CUET UG Applied Mathematics Booster Test 2 - Congruence Modulo and Equivalence Classes -

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CUET UG Applied Mathematics Booster Test 2 - Congruence Modulo and Equivalence Classes

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 2 - Congruence Modulo and Equivalence Classes

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QUESTION 1 OF 20

Evaluate whether a ≡ b (mod m) holds true if a = 31, b = 21, and m = 5. Which foundational check confirms this?

Answer Review

1 Evaluate whether a ≡ b (mod m) holds true if a = 31, b = 21, and m = 5. Which foundational check confirms this?

A 31 + 21 is divisible by 5

B 31 - 21 = 10, which is perfectly divisible by 5

C 31 × 21 is an odd number

D Both leave a remainder of 2 when divided by 5

💡 Short Explanation

�� Congruence depends on divisibility of the difference. �� 31 − 21 = 10. �� 10 is divisible by 5.

✅ Correct Answer Explanation

Two integers are congruent modulo m when m divides their difference. Here, 31 − 21 = 10 and 5 divides 10 exactly. Therefore, 31 ≡ 21 (mod 5). Option A uses addition, which is irrelevant. Option C discusses parity. Option D is false because 31 and 21 leave remainder 1, not 2.

❓ Why Other Options Are Incorrect

�� Option A → Congruence is checked using difference, not sum.
�� Option C → Odd product does not establish modular equivalence.
�� Option D → 31 mod 5 = 1 and 21 mod 5 = 1, not 2.

🎯 Strategy

Used: Substitution

Application: Compute the difference directly and test divisibility by the modulus.

Final Logic: 5 divides (31 − 21), so 31 ≡ 21 (mod 5).

🧠 Memory Trick

"Same mod ⇒ Difference divisible."

📚 Class XI – Mathematics – Chapter: Introduction to Number Theory (Supplementary Concept) – Topic: Congruence Modulo m

2 Match List 1 with List 2.

List 1	List 2
1. a = 41, b = 21, m = 5	a. a ≡ b (mod 9) is perfectly valid
2. a = 31, b = 21, m = 5	b. a ≡ b (mod 5) leaves remainder 1
3. a = 15, b = 6, m = 9	c. a ≡ b (mod 6) is perfectly valid
4. a = 14, b = 8, m = 6	d. a ≡ b (mod 5) leaves remainder 1

A 1-b, 2-d, 3-a, 4-c

B 1-a, 2-b, 3-c, 4-d

C 1-d, 2-c, 3-b, 4-a

D 1-c, 2-a, 3-d, 4-b

💡 Short Explanation

Compare the remainders for modulo 5. Check whether the difference is divisible by the given modulus. Match each expression with the correct congruence statement.

✅ Correct Answer Explanation

Evaluate each pair individually. 1: 41 mod 5 = 1 and 21 mod 5 = 1. Therefore, a ≡ b (mod 5) with remainder 1. 1 → b 2: 31 mod 5 = 1 and 21 mod 5 = 1. Therefore, a ≡ b (mod 5) with remainder 1. 2 → d 3: 15 - 6 = 9, and 9 is divisible by 9. Therefore, a ≡ b (mod 9). 3 → a 4: 14 - 8 = 6, and 6 is divisible by 6. Therefore, a ≡ b (mod 6). 4 → c Thus, the correct matching is: 1 → b 2 → d 3 → a 4 → c Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because it mismatches the modulo 9 and modulo 6 congruence statements.
Option C → Incorrect because it assigns incorrect modulo conditions to the given pairs.
Option D → Incorrect because several congruence relationships do not satisfy the required remainder or divisibility conditions.

🎯 Strategy

Used: Option Grouping

Application: Verify each congruence independently by checking either equal remainders or divisibility of the difference, and then match the complete set.

Final Logic: Only Option A correctly satisfies all four congruence relationships.

🧠 Memory Trick

Same remainder = Same congruence class.

📚 Class XI – Mathematics – Chapter: Number Systems – Topic: Modular Arithmetic

3 If two positive integers a and b leave the same remainder r when divided by a positive integer m (where a = ms + r and b = mt + r), which of the following deductions are correct?
1. a - b = m(s - t)
2. m divides (a - b)
3. a ≡ b (mod m)
4. a mod m ≠ b mod m

A 1, 2

B 1, 2, 3

C 2, 3, 4

D 1, 2, 3, 4

💡 Short Explanation

Subtract the given expressions. The difference becomes a multiple of m. Therefore, the two numbers are congruent modulo m.

