CUET UG Applied Mathematics Booster Test 3 - Congruence Modulo and Equivalence Classes -

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CUET UG Applied Mathematics Booster Test 3 - 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 3 - Congruence Modulo and Equivalence Classes

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QUESTION 1 OF 20

Take two positive integers a = 262 and b = 137, and check their congruence modulo 5. Which of the following accurately describes their relationship?

Answer Review

1 Take two positive integers a = 262 and b = 137, and check their congruence modulo 5. Which of the following accurately describes their relationship?

A 262 ≡ 137 (mod 5) because 262 - 137 = 125, which is a multiple of 5.

B 262 ≢ 137 (mod 5) because they leave different remainders.

C 262 ≡ 137 (mod 5) because 262 + 137 = 399.

D They are not congruent because 125 is not divisible by 5.

💡 Short Explanation

�� Congruence modulo m depends on divisibility of the difference �� 262 − 137 = 125 �� 125 is divisible by 5

✅ Correct Answer Explanation

Two integers a and b are congruent modulo m if m divides (a − b). Here: Since 125 is divisible by 5: Therefore: Option A correctly applies the definition of congruence. Option B is incorrect because both numbers actually leave remainder 2 when divided by 5. Option C incorrectly uses addition instead of difference. Option D is wrong because 125 is divisible by 5.

❓ Why Other Options Are Incorrect

�� Option B → Both integers leave the same remainder when divided by 5.
�� Option C → Congruence is determined using subtraction, not addition.
�� Option D → 125 ÷ 5 = 25, so 125 is divisible by 5.

🎯 Strategy

Used: Elimination

Application: Eliminate options using the definition of congruence modulo m.

Final Logic: Since 5 divides 125, the integers are congruent modulo 5.

🧠 Memory Trick

"Same remainder or divisible difference ⇒ congruent."

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

2 Match List 1 with List 2.

List 1	List 2
1. a = 41, b = 21, m = 5	a. a - b = 125
2. a = 31, b = 21, m = 5	b. a - b = 8
3. a = 262, b = 137, m = 5	c. a - b = 20
4. a = 11, b = 3, m = 8	d. a - b = 10 Note: The original question contains a typographical error in List II, where "d" is repeated. It should read a, b, c, d as shown above.

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

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

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

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

💡 Short Explanation

Find the value of a - b for each pair. Match the difference with the corresponding value in List II. Verify that the difference is divisible by the given modulus.

✅ Correct Answer Explanation

Compute each difference. 1: 41 - 21 = 20 → 1 → c 2: 31 - 21 = 10 → 2 → d 3: 262 - 137 = 125 → 3 → a 4: 11 - 3 = 8 → 4 → b Therefore, the correct matching is: 1 → c 2 → d 3 → a 4 → b Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because it matches all four differences with incorrect values.
Option C → Incorrect because the calculated differences do not correspond to the given pairings.
Option D → Incorrect because several numerical differences are interchanged.

🎯 Strategy

Used: Substitution

Application: Directly subtract b from a for each pair and match the resulting difference.

Final Logic: Exact subtraction gives the sequence 1-c, 2-d, 3-a, 4-b, which appears only in Option A.

🧠 Memory Trick

Subtract first, match later.

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

3 Let
a = ms + r
and
b = mt + r.
Which of the following analytical statements conclusively prove that
a ≡ b (mod m)?
1. a - b = m(s - t)
2. m divides (a - b)
3. a mod m = b mod m
4. a / b = m

A 1, 2

B 2, 3

C 1, 2, 3

D 1, 2, 3, 4

💡 Short Explanation

Equal remainders imply congruence. A difference divisible by the modulus proves congruence. The quotient of two numbers is unrelated to congruence.

✅ Correct Answer Explanation

Given, a = ms + r b = mt + r Subtracting, a - b = (ms + r) - (mt + r) = ms - mt = m(s - t) Thus, a - b = m(s - t) which means m divides (a - b). Therefore, a ≡ b (mod m). Now evaluate each statement. Statement 1: Correct. The difference is a multiple of m. Statement 2: Correct. Divisibility by m is the definition of congruence modulo m. Statement 3: Correct. Equal remainders after division by m also imply congruence. Statement 4: Incorrect. The value of a / b has no connection with modular congruence. Therefore, Statements 1, 2, and 3 are correct. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect because it omits Statement 3, which is also a valid condition for congruence.
Option B → Incorrect because it omits Statement 1, which directly proves congruence.
Option D → Incorrect because Statement 4 is unrelated to modular arithmetic.

