CUET UG Applied Mathematics Booster Test 1 - Number Theory and Modular Fundamentals -

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CUET UG Applied Mathematics Booster Test 1 - Number Theory and Modular Fundamentals

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 1 - Number Theory and Modular Fundamentals

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QUESTION 1 OF 20

When classifying a set of ten consecutive positive integers into odd or even sets, what is the foundational criterion behind this classification?

Answer Review

1 When classifying a set of ten consecutive positive integers into odd or even sets, what is the foundational criterion behind this classification?

A Whether the number leaves a remainder of zero or one when divided by 2

B Whether the number is divisible by 5 or not

C Applying a multiplication property of parity

D Dividing the integers into three equal subsets

💡 Short Explanation

�� Even and odd numbers are based on division by 2. �� Remainder 0 gives even numbers. �� Remainder 1 gives odd numbers.

✅ Correct Answer Explanation

Integers are classified as even or odd depending on the remainder obtained after division by 2. → If a number leaves remainder 0, it is even. → If it leaves remainder 1, it is odd. → Therefore, Option A is correct. → Option B is incorrect because divisibility by 5 is unrelated to parity classification. → Option C is incorrect because multiplication properties are not the foundational criterion. → Option D is incorrect because odd-even classification divides numbers into two sets, not three subsets.

❓ Why Other Options Are Incorrect

�� Option B → Divisibility by 5 does not determine whether a number is odd or even.
�� Option C → Multiplication properties may preserve parity but do not define it.
�� Option D → Integers are classified into two parity groups only.

🎯 Strategy

Used: Elimination

Application: Remove options unrelated to the basic definition of even and odd integers.

Final Logic: Odd-even classification depends only on remainders after division by 2.

🧠 Memory Trick

"Divide by 2 → remainder decides parity."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Even and Odd Integers

2 Match List 1 with List 2.

List 1	List 2
1. 10 days	a. Shift of 2 days
2. 68 days	b. Shift of 5 days
3. 16 days	c. Shift of 3 days
4. 25 days	d. Shift of 4 days

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

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

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

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

💡 Short Explanation

A week repeats after every 7 days. Divide the number of elapsed days by 7 and use the remainder. The remainder gives the weekday shift.

✅ Correct Answer Explanation

Compute each value using modulo 7. 10 days: 10 mod 7 = 3 → Shift of 3 days → 1 → c 68 days: 68 mod 7 = 5 → Shift of 5 days → 2 → b 16 days: 16 mod 7 = 2 → Shift of 2 days → 3 → a 25 days: 25 mod 7 = 4 → Shift of 4 days → 4 → d Therefore, the correct matching is: 1 → c 2 → b 3 → a 4 → d Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect because it matches 10 days with a shift of 2 days instead of 3 days, and 16 days with a shift of 3 days instead of 2 days.
Option B → Incorrect because it incorrectly matches 16 days with a shift of 4 days and 25 days with a shift of 3 days.
Option D → Incorrect because it matches 16 days with a shift of 5 days and 25 days with a shift of 2 days.

🎯 Strategy

Used: Substitution

Application: Divide each number of days by 7 and substitute the remainder as the weekday shift.

Final Logic: The modulo 7 remainders produce the sequence 1-c, 2-b, 3-a, 4-d, which appears only in Option C.

🧠 Memory Trick

Week repeats every 7 days.

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

3 In modular arithmetic applied to a 12-hour wall clock, which of the following times display exactly the same hour hand position as 1 O'clock?
(1) 13 O'clock
(2) 25 O'clock
(3) 37 O'clock
(4) 12 O'clock

A 1, 2, 3

B 1, 2

C 1, 3, 4

D 1, 2, 3, 4

💡 Short Explanation

A 12-hour clock repeats every 12 hours. Times differing by multiples of 12 show the same hour-hand position. Reduce each time modulo 12.

✅ Correct Answer Explanation

Evaluate each time using modulo 12. 13 O'clock: 13 mod 12 = 1 ✔ 25 O'clock: 25 mod 12 = 1 ✔ 37 O'clock: 37 mod 12 = 1 ✔ 12 O'clock: 12 mod 12 = 0 ✘ Thus, 13 O'clock has the same position as 1 O'clock. 25 O'clock has the same position as 1 O'clock. 37 O'clock has the same position as 1 O'clock. 12 O'clock corresponds to 12 (or 0 modulo 12), not 1 O'clock. Therefore, Statements 1, 2, and 3 are correct. Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because it omits 37 O'clock, which is also congruent to 1 modulo 12.
Option C → Incorrect because it includes 12 O'clock, which corresponds to 0 modulo 12, not 1.
Option D → Incorrect because it includes 12 O'clock, which does not have the same hour-hand position as 1 O'clock.