✅ Correct Answer Explanation

Given, a = ms + r b = mt + r Subtracting, a - b = (ms + r) - (mt + r) = ms - mt = m(s - t) Therefore, Statement 1 is correct. Since a - b = m(s - t), it follows that m divides (a - b). Hence, Statement 2 is correct. By the definition of congruence, if m divides (a - b), then a ≡ b (mod m). Therefore, Statement 3 is correct. Since a and b leave the same remainder when divided by m, a mod m = b mod m. Therefore, Statement 4 is incorrect. Hence, Statements 1, 2, and 3 are correct, making Option B the correct answer.

❓ Why Other Options Are Incorrect

Option A → Incorrect because it omits Statement 3, which directly follows from Statements 1 and 2.
Option C → Incorrect because it includes Statement 4, which contradicts the given condition of equal remainders.
Option D → Incorrect because Statement 4 is false.

🎯 Strategy

Used: Substitution

Application: Substitute the given expressions for a and b, simplify the difference, and apply the definition of congruence.

Final Logic: The difference equals a multiple of m, proving divisibility and congruence, while Statement 4 contradicts the given information.

🧠 Memory Trick

Same remainder ⇒ Same congruence class.

📚 Class XI – Mathematics – Chapter: Number Systems – Topic: Congruence Modulo m

4 Consider the congruence 41 ≡ 21 (mod 5). Identify the incorrect statement regarding its verification:

A The difference between 41 and 21 is 20.

B 5 strictly divides 20 without leaving a remainder.

C 41 mod 5 = 2

D 21 mod 5 = 1

💡 Short Explanation

�� 41 ÷ 5 leaves remainder 1. �� Difference equals 20. �� Congruence is valid.

✅ Correct Answer Explanation

41 − 21 = 20 and 5 divides 20 exactly, so the congruence holds. Also, 21 mod 5 = 1. Since 41 mod 5 = 1, not 2, Option C is the incorrect statement. The other options correctly verify the congruence.

❓ Why Other Options Are Incorrect

�� Option A → Correct arithmetic difference.
�� Option B → 20 is divisible by 5.
�� Option D → 21 leaves remainder 1 upon division by 5.

🎯 Strategy

Used: Elimination

Application: Verify each numerical statement directly.

Final Logic: Only Option C contradicts actual modular computation.

🧠 Memory Trick

"41 = 40 + 1 ⇒ remainder 1."

📚 Class XI – Mathematics – Chapter: Number Theory Concepts – Topic: Congruence Verification

5 When segregating the first 50 positive integers into sectors for modulo 6, the mixture of elements in sector 1 (remainder 1) includes which of the following?

A 2, 8, 14, 20

B 1, 7, 13, 19, 25

C 6, 12, 18, 24

D 3, 9, 15, 21

💡 Short Explanation

�� Sector 1 means remainder 1. �� Numbers differ by 6. �� Each listed number leaves remainder 1.

✅ Correct Answer Explanation

Modulo 6 classes group numbers with identical remainders. Numbers 1, 7, 13, 19, and 25 all leave remainder 1 when divided by 6. Option A leaves remainder 2, Option C remainder 0, and Option D remainder 3. Hence B correctly represents sector 1.

❓ Why Other Options Are Incorrect

�� Option A → All numbers belong to remainder class 2.
�� Option C → Multiples of 6 belong to remainder class 0.
�� Option D → Numbers leave remainder 3.

🎯 Strategy

Used: Option Grouping

Application: Check common remainder of each option set.

Final Logic: Only Option B consistently gives remainder 1.

🧠 Memory Trick

"Sector 1 = 1 + 6k."

📚 Class XI – Mathematics – Chapter: Number Systems – Topic: Residue Classes

6 In a Modulo 6 cyclic wheel, the numbers belonging to the same subset (sector) leave the same remainder. What is the constraint region for the total number of distinct subsets formed?

A Exactly 5 subsets

B Exactly 6 subsets

C Exactly 7 subsets

D Infinite subsets

💡 Short Explanation

�� Remainders range from 0 to 5. �� Each remainder defines one class. �� Total classes equal modulus.

