar with a total participants, b attended a session on cybersecurity, and c attended a session on cloud computing. If exactly d participants attended neither session, then in terms of a, b, c, and d, what fraction of the participants attended both sessions?

A. \(\frac{a - b - c + d}{a}\)

B. \(\frac{a - b - c - d}{a}\)

C. \(\frac{b + c + d - a}{a}\)

D. \(\frac{a - b - c + 2d}{a}\)

E. \(\frac{b + c - a + d}{a-d}\)


 


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Let a = 10; b= 3; c=2; d=3;

And participants that attended both sessions = 10-3-2-3 = 2
Fraction = 2/10= 1/5

Plug in the answer choices

A) (10-3-2+3)/10 = 8/10 = 4/5 incorrect
B) (10-3-2-3)/10 = 2/10 = 1/5 correct

You can skip checking the rest since there is just 1 correct answer
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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:21
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let a be 10
b=6 , c=5 , d= 3 and both will be 4
-----CS---NCS----total
CC-----4------1------5
NCC--2------3-------5
total--6---4-------10
4/10

\(\frac{b + c + d - a}{a}\)
OPTION C
Bunuel
At a tech seminar with a total participants, b attended a session on cybersecurity, and c attended a session on cloud computing. If exactly d participants attended neither session, then in terms of a, b, c, and d, what fraction of the participants attended both sessions?

A. \(\frac{a - b - c + d}{a}\)

B. \(\frac{a - b - c - d}{a}\)

C. \(\frac{b + c + d - a}{a}\)

D. \(\frac{a - b - c + 2d}{a}\)

E. \(\frac{b + c - a + d}{a-d}\)


 


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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:22
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  • Total people = a
  • Cybersecurity = b
  • Cloud computing = c
  • Neither = d
We want to find how many went to both sessions.

- People who went to at least one session = a − d


- Add both groups: b + c


- Since b + c counts everyone who went to at least one session, but counts the "both" people twice:
b + c = (people who went to at least one) + (people who went to both)
b + c = (a - d) + (people who went to both)
b+c−(a−d)=people who went to both


    So, number who went to both = b + c + d − a
    Fraction who went to both = b + c + d − a/ a


Bunuel
At a tech seminar with a total participants, b attended a session on cybersecurity, and c attended a session on cloud computing. If exactly d participants attended neither session, then in terms of a, b, c, and d, what fraction of the participants attended both sessions?

A. \(\frac{a - b - c + d}{a}\)

B. \(\frac{a - b - c - d}{a}\)

C. \(\frac{b + c + d - a}{a}\)

D. \(\frac{a - b - c + 2d}{a}\)

E. \(\frac{b + c - a + d}{a-d}\)


 


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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:23
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n(total participants) = a
n(cybersecurity) = b
n(cloud computing) = c
n(neither) = d

Let n(cybersecurity U cloud computing) = x

So by formula,
n(total participants) = n(cybersecurity) + n(cloud computing) - n(cybersecurity U cloud computing) - n(neither)
=> a = b + c - x + d
=> x = b + c + d - a

Therefore, fraction of participants who attended both,
\([x][/a] = [b + c + d - a][/a]\)
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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:25
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On paper solution
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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:25
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Lets consider given conditions as

a = Total number of participants
b = no of participants who attended the cybersecurity session (C)
c = no of participants who attended the cloud computing session (L)
d = no of participants who attended neither session (Neither)

We have to find the fraction of participants who attended both sessions (Both) = ?

The total number of participants can be expressed as:
Total = (Only C) + (Only L) + (Both) + (Neither)

formula : union of two sets - C U L=C+L−C ∩ L
C U L=a−d
after substitute we can see : a−d=b+c−x
x=b+c−(a−d)
x=b+c−a+d

Therefore, the fraction of participants who attended both sessions is (b+c−a+d)/a
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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:26
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Bunuel
At a tech seminar with a total participants, b attended a session on cybersecurity, and c attended a session on cloud computing. If exactly d participants attended neither session, then in terms of a, b, c, and d, what fraction of the participants attended both sessions?

