GMAT Club Olympics 2025 (Day 2): At a tech seminar with a total : Problem Solving (PS)

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01 Jul 2025, 10:30

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	Cyber security	No cyber security	Total
Cloud Computing	c - (a - b - d) (Step 3)	a - b - d (Step 2)	c (given)
No cloud computing		d(given)
Total	b (given)	a - b (Step 1)	a (given)

c - (a - b - d) = c - a + b + d

Therefore, fraction of the participants who attended both sessions = $\frac{ c - a + b + d}{a }$

Or, (C) $\frac{b + c + d - a}{a}$

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01 Jul 2025, 10:30

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Cybersecurity = X, Cloud Computing - Y

X. X Not
Y (b-a+c+d) C
Y Not. (a-c-d) d. a-c
b b-a a

According to this table, we can find out where both = (b-a+c+d)/a

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01 Jul 2025, 10:31

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Correct answer is option C

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Screenshot 2025-07-01 at 10.59.00 PM.png [ 2 MiB | Viewed 368 times ]

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01 Jul 2025, 10:35

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we can solve using simple formula: AUB=A+B-AB

putting values: a-d=b+c- AB
thus AB= b+c-a+d

answer asked in fraction so divide by total=
b+c-a+d/a- hence option E

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01 Jul 2025, 10:37

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Correct Answer: Option C
Use option checking approach using Venn Diagram example,
Check option 1, this we can easily eliminate as a-b-c+d this case might be in negative because total number of participants might be less than Cyber & cloud participant.
Check option 2, Similarly this option can be eliminate too.
Check option 3, This is correct option, check with taking examples.
Check option 4, This option is incorrect too.
Check option 5: this will be wrong if you take example by minimising cyber & cloud participants and maximising no participants.

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01 Jul 2025, 10:39

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Its a set problem so i use Indian way of approaching this problem

Total Participants (a) = Participated (CC \cup CS) + Not Participated (d)
(CC \cup CS) = a - d ........ eq(1)
(CC U CS) indicates participants that attended at least one program between CC (cloud computing) and CS (Cyber Security)

Set of two
\[
(CC \cup CS) = CC + CS - (CC \cap CS)
\]
a-d = b+c - (CC \cap CS)

(CC \cap CS) = b+c+d-a

In terms of fraction = (CC \cap CS)/total participants = (b+c+d-a)/a

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01 Jul 2025, 10:40

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We know that for venn diagram questions, Total-None=X+Y-Both
=> Both = X+Y-Total+None
Given:
total = a
cybersecurity (X) = b
cloud computing (Y) = c
none = d
=> both = b+c-a+d

Total participants = a
Fraction of the participants attended both sessions = (b+c-a+d)/a

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01 Jul 2025, 10:46

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Let's say people who participated in both be X
From the basics of set theory:

a = b + c - X + d

We have subtracted X because it is double counted in b and c

Let's say the fraction is f

f*a = a - b - c - d

f = (a - b - c - d)/a
Option B

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01 Jul 2025, 10:54

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If x people attended both sessions, b-x went to only cybersecurity and c-x went to cloud computing only. (b-x) +(c-x) + x + d = a. Solve for x then divide by A gives option C as the answer.

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01 Jul 2025, 10:57

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let participants attended both be x
equation -> a= b+c -x + d
x = b + c + d - a
fraction -> x/a -> $\frac{b + c + d - a}{a}$

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01 Jul 2025, 11:00

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Answer: C.

Concept
Let's think of a Venn Diagram. And we want to count the total number of unique terms in 2 sets that have some overlap.
We can't simply add their counts, because we would be counting them again.

For the case of 2 sets, this overlap is counted twice, so we have to substract it once to count each unique terms only once.
That's why the formula is T = A + B - (A&B).