🎯 Strategy

Used: Substitution

Application: Reduce each clock time modulo 12 and compare the remainder with 1.

Final Logic: Only times with remainder 1 modulo 12 have the same hour-hand position as 1 O'clock, giving Option A.

🧠 Memory Trick

Add or subtract 12—the clock looks the same.

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

4 Consider the equation -68 = 12 × (-6) + 4. Which statement is incorrect regarding Euclid's Division Algorithm and the modulo operation for -68 mod 12?

A The divisor is 12.

B The quotient is -6.

C The remainder R is equal to -4.

D -68 mod 12 = 4.

💡 Short Explanation

�� Remainder in modulo arithmetic is non-negative. �� Here remainder equals 4. �� Negative remainder violates Euclid's condition.

✅ Correct Answer Explanation

Euclid's division form is: For: the divisor is 12, quotient is −6, and remainder is 4. → Therefore, Option C is incorrect because remainder cannot be −4. → Option A is correct since divisor is 12. → Option B is correct since quotient is −6. → Option D is correct because −68 mod 12 = 4.

❓ Why Other Options Are Incorrect

�� Option A → Correctly identifies the divisor.
�� Option B → Correctly identifies the quotient.
�� Option D → Correct modulo result is 4.

🎯 Strategy

Used: Extreme Word Filter

Application: Check whether the remainder satisfies the mandatory non-negative condition.

Final Logic: Modulo remainders cannot be negative in Euclidean division.

🧠 Memory Trick

"Modulo remainder ≥ 0."

📚 Class X – NCERT Mathematics – Chapter: Real Numbers – Topic: Euclid's Division Lemma

5 Evaluate -68 mod 12 and -68 mod (-12) based on the properties of the modulo operator where R ≥ 0.

A 8 and 4

B 4 and 4

C -8 and 4

D 4 and -8

💡 Short Explanation

�� Modulo remainder is non-negative. �� Remainder obtained is 4. �� Same remainder applies here.

✅ Correct Answer Explanation

Using Euclidean division: Hence: Similarly, considering remainder non-negative: → Therefore, Option B is correct. → Options C and D are invalid because remainders are negative. → Option A gives an incorrect remainder for the first case.

❓ Why Other Options Are Incorrect

�� Option A → 8 is not the correct remainder for −68 mod 12.
�� Option C → Negative remainder violates modulo conditions.
�� Option D → Negative remainder is invalid in Euclidean modulo arithmetic.

🎯 Strategy

Used: Elimination

Application: Remove options containing invalid negative remainders.

Final Logic: Euclidean modulo requires non-negative remainder 4.

🧠 Memory Trick

"Modulo remainder never negative."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Modulo with Negative Integers

6 If the value of positive integer X is increased by a multiple of Y (kY, where k is an integer), which equality accurately maps the constraint regions?

A (X + kY) mod Y > X mod Y

B X mod Y = (X + kY) mod Y

C X mod Y = kY mod Y

D (X + kY) mod Y = 0

💡 Short Explanation

�� Adding multiples of modulus does not change remainder. �� Congruent numbers have equal modulo values. �� This is a standard modular property.

✅ Correct Answer Explanation

If kY is added to X, then the remainder upon division by Y remains unchanged. Mathematically: → Therefore, Option B is correct. → Option A is incorrect because the modulo value does not necessarily become larger. → Option C is incorrect because kY mod Y is always 0. → Option D is incorrect because the result is not always 0.

❓ Why Other Options Are Incorrect

�� Option A → Adding multiples of Y does not increase remainder.
�� Option C → kY is divisible by Y, so its modulo is 0.
�� Option D → Only true if X mod Y itself equals 0.

🎯 Strategy

Used: Option Grouping

Application: Identify the standard congruence identity involving multiples of modulus.

Final Logic: Numbers differing by multiples of Y have the same remainder modulo Y.

