Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).
Comprehensive Concepts & Principles
Venn Diagrams & Operations on Sets forms an essential building block in CBSE Class 11 Applied Mathematics (Sets). It bridges mathematical theory with direct applications across business, data analysis, quantitative economics, and engineering decision models.
Core Focus Areas:
Union of sets (A ∪ B)
Intersection of sets (A ∩ B) and disjoint sets
Difference of sets (A - B) and symmetric difference (A Δ B)
Complement of a set (A') and properties of complements (De Morgan’s Laws)
Mathematical Framework & Key Formulas
Master the governing relationships and definitions established in this exercise:
$$A \cup B = \{x : x \in A or x \in B\}$$
$$A \cap B = \{x : x \in A and x \in B\}$$
$$A - B = \{x : x \in A and x \notin B\}$$
$$(A \cup B)' = A' \cap B', \quad (A \cap B)' = A' \cup B'$$
Worked Benchmark Examples
Example 1 (Market Research & Two-Set Venn Diagram): In a market survey of $500$ consumers, it was found that $280$ purchase Brand A, $220$ purchase Brand B, and $100$ purchase both brands. Determine:
(i) How many consumers purchase at least one of the two brands?
(ii) How many consumers purchase neither Brand A nor Brand B?
(iii) How many consumers purchase only Brand A?
Show Step-by-Step Solution
Given Data:
Total consumers $n(U) = 500$
Consumers purchasing Brand A: $n(A) = 280$
Consumers purchasing Brand B: $n(B) = 220$
Consumers purchasing both brands: $n(A \cap B) = 100$
Part (i) Consumers purchasing at least one brand $n(A \cup B)$:
Using the addition principle of sets:
$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$
$$n(A \cup B) = 280 + 220 - 100 = 400$$
Hence, $400$ consumers purchase at least one brand.
Part (ii) Consumers purchasing neither brand $n((A \cup B)')$:
$$n((A \cup B)') = n(U) - n(A \cup B) = 500 - 400 = 100$$
Thus, $100$ consumers purchase neither brand.
Part (iii) Consumers purchasing only Brand A $n(A - B)$:
$$n(A - B) = n(A) - n(A \cap B) = 280 - 100 = 180$$
Therefore, $180$ consumers purchase Brand A exclusively.
Example 2 (Three-Set Business Student Survey): In a commerce faculty of $120$ students, $65$ study Accounting ($A$), $50$ study Economics ($E$), and $40$ study Business Law ($L$). Further, $25$ study Accounting and Economics, $20$ study Economics and Law, $18$ study Accounting and Law, and $10$ study all three subjects. Find:
(i) How many students study at least one of the three subjects?
(ii) How many students study none of these three subjects?
(iii) How many students study exactly one subject?
Show Step-by-Step Solution
Part (i) Number of students studying at least one subject $n(A \cup E \cup L)$:
Applying the 3-set inclusion-exclusion formula:
$$n(A \cup E \cup L) = n(A) + n(E) + n(L) - [n(A \cap E) + n(E \cap L) + n(A \cap L)] + n(A \cap E \cap L)$$
$$n(A \cup E \cup L) = 65 + 50 + 40 - (25 + 20 + 18) + 10 = 155 - 63 + 10 = 102$$
So, $102$ students study at least one subject.
Part (ii) Students studying none of these subjects:
$$n((A \cup E \cup L)') = n(U) - n(A \cup E \cup L) = 120 - 102 = 18 \text{ students}$$ Part (iii) Students studying exactly one subject:
• Studying only Accounting $= n(A) - [n(A \cap E) + n(A \cap L) - n(A \cap E \cap L)] = 65 - [25 + 18 - 10] = 65 - 33 = 32$
• Studying only Economics $= n(E) - [n(A \cap E) + n(E \cap L) - n(A \cap E \cap L)] = 50 - [25 + 20 - 10] = 50 - 35 = 15$
• Studying only Business Law $= n(L) - [n(A \cap L) + n(E \cap L) - n(A \cap E \cap L)] = 40 - [18 + 20 - 10] = 40 - 28 = 12$
Total students studying exactly one subject $= 32 + 15 + 12 = 59$.
Example 3 (De Morgan's Laws Verification with Empirical Sets): Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ be the universal set of positive integers $\le 10$. Let $A = \{2, 4, 6, 8, 10\}$ (even numbers) and $B = \{1, 2, 3, 4, 5, 6\}$. Verify De Morgan's First Law: $(A \cup B)' = A' \cap B'$.