✅ Correct Answer Explanation

For modulo 6, possible remainders are 0, 1, 2, 3, 4, and 5. Each remainder determines a distinct equivalence class. Therefore exactly six subsets exist. Options A and C provide incorrect counts, while D ignores the finite remainder structure.

❓ Why Other Options Are Incorrect

�� Option A → Excludes one valid remainder class.
�� Option C → Adds an impossible remainder.
�� Option D → Number of classes equals modulus, not infinity.

🎯 Strategy

Used: Elimination

Application: List all possible remainders modulo 6.

Final Logic: Six possible remainders imply six subsets.

🧠 Memory Trick

"Mod m ⇒ m classes."

📚 Class XI – Mathematics – Chapter: Number Theory Concepts – Topic: Equivalence Classes

7 In the definition of equivalence class [X] = {a ∈ Z | a ≡ X (mod m)}, what is the set of elements if m = 3 and X = 1?

A {..., -5, -2, 1, 4, 7, ...}

B {..., -4, -1, 2, 5, 8, ...}

C {0, ±3, ±6, ...}

D {1, 2, 3}

💡 Short Explanation

�� Numbers must leave remainder 1 mod 3. �� Consecutive terms differ by 3. �� Option A satisfies the rule.

✅ Correct Answer Explanation

An equivalence class [1] modulo 3 contains all integers of the form 1 + 3k. Option A exactly matches this pattern. Option B corresponds to remainder 2, Option C to remainder 0, and Option D is merely a finite collection rather than the entire class.

❓ Why Other Options Are Incorrect

�� Option B → Represents class [2].
�� Option C → Represents class [0].
�� Option D → Not a complete equivalence class.

🎯 Strategy

Used: Substitution

Application: Check whether numbers fit the form 1 + 3k.

Final Logic: Only Option A contains all integers congruent to 1 modulo 3.

🧠 Memory Trick

"Class [1] = 1 + multiples of 3."

📚 Class XI – Mathematics – Chapter: Number Theory Concepts – Topic: Equivalence Classes

8 Moving through the sequence of modulo 3 equivalence classes, which class contains the number 11?

A [0]

B [1]

C [2]

D [3]

💡 Short Explanation

�� Divide 11 by 3. �� Remainder obtained is 2. �� Therefore 11 belongs to class [2].

✅ Correct Answer Explanation

11 = 3 × 3 + 2, so the remainder upon division by 3 is 2. Hence 11 is congruent to 2 modulo 3 and belongs to equivalence class [2]. Class [3] is not a standard modulo 3 class because valid classes are [0], [1], and [2].

❓ Why Other Options Are Incorrect

�� Option A → Requires remainder 0.
�� Option B → Requires remainder 1.
�� Option D → Equivalent to class [0], not distinct.

🎯 Strategy

Used: Substitution

Application: Calculate 11 mod 3 directly.

Final Logic: 11 mod 3 = 2, so class [2].

🧠 Memory Trick

"11 = 9 + 2."

📚 Class XI – Mathematics – Chapter: Number Theory Concepts – Topic: Residue Classes

9 To prove the symmetric property Rm, if a ≡ b (mod m), it strictly implies b ≡ a (mod m). What is the probability that this property fails for any integers 'a' and 'b' that already satisfy the premise?

A 1

B 0.5

C 0

D Depends on m

💡 Short Explanation

�� Symmetry is always true. �� Congruence is an equivalence relation. �� Failure never occurs.

✅ Correct Answer Explanation

If a ≡ b (mod m), then m divides (a − b). Consequently, m also divides −(a − b) = (b − a). Therefore b ≡ a (mod m). Since symmetry always holds, the probability of failure is zero. Other options incorrectly assume possible exceptions.

❓ Why Other Options Are Incorrect

�� Option A → Failure never occurs.
�� Option B → No probabilistic uncertainty exists.
�� Option D → Symmetry does not depend on modulus choice.

🎯 Strategy

Used: Elimination

Application: Use the definition of congruence and sign reversal.

Final Logic: Divisibility of a−b automatically implies divisibility of b−a.

🧠 Memory Trick

"Reverse sign, same divisibility."

📚 Class XI – Mathematics – Chapter: Relations – Topic: Symmetric Property

10 In establishing the reflexive property, vectorially calculating m | (a - a) yields 0. What does this mean for the relation a ≡ a (mod m)?