🎯 Strategy

Used: Option Grouping

Application: Identify all statements that are logically equivalent definitions or consequences of congruence modulo m.

Final Logic: Statements 1, 2, and 3 all prove a ≡ b (mod m), whereas Statement 4 does not.

🧠 Memory Trick

Same remainder = Divisible difference.

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

4 Which of the following is an INCORRECT algebraic derivation for establishing congruence modulo m between integers a and b?

A a = ms + r

B b = mt + r

C a - b = r(s - t)

D a - b = m(s - t)

💡 Short Explanation

�� Correct subtraction eliminates remainder r �� Factor should contain m �� r(s − t) is algebraically incorrect

✅ Correct Answer Explanation

Given: Subtracting: The remainder r cancels out completely. Therefore Option D is correct derivation, while Option C is incorrect because r is not factored from the expression. Hence the incorrect statement is Option C.

❓ Why Other Options Are Incorrect

�� Option A → Standard representation of division algorithm.
�� Option B → Correct modular representation.
�� Option D → Correct subtraction result proving congruence.

🎯 Strategy

Used: Elimination

Application: Check algebraic subtraction carefully.

Final Logic: The remainder cancels, so factor must involve m, not r.

🧠 Memory Trick

"Remainders cancel; modulus remains."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Congruence Derivation

5 In the classification of 50 consecutive positive integers mapped to modulo 6 sectors, how many integers fall into the mixture subset?

A 7

B 8

C 9

D 10

💡 Short Explanation

�� Integers are grouped into modulo 6 classes �� Each class repeats after every 6 numbers �� Counting required class gives 8 elements

✅ Correct Answer Explanation

Among 50 consecutive positive integers, numbers distribute almost equally among 6 equivalence classes modulo 6. Since: Each class gets at least 8 integers. The "mixture subset" here corresponds to one modulo class receiving 8 elements. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Too small for equal modulo distribution.
�� Option C → Would require more repetitions.
�� Option D → Not possible since 50 numbers spread over 6 classes.

🎯 Strategy

Used: Substitution

Application: Divide total integers by modulus to estimate class size.

Final Logic: 50 divided by 6 gives quotient 8, so each class has about 8 members.

🧠 Memory Trick

"Class size ≈ total ÷ modulus."

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

6 If the constraint region is limited to positive integers strictly ≤ 30, what is the sum of all elements found within the modulo 6 equivalence class ?

A 90

B 60

C 30

D 120

💡 Short Explanation

�� Consider numbers congruent to required residue modulo 6 �� Add all valid numbers ≤ 30 �� Sum becomes 90

✅ Correct Answer Explanation

The intended equivalence class is: Adding: Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incomplete addition.
�� Option C → Only one element considered.
�� Option D → Exceeds actual total.

🎯 Strategy

Used: Substitution

Application: List all class members satisfying the bound.

Final Logic: Correct arithmetic sum gives 90.

🧠 Memory Trick

"Multiples of 6 till 30 → add quickly."

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

7 If the equivalence class is denoted as [X] = {a ∈ Z | a ≡ X (mod m)}, which of the following sets correctly represents modulo 3?

A {..., -4, -1, 2, 5, 8, 11, ...}

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

C {0, ±3, ±6, ...}

D {2, 4, 6, 8, 10}

💡 Short Explanation

�� All elements must leave same remainder mod 3 �� Each term in Option A leaves remainder 2 �� Hence they belong to same equivalence class

✅ Correct Answer Explanation

Check elements of Option A modulo 3: -4 ≡ 2 -1 ≡ 2 2 ≡ 2 5 ≡ 2 All belong to the same equivalence class modulo 3. Hence: Option A correctly represents an equivalence class modulo 3.

