Name: Module 1 MCQ + Assertion Reason Mastery
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Index

80/80 Maths roadmap

138

MCQs

Chapters

Papers

Chapter-wise weighted practice

PYQ-pattern and board-trend MCQs

Chapter quiz with score tracking

Real Numbers

MCQ practice, detailed solution, chapter-end quiz

10 QuestionsHigh

Polynomials

MCQ practice, detailed solution, chapter-end quiz

10 QuestionsHigh

Pair of Linear Equations

MCQ practice, detailed solution, chapter-end quiz

12 QuestionsHigh

Quadratic Equations

MCQ practice, detailed solution, chapter-end quiz

12 QuestionsHigh

Arithmetic Progressions

MCQ practice, detailed solution, chapter-end quiz

10 QuestionsHigh

Triangles

MCQ practice, detailed solution, chapter-end quiz

12 QuestionsHigh

Coordinate Geometry

MCQ practice, detailed solution, chapter-end quiz

10 QuestionsMedium

Introduction to Trigonometry

MCQ practice, detailed solution, chapter-end quiz

12 QuestionsHigh

Applications of Trigonometry

MCQ practice, detailed solution, chapter-end quiz

6 QuestionsMedium

Circles

MCQ practice, detailed solution, chapter-end quiz

8 QuestionsMedium

Areas Related to Circles

MCQ practice, detailed solution, chapter-end quiz

8 QuestionsMedium

Surface Areas and Volumes

MCQ practice, detailed solution, chapter-end quiz

12 QuestionsHigh

Statistics

MCQ practice, detailed solution, chapter-end quiz

8 QuestionsMedium

Probability

MCQ practice, detailed solution, chapter-end quiz

8 QuestionsMedium

Final sample paper zone

3 Section A sample papers, 18 MCQs each, with answer key and score analysis for final exam confidence.

Class 10Complete this module for Section A mastery

80/80 Module

Page 1

Chapter 1 Real Numbers

Real Numbers is a high-scoring chapter because most questions are logic-based, not lengthy. In Section A, questions usually test Euclid's division lemma, HCF-LCM relation, prime factorisation and irrationality.

Before solving, revise these four tools: Euclid's division algorithm, Fundamental Theorem of Arithmetic, HCF x LCM = product of two numbers, and decimal expansion rules.

Chapter target: 10 MCQs in Module 1

Practice style: concept MCQ, calculation MCQ and assertion-reason

Weightage signal: High for basic number theory and decimal expansion

Attempt rule: solve first, then check solution direction

Class 10Amit Shrivastava

80/80 Module

Page 2

Real Numbers MCQ Set 1

Q1. For positive integers $a$ and $b$ , Euclid's division lemma states that $a = b q + r$ . Which condition is correct for $r$ ?

0 < r < b

0 \leq r < b

0 \leq b < r

r = b

Answer: B

Solution: In Euclid's division lemma, the remainder may be zero, but it must always be smaller than the divisor. Therefore,

0 \leq r < b

Q2. The HCF of $96$ and $404$ is:

2

4

8

12

Answer: B

Solution:

404 = 96 \times 4 + 20

96 = 20 \times 4 + 16

20 = 16 \times 1 + 4

16 = 4 \times 4 + 0

. The last non-zero remainder is

4

, so HCF =

4

Q3. If $H C F (26, 91) = 13$ , then $L C M (26, 91)$ is:

182

169

117

91

Answer: A

Solution: For two numbers,

H C F \times L C M = P r o d u c t o f n u m b e r s

. So

L C M = (26 \times 91) / 13 = 182

Accuracy checkpoint: If Q2 is wrong, revise Euclid's division algorithm.

Speed checkpoint: Q3 should take less than 45 seconds.

Class 10Amit Shrivastava

80/80 Module

Page 3

Real Numbers MCQ Set 2

Q4. The prime factorisation of $156$ is:

2 \times 2 \times 3 \times 13

2 \times 3 \times 3 \times 13

2 \times 2 \times 3 \times 7

3 \times 4 \times 13

Answer: A

Solution:

156 = 12 \times 13 = 2 \times 2 \times 3 \times 13

. Option D is not prime factorisation because

4

is not prime.

Q5. If $p$ is a prime number and $p$ divides $a 2$ , then $p$ divides:

a

2 a

p 2

a + p

Answer: A

Solution: This result is used in irrationality proofs. If a prime number divides the square of a number, it must divide the number itself.

Q6. Which of the following is irrational?

0.75

227

2

0.333 . . .

Answer: C

Solution: Terminating decimals and recurring decimals are rational.

227

is also rational because it is in

p q

form.

2

is irrational.

PYQ pattern: Irrationality questions often test $2$ , $3$ , $5$ or their simple combinations.

Common trap: $227$ is rational because it is in $p q$ form.