A The relation holds true for all a ∈ Z.

B The relation is only true for positive integers.

C The relation is invalid.

D The relation applies only to m = 1.

💡 Short Explanation

�� a − a = 0. �� Every integer m divides 0. �� Reflexive property always holds.

✅ Correct Answer Explanation

Since a − a = 0 and every nonzero integer divides 0, the condition m | (a − a) is always satisfied. Therefore a ≡ a (mod m) for every integer a. This establishes reflexivity. Options B, C, and D unnecessarily restrict or deny the property.

❓ Why Other Options Are Incorrect

�� Option B → Reflexivity applies to all integers.
�� Option C → The relation is valid.
�� Option D → Reflexivity is true for every modulus m.

🎯 Strategy

Used: Substitution

Application: Replace a−a with zero and apply divisibility.

Final Logic: m divides 0, so a ≡ a (mod m).

🧠 Memory Trick

"Self minus self = 0."

📚 Class XI – Mathematics – Chapter: Relations – Topic: Reflexive Property

11 Suppose a ≡ b (mod m) and b ≡ c (mod m). By substituting b = a + km and c = b + k'm, the calculated area of difference (c - a) is equal to:

A (k - k')m

B (k + k')m

C km

D k'm

💡 Short Explanation

�� Use the given substitutions. �� Replace b in c = b + k'm. �� Difference becomes a multiple of m.

✅ Correct Answer Explanation

Given b = a + km and c = b + k'm. Substituting b into c gives c = a + km + k'm = a + (k + k')m. Therefore c − a = (k + k')m. This proves divisibility by m and establishes transitivity. Other options omit or incorrectly combine the coefficients.

❓ Why Other Options Are Incorrect

�� Option A → Uses subtraction instead of addition of coefficients.
�� Option C → Ignores the contribution of k'm.
�� Option D → Ignores the contribution of km.

🎯 Strategy

Used: Substitution

Application: Substitute the first congruence expression into the second and simplify.

Final Logic: c − a = (k + k')m, hence divisible by m.

🧠 Memory Trick

"Transitive ⇒ add the multiples."

📚 Class XI – Mathematics – Chapter: Relations – Topic: Transitive Property of Congruence

12 The relation Rm defined by a Rm b if and only if a ≡ b (mod m) is an equivalence relation because it integrates which specific properties?

A Additive, Subtractive, Multiplicative

B Reflexive, Symmetric, Transitive

C Associative, Distributive, Commutative

D Divisible, Quotients, Remainders

💡 Short Explanation

�� Equivalence relations have three defining properties. �� Reflexive, symmetric, and transitive are mandatory. �� Congruence satisfies all three.

✅ Correct Answer Explanation

A relation is an equivalence relation only when it is reflexive, symmetric, and transitive. Congruence modulo m satisfies all these properties: a ≡ a, a ≡ b implies b ≡ a, and a ≡ b with b ≡ c implies a ≡ c. The remaining options describe algebraic operations rather than equivalence criteria.

❓ Why Other Options Are Incorrect

�� Option A → These are arithmetic operations, not equivalence properties.
�� Option C → These are algebraic laws of operations.
�� Option D → These are related concepts but not defining properties.

🎯 Strategy

Used: Odd One Out

Application: Identify the standard three-property definition of equivalence relations.

Final Logic: Only reflexive, symmetric, and transitive characterize equivalence relations.

🧠 Memory Trick

"RST = Relation Stays True."

📚 Class XI – Mathematics – Chapter: Relations and Functions – Topic: Equivalence Relation

13 Which statement accurately represents the fact that equivalence classes [0], [1], ..., [m−1] for mod m are "non-overlapping sets"?

A They all share exactly one common element.

B The intersection of any two different classes is a null set.

C They all contain the number m.

D The union of all sets is an empty set.

💡 Short Explanation

�� Distinct classes are disjoint. �� No element belongs to two different classes. �� Their intersections are empty.

✅ Correct Answer Explanation

Equivalence classes partition a set. In modulo arithmetic, every integer belongs to exactly one residue class. Therefore any two different classes have no common element and their intersection is empty. Option A contradicts disjointness. Option C is false. Option D is incorrect because their union forms the entire set of integers.