Show Step-by-Step Solution
Step 1: Compute LHS $(A \cup B)'$
Find $A \cup B$ by listing all unique elements in $A$ or $B$:
$$A \cup B = \{1, 2, 3, 4, 5, 6, 8, 10\}$$
Take the complement of $(A \cup B)$ relative to $U$:
$$(A \cup B)' = U - (A \cup B) = \{7, 9\}$$ Step 2: Compute RHS $A' \cap B'$
Find individual complements:
$$A' = U - A = \{1, 3, 5, 7, 9\}$$
$$B' = U - B = \{7, 8, 9, 10\}$$
Find the intersection of $A'$ and $B'$ (elements common to both):
$$A' \cap B' = \{7, 9\}$$ Conclusion:
Since $\text{LHS} = \{7, 9\}$ and $\text{RHS} = \{7, 9\}$, we have verified that $(A \cup B)' = A' \cap B'$.
CHECK YOUR PROGRESS 4.4
Graded textbook exercises based directly on the CBSE Applied Mathematics support material for Venn Diagrams & Operations on Sets. Solve each question independently before toggling the step-by-step solutions.
Problem 1Textbook Drill • Set Operations & Identities
Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$ be the universal set. Let $A = \{x : x \text{ is an even natural number } \le 12\}$ and $B = \{x : x \text{ is a multiple of } 3 \le 12\}$.
Determine:
(i) $A \cup B$ and $A \cap B$
(ii) $A - B$ and $B - A$
(iii) Symmetric Difference $A \Delta B$
(iv) $(A \cup B)'$ and verify that $(A \cup B)' = A' \cap B'$.
(i) Union and Intersection:
$$A \cup B = \{2, 3, 4, 6, 8, 9, 10, 12\}$$
$$A \cap B = \{6, 12\}$$ (ii) Differences ($A - B$ and $B - A$):
• $A - B$ consists of elements in $A$ that are not in $B$:
$$A - B = \{2, 4, 8, 10\}$$
• $B - A$ consists of elements in $B$ that are not in $A$:
$$B - A = \{3, 9\}$$ (iii) Symmetric Difference ($A \Delta B$):
$$A \Delta B = (A - B) \cup (B - A) = \{2, 4, 8, 10\} \cup \{3, 9\} = \{2, 3, 4, 8, 9, 10\}$$
Alternative verification: $A \Delta B = (A \cup B) - (A \cap B) = \{2, 3, 4, 6, 8, 9, 10, 12\} - \{6, 12\} = \{2, 3, 4, 8, 9, 10\}$.
(iv) De Morgan's Law Verification:
• LHS: $(A \cup B)' = U - (A \cup B) = \{1, 5, 7, 11\}$
• RHS: $A' = U - A = \{1, 3, 5, 7, 9, 11\}$ and $B' = U - B = \{1, 2, 4, 5, 7, 8, 10, 11\}$
$$A' \cap B' = \{1, 5, 7, 11\}$$
Since $\text{LHS} = \text{RHS} = \{1, 5, 7, 11\}$, De Morgan's First Law is confirmed.
Problem 2Applied Business Survey • Two Sets
A financial wealth management consultancy audited the portfolios of $200$ high-net-worth investors. The audit reported that $120$ clients invest in Equity Mutual Funds ($M$), $90$ invest in Sovereign Gold Bonds ($G$), and $30$ invest in neither instrument.
Calculate:
(i) How many clients invest in both Equity Mutual Funds and Gold Bonds?
(ii) How many clients invest strictly in Sovereign Gold Bonds only?
(iii) How many clients invest in at most one of the two asset classes?
Step 1: Total clients investing in at least one asset class:
$$n(M \cup G) = n(U) - n((M \cup G)') = 200 - 30 = 170$$ Part (i) Clients investing in both instruments $n(M \cap G)$:
Using the principle of inclusion-exclusion for 2 sets:
$$n(M \cup G) = n(M) + n(G) - n(M \cap G)$$
$$170 = 120 + 90 - n(M \cap G) = 210 - n(M \cap G)$$
$$n(M \cap G) = 210 - 170 = 40$$
Thus, $40$ clients invest in both Equity Mutual Funds and Sovereign Gold Bonds.
