JEE Main 2023 (11 Apr Shift 1)

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Animated Solution for Mathematics - Sets and Relations: An organization awarded 48 medals in event 'A', 25 in event 'B' and 18 in event 'C'. If these medals went to total 60 men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?

Select Answer:

Progressive Hints

Hint 1

Visualize the Venn diagram. The region representing 'exactly two events' is distinct from the region representing 'all three events'. The total sum of pairwise intersections,

\sum n (A \cap B)

, actually hides the central intersection region three times. Can you see why?

Hint 2

You have the total union,

n (A \cup B \cup C) = 60

, and the individual counts. The Principle of Inclusion-Exclusion is your best friend here:

n (A \cup B \cup C) = n (A) + n (B) + n (C) - [n (A \cap B) + n (B \cap C) + n (C \cap A)] + n (A \cap B \cap C)

. Use this to isolate the sum of pairwise intersections.

Hint 3

Once you have the sum of pairwise intersections, remember that this sum includes the central region (where all three events overlap) three times. To isolate only the 'exactly two' regions, use the relation:

Exactly 2 = [n (A \cap B) + n (B \cap C) + n (C \cap A)] - 3 \times n (A \cap B \cap C)

. Now, plug in your values.

Visualized Solution

Visualizing the Three Events

Let events $A$ , $B$ , and $C$ represent the sets of men who received medals.
The total number of unique men who received at least one medal is the union of these sets.
$n (A \cup B \cup C) = 60$

Identifying Given Data

$n (A) = 48$ (Medals in event A)
$n (B) = 25$ (Medals in event B)
$n (C) = 18$ (Medals in event C)

The 'All Three' Intersection

Only $5$ men got medals in all three events.
This is the intersection of all three sets.
$n (A \cap B \cap C) = 5$

The Principle of Inclusion-Exclusion

To relate the union to individual sets and intersections, we use the Inclusion-Exclusion Principle.
$n (A \cup B \cup C) = n (A) + n (B) + n (C) - [n (A \cap B) + n (B \cap C) + n (C \cap A)] + n (A \cap B \cap C)$

Substituting the Values

Let's substitute the known values into our formula.
$60 = 48 + 25 + 18 - [n (A \cap B) + n (B \cap C) + n (C \cap A)] + 5$

Simplifying the Constants

Add the individual set totals and the all-three intersection.
$48 + 25 + 18 + 5 = 96$
$60 = 96 - [n (A \cap B) + n (B \cap C) + n (C \cap A)]$

Solving for Pairwise Intersections

Rearrange the equation to solve for the sum of pairwise intersections.
$[n (A \cap B) + n (B \cap C) + n (C \cap A)] = 96 - 60$
$\sum n (A \cap B) = 36$

Defining 'Exactly Two Events'

We need the number of men in exactly two events.
Visually, these are the regions where exactly two circles overlap, excluding the center.
Formula: $Exactly 2 = \sum n (A \cap B) - 3 n (A \cap B \cap C)$

Why Subtract Three Times?

The sum of pairwise intersections $\sum n (A \cap B)$ counts the center region $n (A \cap B \cap C)$ three times (once for each pair).
Since we want exactly two, we must remove the center region completely from this sum.
Therefore, we subtract $3 \times n (A \cap B \cap C)$ .

Final Substitution

Substitute the values we found into the 'Exactly 2' formula.
$Exactly 2 = 36 - 3 (5)$

Final Calculation

$Exactly 2 = 36 - 15$
$Exactly 2 = 21$
Final Answer: $21$ men received medals in exactly two events.

00:00 / 00:00

The Sigma Insight: Types of Sets and Set Operations

Solution Diagram

Analyzing the Setup

Imagine a grand hall hosting three events:

A

B

, and

C

. We are given that

60

men have won at least one medal, which in set notation is the union of the three sets:

n (A \cup B \cup C) = 60

The individual counts for each event are

n (A) = 48

n (B) = 25

, and

n (C) = 18

. We are also given the central anchor, representing the men who won medals in all three events:

n (A \cap B \cap C) = 5

The Power of Inclusion-Exclusion

When we sum

n (A) + n (B) + n (C)