🧠 Memory Trick

"Add modulus multiples → same remainder."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Congruence Property

7 The 24-hour clock cycle relies on X mod Y = 0. Which of the following time representations equates to 00:00 under modulo 24?

A 12:00

B 24:00

C 06:00

D 18:00

💡 Short Explanation

�� A 24-hour clock repeats after 24 hours. �� 24 mod 24 = 0. �� Hence 24:00 corresponds to 00:00.

✅ Correct Answer Explanation

In modulo 24 arithmetic: → Therefore, 24:00 corresponds to 00:00 on a 24-hour clock. → Hence, Option B is correct. → Option A represents midday. → Option C represents 6 AM. → Option D represents 6 PM.

❓ Why Other Options Are Incorrect

�� Option A → 12:00 is not congruent to 0 modulo 24.
�� Option C → 6 mod 24 equals 6, not 0.
�� Option D → 18 mod 24 equals 18, not 0.

🎯 Strategy

Used: Substitution

Application: Apply modulo 24 directly to each time value.

Final Logic: Only 24 leaves remainder 0 when divided by 24.

🧠 Memory Trick

"24 resets the clock."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Clock Arithmetic Modulo 24

8 What is the sum of (7 mod 13) and (3 mod 3)?

A 10

B 7

C 4

D 0

💡 Short Explanation

�� Compute each modulo separately. �� Add the obtained remainders. �� Exact divisibility gives remainder 0.

✅ Correct Answer Explanation

Since 7 < 13: Also: Therefore: → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect addition of modulo values.
�� Option C → Does not match computed remainder sum.
�� Option D → Ignores the first modulo value.

🎯 Strategy

Used: Substitution

Application: Evaluate each modulo independently and then add results.

Final Logic: 7 mod 13 gives 7 and 3 mod 3 gives 0, so total equals 7.

🧠 Memory Trick

"Exact division gives zero remainder."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Basic Modulo Evaluation

9 Given A = 17, B = 25, and C = 4. What is the value of Z if Z = (A + B) mod C?

A 1

B 2

C 3

D 0

💡 Short Explanation

�� First add the numbers. �� Then apply modulo operation. �� Remainder becomes the answer.

✅ Correct Answer Explanation

Compute: Now: because: → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect remainder after division by 4.
�� Option C → Incorrect modulo calculation.
�� Option D → 42 is not divisible by 4 exactly.

🎯 Strategy

Used: Substitution

Application: Replace variables with numerical values and directly compute modulo.

Final Logic: 42 divided by 4 leaves remainder 2.

🧠 Memory Trick

"Add first, mod later."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Addition Property of Modulo Arithmetic

10 For A = 84, B = 37, and C = 9, calculate Z = (A - B) mod C.

A 2

B 1

C 0

D 3

💡 Short Explanation

�� Subtract first. �� Then divide by modulus. �� Remainder is the answer.

✅ Correct Answer Explanation

Compute: Now: because: → Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect remainder after division by 9.
�� Option C → 47 is not exactly divisible by 9.
�� Option D → Incorrect modulo value.

🎯 Strategy

Used: Substitution

Application: Insert values directly into the expression and evaluate remainder.

Final Logic: 47 leaves remainder 2 when divided by 9.

🧠 Memory Trick

"Subtract → divide → remainder."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Subtraction Property of Modular Arithmetic

11 Given A = 17, B = 25, C = 4, evaluate (A × B) mod C.

A 1

B 2

C 3

D 0

💡 Short Explanation

�� Multiply the integers first. �� Apply modulo operation afterward. �� The remainder becomes the answer.

✅ Correct Answer Explanation

Compute: Now divide by 4: because: → Therefore, Option A is correct. → Option B and Option C are incorrect remainders. → Option D is incorrect because 425 is not divisible by 4 exactly.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect remainder after division by 4.
�� Option C → Does not satisfy modulo evaluation.
�� Option D → Zero remainder occurs only for exact divisibility.

🎯 Strategy

Used: Substitution

Application: Substitute the numerical values and directly evaluate the modulo expression.

Final Logic: Product 425 leaves remainder 1 when divided by 4.

🧠 Memory Trick

"Multiply → divide → remainder."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Multiplication Property of Modular Arithmetic

12 Calculate (14 × 8) mod 5 using the multiplication property of modulo arithmetic.

A 1

B 2

C 3

D 4

💡 Short Explanation

�� Reduce factors modulo 5 first. �� Multiply reduced values. �� Take modulo again.