Part (ii) Clients investing strictly in Sovereign Gold Bonds only $n(G - M)$:
$$n(G - M) = n(G) - n(M \cap G) = 90 - 40 = 50 \text{ clients}$$ Part (iii) Clients investing in at most one asset class:
This includes clients investing in neither, only $M$, or only $G$:
• Clients with only $M$: $n(M - G) = 120 - 40 = 80$
• Clients with only $G$: $50$
• Clients with neither: $30$
$$\text{Total} = 80 + 50 + 30 = 160 \text{ clients}$$
Direct formula: $n(U) - n(M \cap G) = 200 - 40 = 160$ clients.
Problem 3Corporate Market Analysis • Three Sets
A consumer research agency surveyed $500$ smartphone users in a metro city regarding active payment applications: PhonePe ($P$), Google Pay ($G$), and Paytm ($T$).
The findings were:
$n(P) = 260, \quad n(G) = 220, \quad n(T) = 180$
$n(P \cap G) = 110, \quad n(G \cap T) = 80, \quad n(P \cap T) = 70$
Exactly $40$ respondents reported using all three payment apps.
Find:
(i) The number of users who use at least one of the three payment apps.
(ii) The number of users who use none of these three apps.
(iii) The number of users who use exactly two payment apps.
Part (i) Users using at least one app $n(P \cup G \cup T)$:
Using the 3-set inclusion-exclusion formula:
$$n(P \cup G \cup T) = [n(P) + n(G) + n(T)] - [n(P \cap G) + n(G \cap T) + n(P \cap T)] + n(P \cap G \cap T)$$
$$n(P \cup G \cup T) = [260 + 220 + 180] - [110 + 80 + 70] + 40$$
$$n(P \cup G \cup T) = 660 - 260 + 40 = 440 \text{ users}$$ Part (ii) Users using none of these apps:
$$n((P \cup G \cup T)') = n(U) - n(P \cup G \cup T) = 500 - 440 = 60 \text{ users}$$ Part (iii) Users using exactly two apps:
For each pair, subtract those who use all three ($n(P \cap G \cap T) = 40$):
• Only PhonePe and Google Pay: $110 - 40 = 70$
• Only Google Pay and Paytm: $80 - 40 = 40$
• Only PhonePe and Paytm: $70 - 40 = 30$
$$\text{Total using exactly two apps} = 70 + 40 + 30 = 140 \text{ users}$$
Exercise 4.4 Revision Worksheet
Printable classroom and homework worksheet for Venn Diagrams & Operations on Sets. Covers objective drills, intermediate calculations, and high-yield applications.
Section A: Conceptual & Objective Drill3 Questions • 1 Mark Each
Q1. Disjoint Sets Cardinality1 Mark
If $A$ and $B$ are two disjoint sets such that $n(A) = 18$ and $n(B) = 27$, state the values of $n(A \cap B)$ and $n(A \cup B)$.
Show Model Solution
Model Solution:
By definition, disjoint sets have no common elements, meaning $A \cap B = \emptyset \implies n(A \cap B) = 0$.
Therefore, $n(A \cup B) = n(A) + n(B) = 18 + 27 = 45$.
Q2. Partition Property1 Mark
State whether the statement is True or False and justify: "For any two non-empty sets $A$ and $B$, the subsets $(A - B)$, $(A \cap B)$, and $(B - A)$ are mutually pairwise disjoint."
Show Model Solution
Answer: True. Justification: Any element $x \in (A - B)$ satisfies $x \in A$ and $x \notin B$. Any element in $(A \cap B)$ satisfies $x \in A$ and $x \in B$. Any element in $(B - A)$ satisfies $x \in B$ and $x \notin A$. No element can simultaneously satisfy any two conditions, so their mutual intersections are empty (pairwise disjoint), and their union forms $A \cup B$.
Q3. Complement Identity1 Mark
Using De Morgan's Laws and set algebra, simplify the expression: $[(A \cup B)' \cup (A' \cap B)]'$.
Show Model Solution
Model Solution:
Apply De Morgan's Law $(X \cup Y)' = X' \cap Y'$ where $X = (A \cup B)'$ and $Y = (A' \cap B)$:
$$[(A \cup B)']' \cap [A' \cap B]' = (A \cup B) \cap [A'' \cup B'] = (A \cup B) \cap (A \cup B')$$
By distributive law:
$$(A \cup B) \cap (A \cup B') = A \cup (B \cap B') = A \cup \emptyset = A$$
Answer: $A$.
Section B: Short Numerical Problems2 Questions • 2 Marks Each
Q4. Difference Cardinality Drill2 Marks
Let $A$ and $B$ be two finite sets such that $n(A - B) = 34$, $n(B - A) = 22$, and $n(A \cap B) = 16$.