But for the cases that there are also terms outside the 2 sets, let's call them N, the formula becomes: T = A + B - (A&B) + N.
Or also: (A&B) = A + B - T + N

Solution
Now, let's address the case where both participants attended the cyber and cloud sessions, by inputting the terms given:
x = b + c - a + d (I)

However, notice that's not what the question asks. They specifically ask for what fraction of the participants attended both sessions.
In other words, what fraction of a, is x?
The answer is: x / a (II)

Okay, let's combine (I) and (II) to arrive at our final destination:

$\frac{x }{ a} = \frac{(b + c - a + d ) }{ a}$
Option C.

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01 Jul 2025, 11:09

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Given,
A-total number of participants
B-participants attended cyber security session
C-participants attended cloud computing
D-participants attended neither of two sessions

Formula: A = B + C + D - Both
From the above, participants attended both sessions will be equal to total - neither

Both = A-D
IMO : option E

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01 Jul 2025, 11:09

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Bunuel

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Consider the set diagram for the question where we have to calculate B intersection C - Let us consider required intersection as X.

Total set (Inclusive of intersection ) = C+B +D

Now To find of X = B+C+D-A (Since A cancels D and cancels Only B and Only D and intersection remains)

Hence now we require fraction of participants who attended both session i.e X/A = (B+C+D-A)/A

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01 Jul 2025, 11:15

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Bunuel

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Let 'x' be the no. Of participants in both sessions,
Then total participants a= b+c - x +d → x=b+c+d-a (rearranging the prev equation)
Finally, fraction of participants who attended both sessions is (b+c+d-a)/a

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01 Jul 2025, 11:21

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We are given:
• a: total participants
• b: attended cybersecurity
• c: attended cloud computing
• d: attended neither

We are asked: what fraction of the participants attended both sessions?

Step 1: Use the principle of inclusion and exclusion.

Let:
• Let x be the number of people who attended both sessions.

From set theory:
Number who attended at least one session = b + c - x

We are also told:
*Number who attended neither session = d
*Number who attended at least one} = a - d

So:
b + c - x = a - d
=> x = b + c - (a - d) = b + c + d - a

⸻

Step 2: Find the fraction who attended both:

Fraction =x/a = (b + c + d - a)/a

✅ Final Answer:

Option C: \boxed{\frac{b + c + d - a}{a}}

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01 Jul 2025, 11:30

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cyber meeting + cloud meeting - both + none = total
or: b+c+d-n = a; where n equals both
in this question, you want to find n/a
rearranging, you get: n = b+c+d-a and then, dividing n by a, you get n/a = (b+c+d-a)/a

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01 Jul 2025, 11:34

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variables in answer choices and problem --> PLUG IN
each variable must be an integer as they represent numbers of people within groups

Let A= 20 and apply the Group formula for 2 Groups: i.e., Total = G1 + G2 - Both + Neither

A= Total; B= G1; C = G2, D= NEITHER, and we are asked to determine the fraction of Both/Total

allowing A= 20, b= 10, c= 5, and = 2,

20= 10+ 5 - Both + 2
20= 17 - Both
B = 3

plugging in each value into the answer choices results in B being the correct answer choice: a - b - c - d = 20-10-5-2=3/20

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01 Jul 2025, 11:40

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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?

we need to find (b intersection c) ;
as: b+ c = ( only b + only c+ 2 times (b intersection c) ... eq-1
and as : a = only b + only c + (b intersection c) + d -----eq-2
so , only b + only c = a-(b intersection c) - d ---eq-3
put eq-3 in eq-1

b+c= a -(b intersection c) - d + 2 times (b intersection c)

so , (b intersection c)= b+c+d-a , Hence the required fraction.

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01 Jul 2025, 11:42

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so there are total a participants. here a is the constitute of people who attended b session, c session and d neither session and also both session

in terms of equation a = b + c + d + both

now both = b + c + d - a /a

Bunuel

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01 Jul 2025, 11:45

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A is total number of participants.
A-D = total number of participants that attended at least on session.
B+C = participants who attend either session (this includes a overlap of people who attended both sessions)

To find the overlap
B+C - (A-D)= B+C+D-A

Divided by A to get the fraction.

Bunuel

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