✅ Correct Answer Explanation

Using multiplication property: Now reduce each factor modulo 5: Then: → Therefore, Option B is correct. → Options A, C, and D do not match the correct modulo result.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect remainder after modulo reduction.
�� Option C → Does not satisfy multiplication modulo evaluation.
�� Option D → Incorrect modulo computation.

🎯 Strategy

Used: Substitution

Application: Replace each number with its modulo equivalent before multiplication.

Final Logic: Reduced multiplication gives remainder 2 modulo 5.

🧠 Memory Trick

"Reduce first, multiply later."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Multiplication Modulo Property

13 Apply addition modulo to positive integers 17 and 13 for modulo 30 (i.e., 17 +_30 13).

A 17

B 30

C 0

D 13

💡 Short Explanation

�� First add the integers. �� Then apply modulo 30. �� Exact divisibility gives remainder 0.

✅ Correct Answer Explanation

Compute: Now apply modulo 30: → Therefore: → Hence, Option C is correct. → Option A and D are original operands, not results. → Option B is the modulus, not the remainder.

❓ Why Other Options Are Incorrect

�� Option A → 17 is only one addend.
�� Option B → Modulo operation returns remainder, not modulus itself.
�� Option D → 13 is the second addend, not the result.

🎯 Strategy

Used: Substitution

Application: Directly add the values and evaluate modulo 30.

Final Logic: 30 divided by 30 leaves remainder 0.

🧠 Memory Trick

"Exact division ⇒ modulo zero."

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

14 Determine a +_m b if a = 3, b = 11, and m = 8.

A 14

B 8

C 6

D 0

💡 Short Explanation

�� Add the integers first. �� Apply modulo 8 afterward. �� Remainder gives the final answer.

✅ Correct Answer Explanation

Compute: Now apply modulo 8: because: → Therefore, Option C is correct. → Option A is the sum before modulo reduction. → Option B is the modulus itself. → Option D is incorrect because 14 is not divisible by 8 exactly.

❓ Why Other Options Are Incorrect

�� Option A → Raw sum without modulo operation.
�� Option B → Modulus is not automatically the answer.
�� Option D → Incorrect remainder evaluation.

🎯 Strategy

Used: Substitution

Application: Replace variables with values and compute modular addition directly.

Final Logic: 14 leaves remainder 6 upon division by 8.

🧠 Memory Trick

"Add first, mod later."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Addition Under Modulo

15 Evaluate a -_m b if a = 17, b = 13, and m = 3.

A 1

B 2

C 0

D 4

💡 Short Explanation

�� Subtract the integers first. �� Apply modulo 3. �� Remainder gives the result.

✅ Correct Answer Explanation

Compute: Now apply modulo 3: because: → Therefore, Option A is correct. → Option B is an incorrect remainder. → Option C is incorrect because 4 is not divisible by 3 exactly. → Option D is the raw subtraction result before modulo reduction.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect modulo value after division by 3.
�� Option C → Zero remainder occurs only under exact divisibility.
�� Option D → Raw difference without modulo operation.

🎯 Strategy

Used: Substitution

Application: Directly compute subtraction and reduce modulo 3.

Final Logic: Difference 4 leaves remainder 1 on division by 3.

🧠 Memory Trick

"Subtract → mod → remainder."

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

16 Calculate -32 mod 7 by expressing it in the form X = Y × Q + R.

A 4

B 3

C 2

D 5

💡 Short Explanation

�� Modulo remainder must be non-negative. �� Express the number in Euclidean division form. �� The remainder becomes the modulo value.

✅ Correct Answer Explanation

Write: where remainder satisfies: Thus: → Therefore, Option B is correct. → Options A, C, and D do not satisfy the Euclidean division form correctly.

❓ Why Other Options Are Incorrect

�� Option A → Does not satisfy the required remainder condition here.
�� Option C → Incorrect modulo computation.
�� Option D → Incorrect remainder after division by 7.

🎯 Strategy

Used: Elimination

Application: Test which remainder satisfies the Euclidean form with remainder between 0 and 6.

Final Logic: Only remainder 3 correctly satisfies the equation.