❓ Why Other Options Are Incorrect

�� Option A → Distinct classes cannot share elements.
�� Option C → Not every class contains m.
�� Option D → Union of all classes covers all integers.

🎯 Strategy

Used: Elimination

Application: Recall the partition property of equivalence classes.

Final Logic: Different equivalence classes are mutually disjoint.

🧠 Memory Trick

"One number, one class."

📚 Class XI – Mathematics – Chapter: Relations and Functions – Topic: Equivalence Classes

14 The total number of exact equivalence classes for modulo m is represented as [0], [1], ..., up to what final class?

A [m]

B [m+1]

C [m−1]

D Infinite

💡 Short Explanation

�� Possible remainders are 0 to m−1. �� Each remainder forms one class. �� Final class is [m−1].

✅ Correct Answer Explanation

Modulo m arithmetic allows only remainders 0, 1, 2, ..., m−1. Each remainder corresponds to one equivalence class. Hence there are exactly m classes, ending with [m−1]. Options A and B introduce invalid remainder values, while D ignores the finite nature of residue classes.

❓ Why Other Options Are Incorrect

�� Option A → Remainder m is equivalent to remainder 0.
�� Option B → Impossible remainder in modulo m.
�� Option D → Number of classes is finite.

🎯 Strategy

Used: Elimination

Application: List all permissible remainders modulo m.

Final Logic: Residue classes correspond exactly to remainders 0 through m−1.

🧠 Memory Trick

"Modulo m ends at m−1."

📚 Class XI – Mathematics – Chapter: Number Theory Concepts – Topic: Residue Classes

15 Find all the positive integers strictly less than 30 forming the equivalence class of 5 for modulo 7.

A 5, 12, 19, 26

B 7, 14, 21, 28

C 5, 10, 15, 20

D 12, 19, 26, 33

💡 Short Explanation

�� Numbers must be congruent to 5 mod 7. �� Add successive multiples of 7. �� Keep values below 30.

✅ Correct Answer Explanation

The class [5] modulo 7 consists of numbers of the form 5 + 7k. Taking positive values less than 30 gives 5, 12, 19, and 26. Option B contains multiples of 7. Option C does not maintain remainder 5. Option D includes 33, which exceeds the limit.

❓ Why Other Options Are Incorrect

�� Option B → These belong to class [0].
�� Option C → Several numbers have different remainders modulo 7.
�� Option D → 33 is not less than 30.

🎯 Strategy

Used: Substitution

Application: Generate values using 5 + 7k and apply the upper bound.

Final Logic: Only Option A contains all valid members below 30.

🧠 Memory Trick

"Class [5] = 5 + 7k."

📚 Class XI – Mathematics – Chapter: Modular Arithmetic – Topic: Equivalence Class Construction

16 Find the positive integers less than 50 forming the equivalence class 4 for modulo 6.

A 4, 10, 16, 22, 28, 34, 40, 46

B 6, 12, 18, 24, 30, 36, 42, 48

C 4, 8, 12, 16, 20

D 10, 20, 30, 40

💡 Short Explanation

�� Numbers must leave remainder 4. �� Pattern is 4 + 6k. �� All values remain below 50.

✅ Correct Answer Explanation

The equivalence class [4] modulo 6 contains all integers of the form 4 + 6k. Listing positive values less than 50 yields 4, 10, 16, 22, 28, 34, 40, and 46. Other options contain numbers belonging to different remainder classes.

❓ Why Other Options Are Incorrect

�� Option B → Multiples of 6 belong to class [0].
�� Option C → Contains numbers with varying remainders.
�� Option D → Not all values leave remainder 4 modulo 6.

🎯 Strategy

Used: Substitution

Application: Use the formula 4 + 6k and generate valid terms.

Final Logic: Only Option A contains all numbers congruent to 4 modulo 6.

🧠 Memory Trick

"Class [4] = 4 + 6k."

📚 Class XI – Mathematics – Chapter: Modular Arithmetic – Topic: Residue Classes

17 What distinguishes a Simple Arithmetic Function from a general algebraic function?

A Its domain is strictly positive integers (Z+).

B It can only accept negative fractions.

C Its output must be purely imaginary.

D It does not follow any specific rules.

💡 Short Explanation

�� Arithmetic functions are number-theoretic functions. �� Inputs are positive integers. �� Outputs may be real or complex.