Find: (i) $n(A)$, (ii) $n(B)$, and (iii) $n(A \cup B)$.
Show Model Solution
Step 1: Compute $n(A)$ and $n(B)$:
$$n(A) = n(A - B) + n(A \cap B) = 34 + 16 = 50$$
$$n(B) = n(B - A) + n(A \cap B) = 22 + 16 = 38$$ Step 2: Compute $n(A \cup B)$:
$$n(A \cup B) = n(A - B) + n(B - A) + n(A \cap B) = 34 + 22 + 16 = 72$$
Check: $n(A) + n(B) - n(A \cap B) = 50 + 38 - 16 = 72$.
Q5. School Subject Enrollment2 Marks
In a senior secondary school with $350$ students in the Commerce stream, $210$ study Accountancy ($A$), $160$ study Economics ($E$), and $45$ study neither. Determine how many students study both Accountancy and Economics.
Section C: High-Yield Applied Word Problems1 Question • 4 Marks
Q6. E-Commerce Cart Abandonment Case4 Marks
A leading direct-to-consumer e-commerce portal evaluated cart abandonment reasons among $1,000$ online shoppers:
• $520$ cited High Delivery Fees ($D$)
• $450$ cited Forced Account Creation ($A$)
• $380$ cited Slow Payment Gateway ($S$)
• $210$ cited both $D$ and $A$
• $170$ cited both $A$ and $S$
• $150$ cited both $D$ and $S$
• $80$ cited all three friction factors.
Calculate:
(i) The number of shoppers who abandoned carts due to at least one of these three factors. (1 Mark)
(ii) The number of shoppers who experienced none of these three friction points. (1 Mark)
(iii) The number of shoppers who abandoned carts solely due to High Delivery Fees. (1 Mark)
(iv) The number of shoppers who cited exactly two friction factors. (1 Mark)
Show Model Solution
(i) At least one factor $n(D \cup A \cup S)$:
$$n(D \cup A \cup S) = [520 + 450 + 380] - [210 + 170 + 150] + 80 = 1350 - 530 + 80 = 900$$
So, $900$ shoppers were affected by at least one friction factor. [1 Mark]
(ii) Shoppers experiencing none of the three:
$$n((D \cup A \cup S)') = n(U) - n(D \cup A \cup S) = 1000 - 900 = 100 \text{ shoppers. [1 Mark]}$$ (iii) Shoppers citing only High Delivery Fees:
$$n(\text{Only } D) = n(D) - [n(D \cap A) + n(D \cap S) - n(D \cap A \cap S)]$$
$$n(\text{Only } D) = 520 - [210 + 150 - 80] = 520 - 280 = 240 \text{ shoppers. [1 Mark]}$$ (iv) Shoppers citing exactly two friction factors:
• Only $D$ and $A$: $210 - 80 = 130$
• Only $A$ and $S$: $170 - 80 = 90$
• Only $D$ and $S$: $150 - 80 = 70$
$$\text{Total} = 130 + 90 + 70 = 290 \text{ shoppers. [1 Mark]}$$
Exercise 4.4 Comprehensive Revision Sheet
Printable high-yield summary for last-minute revision and board examination preparation on Venn Diagrams & Operations on Sets.
Concept Architecture & Key Takeaways
Union ($A \cup B$): The set of all elements belonging to $A$, or $B$, or both: $\{x : x \in A \lor x \in B\}$. Represents logical OR.
Intersection ($A \cap B$): The set of all elements belonging to both $A$ and $B$: $\{x : x \in A \land x \in B\}$. If $A \cap B = \emptyset$, the sets are called disjoint.
Difference ($A - B$): The relative complement of $B$ in $A$, containing elements in $A$ but strictly not in $B$: $\{x : x \in A \land x \notin B\} = A \cap B'$.
Symmetric Difference ($A \Delta B$): Elements belonging to either $A$ or $B$ but not both: $(A - B) \cup (B - A) = (A \cup B) - (A \cap B)$.
Complement ($A'$): Elements in the Universal set $U$ that do not belong to $A$: $U - A$.
De Morgan's Laws: $(A \cup B)' = A' \cap B'$ and $(A \cap B)' = A' \cup B'$. The complement of a union is the intersection of complements.