A. \(\frac{a - b - c + d}{a}\)

B. \(\frac{a - b - c - d}{a}\)

C. \(\frac{b + c + d - a}{a}\)

D. \(\frac{a - b - c + 2d}{a}\)

E. \(\frac{b + c - a + d}{a-d}\)


 


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Fraction of participants who attended session is in cybesecurity is a-b, cloud computing is a-c, fraction who didn't attend session is -a+d therefore the fraction of those who didn't attend session is a-b+a-c-a+d ie 2a-a-b-c+d=
(a-b-c+d)/a
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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:26
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Plug-in the values for a, b, c, and d.
Say a = 100;
b= 60;
c=50;
d=10;
The formula is a = b + c - Both + d;
By applying the values, we get Both = 20;
Therefore the fraction is 20/100;

Option A = 0; Incorrect
Option B = -(20/100); Incorrect
Option C = 20/100; Correct
Option D = 10/100; Incorrect
Option E = 20/90; Incorrect
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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:27
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CYBERSECURITYNOT CYBERSECURITY
CLOUD COMPUTINGc
NOT CLOUD COMPUTINGHd
TOTALba

Can be solved by 2x2 table as given.......Total = a
Cyber Security Attendees total = b
Cloud Computing total = c
and Neither = d ( Not Cloud Computing and Not Cyber Security)

Filling the Table....Not Cyber Security is a-b
not Cloud Computing is a-c
so Not Cloud computing and Cybersecurity is a-c-d
And then
Both Cloud Computing and Cybersecurity is b-(a-c-d) ...Be careful of the Signs..if we do b-a-c-d then its wrong.....
so b-(a-c-d) = (b - a + c + d)... divide by total participants...a ...so matches with C ...(b+c+d-a)/a


CYBERSECURITYNOT CYBERSECURITY
CLOUD COMPUTINGb-(a-c-d)c
NOT CLOUD COMPUTINGa-c-dda-c
TOTALba-ba





Bunuel
At a tech seminar with a total participants, b attended a session on cybersecurity, and c attended a session on cloud computing. If exactly d participants attended neither session, then in terms of a, b, c, and d, what fraction of the participants attended both sessions?

A. \(\frac{a - b - c + d}{a}\)

B. \(\frac{a - b - c - d}{a}\)

C. \(\frac{b + c + d - a}{a}\)

D. \(\frac{a - b - c + 2d}{a}\)

E. \(\frac{b + c - a + d}{a-d}\)


 


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Bunuel
At a tech seminar with a total participants, b attended a session on cybersecurity, and c attended a session on cloud computing. If exactly d participants attended neither session, then in terms of a, b, c, and d, what fraction of the participants attended both sessions?

A. \(\frac{a - b - c + d}{a}\)

B. \(\frac{a - b - c - d}{a}\)

C. \(\frac{b + c + d - a}{a}\)

D. \(\frac{a - b - c + 2d}{a}\)

E. \(\frac{b + c - a + d}{a-d}\)


 


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Total number of participants = a

number of participants in cybersecurity = b
number of participants in cloud computing = c
And neither = d

Hence using the concept, total = A + B - both + neither


Both = b+c+d-a

Then the fraction of Bothe sessions attended = Both/ a = (b+c+d-a)/a
Hence Option C
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Consider x as the number who attended both the sessions.
We know that, Total = People attending session on Cybersecurity + People attending session on Cloud Computing - People attending both + People attending neither
This means, a = b + c - x + d => x = b + c + d - a
Since the question asks for the fraction, we want x/a => (b + c + d - a)/a
Bunuel
At a tech seminar with a total participants, b attended a session on cybersecurity, and c attended a session on cloud computing. If exactly d participants attended neither session, then in terms of a, b, c, and d, what fraction of the participants attended both sessions?

A. \(\frac{a - b - c + d}{a}\)

B. \(\frac{a - b - c - d}{a}\)

C. \(\frac{b + c + d - a}{a}\)

D. \(\frac{a - b - c + 2d}{a}\)

E. \(\frac{b + c - a + d}{a-d}\)


 


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GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total 01 Jul 2025, 08:27
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We know that a= b + c – X, where X is the number of people who attended both sessions.

Also, we know that:
Number of people who attended neither session = d
So number of people who attended **at least one** session = a – d

Equating both expressions for those who attended at least one session:
b + c – x = a – d

Now solve for x:
x = b + c – (a – d) = b + c + d – a

So the number of people who attended both sessions is:
**x = b + c + d – a**


Final Answer:
(C) (b + c + d – a) / a
Bunuel
At a tech seminar with a total participants, b attended a session on cybersecurity, and c attended a session on cloud computing. If exactly d participants attended neither session, then in terms of a, b, c, and d, what fraction of the participants attended both sessions?

A. \(\frac{a - b - c + d}{a}\)

B. \(\frac{a - b - c - d}{a}\)

C. \(\frac{b + c + d - a}{a}\)

D. \(\frac{a - b - c + 2d}{a}\)

E. \(\frac{b + c - a + d}{a-d}\)


 


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Bunuel
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