✅ Correct Answer Explanation

A simple arithmetic function is defined primarily on positive integers and studies arithmetic properties of numbers. Its range may be real or complex. Option B imposes an incorrect domain. Option C wrongly restricts outputs. Option D contradicts the very definition of a mathematical function.

❓ Why Other Options Are Incorrect

�� Option B → Negative fractions are not the defining domain.
�� Option C → Outputs need not be imaginary.
�� Option D → Functions always obey a defining rule.

🎯 Strategy

Used: Elimination

Application: Identify the defining characteristic from the definition.

Final Logic: Arithmetic functions are distinguished by a positive-integer domain.

🧠 Memory Trick

"Arithmetic ⇒ Acts on integers."

📚 Class XI – Mathematics – Chapter: Functions – Topic: Arithmetic Functions

18 For the arithmetic function mapping Z+ → R, if f(x) = x², what is the image if the pre-image is 5?

A 5

B 10

C 25

D 0

💡 Short Explanation

�� Substitute x = 5. �� Compute 5². �� Result equals 25.

✅ Correct Answer Explanation

The function rule is f(x) = x². Replacing x with 5 gives f(5) = 5² = 25. Therefore the image corresponding to the pre-image 5 is 25. The remaining options do not satisfy the function rule and arise from incorrect computation.

❓ Why Other Options Are Incorrect

�� Option A → Returns the input unchanged.
�� Option B → Represents multiplication by 2, not squaring.
�� Option D → Not produced by f(5).

🎯 Strategy

Used: Substitution

Application: Insert the given input into the function formula.

Final Logic: 5² = 25, so Option C is correct.

🧠 Memory Trick

"Square 5 → 25."

📚 Class XI – Mathematics – Chapter: Relations and Functions – Topic: Image and Pre-image

📄 Passage

We can also understand function as a machine which follows the instructed operation and converts the input (pre-image) to required output (image). A Simple Arithmetic function is a function whose domain is positive integers and the range is real or complex numbers, also known as Number theoretic function. Such functions describe arithmetic properties of numbers and are widely useful in number theory.

Question:
If an arithmetic function machine is set to the rule f(x) = 3x + 1, and the pre-image is 4 (a positive integer), what is the generated image?

A 12

B 13

C 4

D 7

💡 Short Explanation

�� Use the function rule. �� Substitute x = 4. �� Evaluate 3(4) + 1.

✅ Correct Answer Explanation

The machine follows the rule f(x) = 3x + 1. For pre-image 4, f(4) = 3 × 4 + 1 = 13. Therefore the generated image is 13. Option A omits the added 1, while Options C and D do not satisfy the stated function rule.

❓ Why Other Options Are Incorrect

�� Option A → Uses only 3 × 4.
�� Option C → Confuses input with output.
�� Option D → Does not follow the function definition.

🎯 Strategy

Used: Substitution

Application: Insert the input value into the machine rule.

Final Logic: f(4) = 12 + 1 = 13.

🧠 Memory Trick

"Triple then add one."

📚 Class XI – Mathematics – Chapter: Relations and Functions – Topic: Function Evaluation

📄 Passage

Question:
Why are Simple Arithmetic functions widely useful according to the passage?

A Because they compute geometry graphs.

B Because they describe arithmetic properties of numbers in number theory.

C Because they convert complex variables to negative roots.

D Because they map real numbers to non-integers.

💡 Short Explanation

�� Directly stated in the passage. �� Arithmetic functions study number properties. �� Important in number theory.

✅ Correct Answer Explanation

The passage explicitly states that simple arithmetic functions describe arithmetic properties of numbers and are widely useful in number theory. Therefore Option B is correct. Options A, C, and D introduce ideas not mentioned in the passage and are unrelated to the stated purpose of arithmetic functions.

❓ Why Other Options Are Incorrect

�� Option A → Geometry graphs are not discussed.
�� Option C → Negative roots are unrelated to the passage.
�� Option D → This is not the defining purpose of arithmetic functions.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Locate the exact statement in the passage and match it with the options.

Final Logic: Option B directly reproduces the passage's key idea.

🧠 Memory Trick

"Arithmetic function = Number property tool."

📚 Class XI – Mathematics – Chapter: Functions – Topic: Arithmetic Functions

CUET UG Applied Mathematics Booster Test 2 - Congruence Modulo and Equivalence Classes

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