❓ Why Other Options Are Incorrect

�� Option B → Represents remainder 1 class, not the required one.
�� Option C → Represents multiples of 3 only.
�� Option D → Elements do not share a common remainder modulo 3.

🎯 Strategy

Used: Option Grouping

Application: Verify whether all numbers produce same remainder.

Final Logic: Only Option A has identical remainder for every element.

🧠 Memory Trick

"Same remainder → same class."

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

8 Moving across the integer domain, which element belongs to the intersection of the equivalence classes mod 3 and mod 3?

A 1

B 2

C 0

D Null Set (∅)

💡 Short Explanation

�� Distinct equivalence classes do not overlap �� No common element exists �� Intersection is empty

✅ Correct Answer Explanation

Distinct equivalence classes modulo the same modulus are disjoint. For example: An integer cannot simultaneously leave two different remainders modulo 3. Hence the intersection is the null set. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Cannot belong to two distinct classes simultaneously.
�� Option B → Same contradiction occurs.
�� Option C → 0 belongs only to one equivalence class.

🎯 Strategy

Used: Odd One Out

Application: Distinct equivalence classes are non-overlapping.

Final Logic: Their intersection must be empty.

🧠 Memory Trick

"Different remainder classes never mix."

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

9 When evaluating the reflexive relation a Rm b if and only if a ≡ b (mod m), what is the mathematical probability that an integer 'a' fails to be congruent to itself modulo m?

A 1

B 0.5

C 0

D -1

💡 Short Explanation

�� Every integer is congruent to itself �� a − a = 0 �� 0 is divisible by every m

✅ Correct Answer Explanation

For any integer a: Since every integer m divides 0: Therefore: So the probability that a fails to be congruent to itself is 0.

❓ Why Other Options Are Incorrect

�� Option A → Reflexive property always holds.
�� Option B → No partial probability exists here.
�� Option D → Probability cannot be negative.

🎯 Strategy

Used: Elimination

Application: Use reflexive property directly.

Final Logic: Every integer is always congruent to itself.

🧠 Memory Trick

"Self congruence never fails."

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

10 If m | (a - b) establishes the directional vector from a to b in symmetric relations, reversing the direction yields m | (-1)(a - b). What does this establish?

A a ≡ b (mod m) implies b ≡ a (mod m)

B a ≡ b (mod m) implies a = b

C m | (a + b)

D a - b = 0

💡 Short Explanation

�� Multiplying by −1 preserves divisibility �� Symmetry property is established �� Congruence works both ways

✅ Correct Answer Explanation

Given: Then: Hence: This proves the symmetric property of congruence relation.

❓ Why Other Options Are Incorrect

�� Option B → Congruence does not imply equality.
�� Option C → Divisibility of sum is unrelated here.
�� Option D → Difference need not be zero; only divisible by m.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: The statement discusses "reversing direction," indicating symmetry.

Final Logic: Reversing a − b to b − a proves symmetry.

🧠 Memory Trick

"Reverse difference → symmetric congruence."

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

11 Assume a ≡ b (mod m) and b ≡ c (mod m). By expressing b = a + km and c = b + k'm, calculating the aggregate linear area gives c = a + (k + k')m. What is the fundamental conclusion drawn?

A The relation is irreflexive.

B The relation is transitive (a ≡ c (mod m)).

C The relation is asymmetric.

D The relation is invalid for integers.

💡 Short Explanation

�� Congruence relations combine through addition �� Multiples of m remain multiples of m �� This proves transitivity

✅ Correct Answer Explanation

Given: So: Substituting: Thus: Hence: This proves the transitive property of congruence relation.

❓ Why Other Options Are Incorrect

�� Option A → Congruence is reflexive, not irreflexive.
�� Option C → Congruence is symmetric, not asymmetric.
�� Option D → Congruence modulo m is fully valid over integers.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: The statement chains two congruences together, indicating transitivity.

Final Logic: Combining multiples of m again gives a multiple of m.

🧠 Memory Trick

"Chain congruences → transitivity."

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

12 By fully integrating the reflexive, symmetric, and transitive proofs for the relation Rm defined by a Rm b if and only if a ≡ b (mod m), what theorem is conclusively proven?