, we inadvertently over-count individuals who won multiple medals. To correct this, we apply the Principle of Inclusion-Exclusion:

n (A \cup B \cup C) = n (A) + n (B) + n (C) - [n (A \cap B) + n (B \cap C) + n (C \cap A)] + n (A \cap B \cap C)

This formula accounts for the overlaps by subtracting the pairwise intersections and adding back the triple intersection that was removed too many times. Substituting our known values into the equation:

60 = 48 + 25 + 18 - [n (A \cap B) + n (B \cap C) + n (C \cap A)] + 5

Simplifying the expression, we find:

60 = 96 - [n (A \cap B) + n (B \cap C) + n (C \cap A)]

Rearranging the terms, we determine the sum of the pairwise intersections:

\sum n (A \cap B) = 36

The 'Exactly Two' Trap

The sum

\sum n (A \cap B)

represents the total of the pairwise intersections. However, in a Venn diagram, each pairwise intersection region inherently includes the central region where all three events overlap.

Because the central region

n (A \cap B \cap C)

is contained within each of the three pairwise intersections, it has been counted three times in our sum of

36

. To find the number of men who received medals in exactly two events, we must remove this central region three times.

The logic is expressed as follows:

Exactly 2 = \sum n (A \cap B) - 3 \times n (A \cap B \cap C)

Substituting our calculated values:

Exactly 2 = 36 - 3 (5)

Exactly 2 = 36 - 15 = 21

Final Result

By carefully stripping away the triple-counted overlap, we conclude that the number of men who received medals in exactly two events is 21.

Common Traps & FAQs

Wait, why do we subtract 3 times the intersection for 'exactly two'? Shouldn't it be just once?

That is a brilliant question! Think of the Venn diagram. The pairwise intersection

n (A \cap B)

includes the central region

n (A \cap B \cap C)

. When you sum

n (A \cap B) + n (B \cap C) + n (C \cap A)

, you are counting the central region three times—once for each pair. Since we want *exactly* two, we must remove that central region entirely. Subtracting it three times clears it out completely, leaving you with only the regions where exactly two events overlap.

Can I just draw a Venn diagram and fill in the regions with variables

x, y, z

instead of using the formula?

Absolutely! In fact, that is a fantastic way to verify your answer. Using variables for each disjoint region is a powerful technique. However, the formulaic approach is faster and less prone to arithmetic errors in complex problems. Mastering both is the hallmark of a JEE topper.

What if the question asked for 'exactly one' event? How would the formula change?

That would be a wonderful challenge! For 'exactly one', you would take the total union and subtract the 'exactly two' and 'exactly three' regions. Or, more elegantly:

n (A) + n (B) + n (C) - 2 [n (A \cap B) + n (B \cap C) + n (C \cap A)] + 3 n (A \cap B \cap C)

. It is all about keeping track of how many times each region is counted!

Animated Solution for Mathematics - Sets and Relations: An organization awarded 48 medals in event 'A', 25 in event 'B' and 18 in event 'C'. If these medals went to total 60 men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?

Select Answer:

Visualized Solution

Visualizing the Three Events

Identifying Given Data

The 'All Three' Intersection

The Principle of Inclusion-Exclusion

Substituting the Values

Simplifying the Constants

Solving for Pairwise Intersections

Defining 'Exactly Two Events'

Why Subtract Three Times?

Final Substitution

Final Calculation

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

The Power of Inclusion-Exclusion

The 'Exactly Two' Trap

Final Result

Similar Questions

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Animated Solution for Mathematics - Sets and Relations: An organization awarded 48 medals in event 'A', 25 in event 'B' and 18 in event 'C'. If these medals went to total 60 men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?

Select Answer:

Visualized Solution

Visualizing the Three Events

Identifying Given Data

The 'All Three' Intersection

The Principle of Inclusion-Exclusion

Substituting the Values

Simplifying the Constants

Solving for Pairwise Intersections

Defining 'Exactly Two Events'

Why Subtract Three Times?

Final Substitution

Final Calculation

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

The Power of Inclusion-Exclusion

The 'Exactly Two' Trap

Final Result

Similar Questions

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In a school, there are three types of games to be played. Some of the students play two types of games, but none play all the three games. Which Venn diagrams can justify the above statement?

In a college of 300 students every student reads 5 newspapers and every newspaper is read by 60 students. The number of newspapers is

The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48, is

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