Essential Formulae & Identities
• $$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$
• $$n(A - B) = n(A) - n(A \cap B) = n(A \cup B) - n(B)$$
• $$n(A \Delta B) = n(A - B) + n(B - A) = n(A) + n(B) - 2n(A \cap B)$$
• $$n(A \cup B \cup C) = n(A) + n(B) + n(C) - [n(A \cap B) + n(B \cap C) + n(C \cap A)] + n(A \cap B \cap C)$$
• $$\text{Number of elements in exactly two sets} = [n(A \cap B) + n(B \cap C) + n(C \cap A)] - 3n(A \cap B \cap C)$$
• $$\text{Number of elements in exactly one set} = n(A) + n(B) + n(C) - 2[n(A \cap B) + n(B \cap C) + n(C \cap A)] + 3n(A \cap B \cap C)$$
Common Examination Mistakes & Pro-Tips
Common Pitfall: Double-Counting Intersections
When given "elements common to A and B", students often subtract this entire count when finding "only A and B", forgetting that the 3-way intersection $n(A \cap B \cap C)$ is included inside $n(A \cap B)$. Always isolate mutual pairs: $n(\text{only } A \cap B) = n(A \cap B) - n(A \cap B \cap C)$.
Examiner Pro-Tip: Venn Region Labeling Method
In 3-set word problems, draw the 3 circles and label regions from innermost to outermost: begin by filling the center core $n(A \cap B \cap C)$, then fill the two-set leaves, and finally determine the single-set outer areas. This guarantees step marks and prevents arithmetic errors!
Time: 45 Minutes Max Marks: 20 CBSE Board Blueprint
Exact pattern test covering Objective MCQs, Assertion-Reason, Short Answer Types, Long Answer, and a Case-Based Study.
Section A: Objective & Assertion-Reason (Q1 – Q4)4 Questions • 1 Mark Each
Question 1MCQ • 1 Mark
If $n(A) = 32$, $n(B) = 44$, and $n(A \cup B) = 60$, what is the cardinality of the intersection $n(A \cap B)$?
(A) $12$
(B) $16$
(C) $20$
(D) $24$
Solution (B): By the addition theorem: $n(A \cap B) = n(A) + n(B) - n(A \cup B) = 32 + 44 - 60 = 76 - 60 = 16$.
Question 2MCQ • 1 Mark
Which of the following set expressions is equivalent to the Symmetric Difference $A \Delta B$?
(A) $(A \cap B) - (A \cup B)$
(B) $(A \cup B) - (A \cap B)$
(C) $(A - B) \cap (B - A)$
(D) $A' \cup B'$
Solution (B): Symmetric difference is $(A - B) \cup (B - A)$, which is identically $(A \cup B) - (A \cap B)$ (elements in either set but not both).
Question 3MCQ • 1 Mark
According to De Morgan's Laws, for any two universal subsets $A$ and $B$, the complement of their union $(A \cup B)'$ equals:
(A) $A' \cap B'$
(B) $A' \cup B'$
(C) $A - B'$
(D) $(A \cap B)'$
Solution (A): De Morgan's First Law states $(A \cup B)' = A' \cap B'$. The complement of a union is the intersection of their individual complements.
Question 4Assertion-Reason • 1 Mark
Assertion (A): If two sets $A$ and $B$ are disjoint, then $n(A \cup B) = n(A) + n(B)$. Reason (R): For disjoint sets $A$ and $B$, their intersection $A \cap B = \emptyset$, which implies $n(A \cap B) = 0$.
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is not the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.
Solution (A): Both Assertion and Reason are true. The general formula is $n(A \cup B) = n(A) + n(B) - n(A \cap B)$. When disjoint, $n(A \cap B) = 0$, giving $n(A \cup B) = n(A) + n(B)$. Thus, R correctly explains A.
Section B: Short Answer Type I (Q5 – Q6)2 Questions • 2 Marks Each
Question 52 Marks
Let universal set $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, $A = \{2, 3, 5, 7\}$ (prime numbers), and $B = \{1, 3, 5, 7, 9\}$ (odd numbers).
Find: (i) $A - B$ and (ii) $B - A$.
Show Marking Scheme & Solution (2 Marks)
(i) $A - B$ [1 Mark]:
Elements in $A$ which are not in $B$:
$A = \{2, 3, 5, 7\}$ and $B = \{1, 3, 5, 7, 9\} \implies A - B = \{2\}$.
(ii) $B - A$ [1 Mark]:
Elements in $B$ which are not in $A$:
$B - A = \{1, 9\}$.