A Rm is a partial order relation.

B Rm is an equivalence relation.

C Rm is a strictly increasing function.

D Rm is an irrational identity.

💡 Short Explanation

�� Reflexive property holds �� Symmetric property holds �� Transitive property holds

✅ Correct Answer Explanation

A relation is called an equivalence relation if it is: • Reflexive • Symmetric • Transitive For congruence modulo m: (reflexive) (symmetric) (transitive) Hence Rm is an equivalence relation.

❓ Why Other Options Are Incorrect

�� Option A → Partial orders require antisymmetry, which congruence lacks.
�� Option C → Congruence is a relation, not a function.
�� Option D → No irrational identity is involved.

🎯 Strategy

Used: Option Grouping

Application: Match the three standard equivalence properties.

Final Logic: Reflexive + symmetric + transitive ⇒ equivalence relation.

🧠 Memory Trick

"RST → Equivalence Relation."

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

13 If there are exactly m equivalence classes for mod m represented as [0], [1], ..., [m-1], what represents the union of all these distinct classes?

A The set of Natural Numbers (N)

B The set of All Integers (Z)

C The Empty Set

D The set of Real Numbers (R)

💡 Short Explanation

�� Every integer belongs to one remainder class �� Modulo m partitions integers completely �� Union of all classes equals Z

✅ Correct Answer Explanation

The equivalence classes modulo m are: Every integer leaves exactly one remainder when divided by m. Hence all integers belong to one of these classes. Therefore: So, the union gives the set of all integers.

❓ Why Other Options Are Incorrect

�� Option A → Negative integers are excluded.
�� Option C → Classes are nonempty.
�� Option D → Real numbers are not partitioned by integer congruence classes.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Equivalence classes partition the complete integer set.

Final Logic: All modulo classes together cover all integers.

🧠 Memory Trick

"All remainder classes together = Z."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Partition of a Set

14 Which property prevents an integer from belonging to both mod 5 and mod 5?

A Transitive overlap

B Equivalence classes are non-overlapping sets.

C Associative distribution

D Reflexive symmetry

💡 Short Explanation

�� Distinct equivalence classes are disjoint �� One integer cannot have two different remainders �� Classes never overlap

✅ Correct Answer Explanation

Equivalence classes formed under congruence modulo m partition integers into disjoint subsets. Thus: An integer cannot simultaneously belong to two distinct remainder classes modulo 5. Hence equivalence classes are non-overlapping.

❓ Why Other Options Are Incorrect

�� Option A → Not a standard property in equivalence relations.
�� Option C → Associativity is unrelated here.
�� Option D → Reflexive and symmetric properties do not describe disjointness.

🎯 Strategy

Used: Elimination

Application: Identify the property defining equivalence class partitions.

Final Logic: Distinct classes are disjoint subsets.

🧠 Memory Trick

"Different classes never intersect."

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

15 Calculate the sum of all elements within the equivalence class modulo 6, where elements are bounded between 0 < x < 30.

A 62

B 80

C 100

D 40

💡 Short Explanation

�� Consider integers congruent to 2 modulo 6 �� Add all valid members below 30 �� Total becomes 80

✅ Correct Answer Explanation

The intended equivalence class is: Adding: Thus, the provided answer B = 80 is incorrect. The correct sum is 70, which is not present in the options.

❓ Why Other Options Are Incorrect

�� Option A → Does not equal the required class sum.
�� Option B → Arithmetic total is incorrect.
�� Option C → Exceeds actual sum.
�� Option D → Too small.

🎯 Strategy

Used: Substitution

Application: List all numbers satisfying the congruence condition.

Final Logic: Direct addition gives 70, showing option mismatch.

🧠 Memory Trick

"Keep adding every +6 term."

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

16 How many positive integers strictly less than 100 belong to the equivalence class modulo 15?

A 5

B 6

C 7

D 8

💡 Short Explanation

�� Consider one residue class modulo 15 �� Count terms below 100 �� Total count becomes 6

✅ Correct Answer Explanation

Taking the intended class: There are 6 positive integers strictly less than 100. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → One element short.
�� Option C → Includes extra invalid term.
�� Option D → Exceeds valid count below 100.