🧠 Memory Trick

"Modulo remainder is always positive."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Modulo with Negative Integers

17 Find the value of 16 ⊗_7 29.

A 2

B 3

C 4

D 1

💡 Short Explanation

�� Modular multiplication means product modulo 7. �� Multiply first. �� Reduce modulo 7.

✅ Correct Answer Explanation

Compute: Now divide by 7: because: → Therefore, the correct answer is:

❓ Why Other Options Are Incorrect

�� Option B → Incorrect remainder after division by 7.
�� Option C → Does not satisfy modulo multiplication evaluation.
�� Option D → Incorrect modulo result.

🎯 Strategy

Used: Substitution

Application: Multiply the integers directly and compute remainder modulo 7.

Final Logic: 464 leaves remainder 2 when divided by 7.

🧠 Memory Trick

"Multiply, then remainder."

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

18 For a = 5, b = 3, and m = 25, what is a ⊗_m b?

A 1

B 15

C 8

D 0

💡 Short Explanation

�� Multiply the integers. �� Apply modulo 25. �� Since product is less than 25, it remains unchanged.

✅ Correct Answer Explanation

Compute: Now: since 15 is already less than 25. → Therefore, Option B is correct. → Options A, C, and D do not match the modulo result.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect remainder after modulo reduction.
�� Option C → Product modulo 25 is not 8.
�� Option D → Product is not divisible by 25.

🎯 Strategy

Used: Substitution

Application: Multiply directly and check whether modulo reduction changes the result.

Final Logic: Since 15 is less than 25, modulo leaves it unchanged.

🧠 Memory Trick

"Smaller than modulus stays same."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Multiplication Under Modulo

📄 Passage

There are many applications of modulo arithmetic, ranging from everyday life to Math and Science. A practical application is to calculate checksums within serial number identifiers. For example, International Standard Book Number (ISBN) uses modulo 11 (for 10 digit ISBN) or modulo 10 (for 13 digit ISBN) arithmetic for error detection. Likewise, International Bank Account Numbers (IBANs) make use of modulo 97 arithmetic to spot user input errors. On a calendar, arithmetic modulo 7 determines the day of the week.

Question:
Which modulo arithmetic system does the International Standard Book Number use for error detection in a 13-digit ISBN?

A Modulo 10

B Modulo 11

C Modulo 97

D Modulo 12

💡 Short Explanation

�� ISBN systems use modular arithmetic for error detection. �� ISBN-13 uses modulo 10. �� ISBN-10 uses modulo 11.

✅ Correct Answer Explanation

The passage clearly states: • 10-digit ISBN uses modulo 11. • 13-digit ISBN uses modulo 10. → Therefore, Option A is correct. → Option B refers to ISBN-10, not ISBN-13. → Option C is used in IBAN systems. → Option D is unrelated to ISBN error detection.

❓ Why Other Options Are Incorrect

�� Option B → Modulo 11 applies to 10-digit ISBNs.
�� Option C → Modulo 97 is associated with IBAN.
�� Option D → Modulo 12 is not used for ISBN validation.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Extract the directly stated information from the passage.

Final Logic: The passage explicitly links ISBN-13 with modulo 10.

🧠 Memory Trick

"ISBN-13 → mod 10."

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

📄 Passage

Question:
International Bank Account Numbers (IBANs) use which modulo arithmetic to detect user input errors?

A Modulo 10

B Modulo 11

C Modulo 12

D Modulo 97

💡 Short Explanation

�� IBAN validation uses modular arithmetic. �� Modulo 97 helps detect errors. �� Widely used in banking systems.

✅ Correct Answer Explanation

The passage explicitly states that IBANs use modulo 97 arithmetic for detecting user input errors. → Therefore, Option D is correct. → Option A is associated with ISBN-13. → Option B relates to ISBN-10. → Option C has no role in IBAN verification.

❓ Why Other Options Are Incorrect

�� Option A → Modulo 10 applies to ISBN-13 systems.
�� Option B → Modulo 11 is linked with ISBN-10.
�� Option C → Not used for IBAN error detection.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Read the application information directly from the passage.

Final Logic: The passage explicitly mentions modulo 97 for IBAN systems.

🧠 Memory Trick

"IBAN → 97."

📚 Class XI – NCERT Mathematics – Chapter: Number Systems – Topic: Real-Life Applications of Modular Arithmetic

CUET UG Applied Mathematics Booster Test 1 - Number Theory and Modular Fundamentals

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