Question 62 Marks
Using set algebra identities, prove that for any sets $A$ and $B$:
$$(A \cap B) \cup (A - B) = A$$
Show Marking Scheme & Solution (2 Marks)
Step 1: Express $A - B$ as intersection [1 Mark]:
We know $A - B = A \cap B'$. Substituting into LHS:
$$\text{LHS} = (A \cap B) \cup (A \cap B')$$ Step 2: Apply distributive law in reverse [1 Mark]:
$$(A \cap B) \cup (A \cap B') = A \cap (B \cup B')$$
Since $B \cup B' = U$ (Universal set):
$$A \cap U = A = \text{RHS}$$
Hence proved.
Section C: Short Answer Type II (Q7 – Q8)2 Questions • 3 Marks Each
Question 73 Marks
In a class of $70$ applied mathematics students, $45$ students enrolled in Business Statistics ($S$) and $38$ enrolled in Financial Analytics ($F$). Every student enrolled in at least one course.
Find:
(i) How many students enrolled in both courses?
(ii) How many enrolled in Business Statistics only?
(iii) How many enrolled in Financial Analytics only?
Prove that for any two finite sets $A$ and $B$:
$$n(A \Delta B) = n(A) + n(B) - 2n(A \cap B)$$
Show all intermediate steps establishing the mutual disjointness of differences.
Show Marking Scheme & Solution (3 Marks)
Step 1: Definition of Symmetric Difference [1 Mark]:
$$A \Delta B = (A - B) \cup (B - A)$$
Since $(A - B) \cap (B - A) = \emptyset$ (they are disjoint):
$$n(A \Delta B) = n(A - B) + n(B - A)$$ Step 2: Substitute difference formulas [1 Mark]:
We know that $n(A - B) = n(A) - n(A \cap B)$ and $n(B - A) = n(B) - n(A \cap B)$.
Section D: Long Answer Type (Q9)1 Question • 4 Marks
Question 94 Marks
A market research survey of $100$ urban households recorded newspaper reading habits across three newspapers: Times of India ($T$), Economic Times ($E$), and The Hindu ($H$):
• $42$ read Times of India
• $35$ read Economic Times
• $30$ read The Hindu
• $15$ read Times of India and Economic Times
• $12$ read Economic Times and The Hindu
• $10$ read Times of India and The Hindu
• $5$ read all three newspapers.
Find:
(i) How many households read at least one of the three newspapers? (1 Mark)
(ii) How many households read none of these three newspapers? (1 Mark)
(iii) How many households read exactly one newspaper? (1 Mark)
(iv) How many households read Times of India and Economic Times, but not The Hindu? (1 Mark)
Section E: Case-Based / Source-Based Problem (Q10)1 Case Study • 4 Marks
Question 10: Case Study4 Marks Total
Case Background: Fast-Moving Consumer Goods (FMCG) Retail Subscription Channels
A consumer goods manufacturer rolled out an omnichannel loyalty subscription to $600$ verified retail merchants in Northern India. The three promotional incentive bundles offered were:
• Bundle A: Instant Invoice Cashback ($A$) — chosen by $280$ merchants
• Bundle B: Free Shipping & Logistics Waiver ($B$) — chosen by $240$ merchants
• Bundle C: Extended 45-day Credit Line ($C$) — chosen by $200$ merchants
Cross-subscription records indicate that $90$ chose Bundles $A$ and $B$, $70$ chose Bundles $B$ and $C$, $60$ chose Bundles $A$ and $C$, and $30$ merchants enrolled in all three bundles.
(i) [1 Mark]: Find the total number of merchants who subscribed to at least one promotional incentive bundle.
(ii) [1 Mark]: How many merchants opted out of all three promotional bundles?
Show Solution for (ii)
[1 Mark]: Number of opt-out merchants $= n(U) - n(A \cup B \cup C) = 600 - 530 = 70$ merchants.
(iii) [2 Marks]: Calculate:
(a) The number of merchants who chose exactly two bundles. [1 Mark]
(b) The number of merchants who subscribed strictly to Bundle A only. [1 Mark]
Show Solution for (iii)
(a) Exactly two bundles [1 Mark]:
• Only $A$ and $B$: $90 - 30 = 60$
• Only $B$ and $C$: $70 - 30 = 40$
• Only $A$ and $C$: $60 - 30 = 30$
$$\text{Total} = 60 + 40 + 30 = 130 \text{ merchants}$$ (b) Bundle A only [1 Mark]:
$$n(\text{Only } A) = n(A) - [n(A \cap B) + n(A \cap C) - n(A \cap B \cap C)] = 280 - [90 + 60 - 30] = 280 - 120 = 160 \text{ merchants}$$