🎯 Strategy

Used: Substitution

Application: Generate arithmetic progression of class elements.

Final Logic: Counting valid terms gives 6.

🧠 Memory Trick

"Add modulus repeatedly."

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

17 The definition of a Simple Arithmetic Function restricts the domain to Z+. Why does the function f(x) = 1/x fail to be a valid arithmetic function if 0 is included in the domain?

A Because 1/0 is undefined, violating Z+ bounds which start from 1.

B Because it produces real numbers.

C Because it acts as a geometric machine.

D It does not fail.

💡 Short Explanation

�� Division by zero is undefined �� Arithmetic functions use positive integers �� Z+ excludes 0

✅ Correct Answer Explanation

A simple arithmetic function is defined on positive integers: If 0 were included, then: But division by zero is undefined. Hence the function fails when 0 is included.

❓ Why Other Options Are Incorrect

�� Option B → Arithmetic functions may produce real outputs.
�� Option C → "Geometric machine" has no mathematical meaning here.
�� Option D → The function definitely fails at x = 0.

🎯 Strategy

Used: Elimination

Application: Check domain restriction and undefined operations.

Final Logic: 1/0 is undefined, so 0 cannot belong to the domain.

🧠 Memory Trick

"No division by zero."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Domain of Functions

18 If a simple arithmetic function maps Z+ to Real Numbers, which of the following is a completely valid set of images for the pre-images {1, 2, 3} under f(x) = √x?

A {1, 1.414..., 1.732...}

B {1, 4, 9}

C {-1, -2, -3}

D {0, 0, 0}

💡 Short Explanation

�� Compute square roots directly �� √1 = 1 �� √2 and √3 are irrational decimals

✅ Correct Answer Explanation

Given: For inputs {1,2,3}: Thus the image set becomes: {1, 1.414..., 1.732...} Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Represents squares, not square roots.
�� Option C → Square roots of positive numbers are nonnegative.
�� Option D → √1, √2, and √3 are not zero.

🎯 Strategy

Used: Substitution

Application: Evaluate the function directly for each input.

Final Logic: Correct square root values match Option A.

🧠 Memory Trick

"√2 ≈ 1.414, √3 ≈ 1.732."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Functions and Images

📄 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 modeled as f(x) = x^3 - x, what is the output image when the pre-image input is 3?

A 24

B 27

C 9

D 8

💡 Short Explanation

�� Substitute x = 3 �� Compute cube and subtraction �� Final value is 24

✅ Correct Answer Explanation

Given: Substitute x = 3: Compute: Hence the output image is 24.

❓ Why Other Options Are Incorrect

�� Option B → Gives only cube value.
�� Option C → Incorrect subtraction result.
�� Option D → Unrelated computation.

🎯 Strategy

Used: Substitution

Application: Replace x directly into the function.

Final Logic: 3³ − 3 = 24.

🧠 Memory Trick

"Cube minus same number."

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

📄 Passage

Question:
In the context of number theoretic relationships, why can't a simple arithmetic function process the input x = -5?

A Because the machine cannot compute negative numbers.

B Because the domain of a simple arithmetic function is strictly restricted to positive integers (Z+).

C Because the range forbids negative outputs.

D Because it is not an equivalence class.

💡 Short Explanation

�� Arithmetic functions use positive integer inputs �� −5 is not in Z+ �� Hence input is invalid

✅ Correct Answer Explanation

A simple arithmetic function has domain: Since −5 is not a positive integer, it cannot be used as an input. Therefore, the arithmetic function cannot process x = −5.

❓ Why Other Options Are Incorrect

�� Option A → Functions can mathematically compute negatives if domain allows.
�� Option C → Range restriction is unrelated to input validity.
�� Option D → Equivalence classes are unrelated to function domain.

🎯 Strategy

Used: Elimination

Application: Verify the domain definition carefully.

Final Logic: −5 ∉ Z+, so the input is invalid.

🧠 Memory Trick

"Arithmetic function ⇒ positive inputs only."

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Domain and Range

CUET UG Applied Mathematics Booster Test 3 - Congruence Modulo and Equivalence Classes

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