Class 10Amit Shrivastava

80/80 Module

Page 4

Real Numbers Decimal Expansion

Q7. The decimal expansion of $13125$ is:

ATerminating

BNon-terminating repeating

CNon-terminating non-repeating

Answer: A

Solution:

125 = 53

. The denominator has only the prime factor

5

, so the decimal expansion is terminating.

Q8. The decimal expansion of $712$ is:

ATerminating

BNon-terminating repeating

CNon-terminating non-repeating

DInteger

Answer: B

Solution:

12 = 22 \times 3

. Since the denominator has factor

3

apart from

2

and

5

, the decimal expansion is non-terminating repeating.

Q9. If a rational number has a terminating decimal expansion, its denominator in simplest form must be of the form:

2 m 5 n

3 m 5 n

2 m 7 n

3 m 7 n

Answer: A

Solution: A rational number has terminating decimal expansion only when the denominator has no prime factors other than

2

and

5

Formula memory: Terminating decimal denominator = $2 m \times 5 n$ after simplification.

Mistake alert: Always reduce the fraction to simplest form first.

Class 10Amit Shrivastava

80/80 Module

Page 5

Real Numbers Assertion Reason

Q10. Assertion: The number $6 n$ can never end with digit $0$ for any natural number $n$ . Reason: Any number ending with $0$ must have $5$ as a prime factor.

A. Both Assertion and Reason are true, and Reason is the correct explanation. B. Both Assertion and Reason are true, but Reason is not the correct explanation. C. Assertion is true, Reason is false. D. Assertion is false, Reason is true.

Answer: A

Solution:

6 n = 2 n \times 3 n

, so its prime factors are only

2

and

3

. A number ending in

0

must be divisible by

10 = 2 \times 5

. Since factor

5

is absent,

6 n

cannot end with

0

Q11. Assertion: HCF of two different prime numbers is always $1$ . Reason: Prime numbers have only two factors, $1$ and the number itself.

Answer: A

Solution: Two different prime numbers cannot have any common factor except

1

. The reason correctly explains why the HCF is

1

Assertion-reason rule: Check Assertion independently, then Reason independently, then connection.

Exam habit: Do not select option A unless the reason actually explains the assertion.

Class 10Amit Shrivastava

80/80 Module

Page 6

Chapter-End Quiz And Score

Mini Quiz: Attempt these without seeing the solution. Give yourself 8 minutes.

Q12. $H C F (65, 117)$ is:

9

13

17

26

Q13. LCM of $12$ and $18$ is:

24

36

48

72

Q14. Decimal expansion of $29 / 40$ is:

ATerminating

BRepeating

CIrrational

DInteger

Q15. Which number is irrational?

58

0.121212 . . .

7

3.5

Quiz Answer Key: Q12-B, Q13-B, Q14-A, Q15-C.

Score check: 4/4 means strong start. 3/4 means revise one weak subtopic. 2/4 or below means repeat Pages 1 to 5 before moving to Polynomials.

Next chapter after this: Polynomials

Target before next chapter: zero confusion in HCF-LCM and decimal expansion

Class 10Amit Shrivastava

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Index

80/80 Maths roadmap

138

MCQs

Chapters

Papers

Chapter-wise weighted practice

PYQ-pattern and board-trend MCQs

Chapter quiz with score tracking

Real Numbers

MCQ practice, detailed solution, chapter-end quiz

10 QuestionsHigh

Polynomials

MCQ practice, detailed solution, chapter-end quiz

10 QuestionsHigh

Pair of Linear Equations

MCQ practice, detailed solution, chapter-end quiz

12 QuestionsHigh

Quadratic Equations

MCQ practice, detailed solution, chapter-end quiz

12 QuestionsHigh

Arithmetic Progressions

MCQ practice, detailed solution, chapter-end quiz

10 QuestionsHigh

Triangles

MCQ practice, detailed solution, chapter-end quiz

12 QuestionsHigh

Coordinate Geometry

MCQ practice, detailed solution, chapter-end quiz

10 QuestionsMedium

Introduction to Trigonometry

MCQ practice, detailed solution, chapter-end quiz

12 QuestionsHigh

Applications of Trigonometry

MCQ practice, detailed solution, chapter-end quiz

6 QuestionsMedium

Circles

MCQ practice, detailed solution, chapter-end quiz

8 QuestionsMedium

Areas Related to Circles

MCQ practice, detailed solution, chapter-end quiz

8 QuestionsMedium

Surface Areas and Volumes

MCQ practice, detailed solution, chapter-end quiz

12 QuestionsHigh

Statistics

MCQ practice, detailed solution, chapter-end quiz

8 QuestionsMedium

Probability

MCQ practice, detailed solution, chapter-end quiz

8 QuestionsMedium

Final sample paper zone

3 Section A sample papers, 18 MCQs each, with answer key and score analysis for final exam confidence.

Class 10Complete this module for Section A mastery

80/80 Module

Page 1

Chapter 1 Real Numbers

Before solving, revise these four tools: Euclid's division algorithm, Fundamental Theorem of Arithmetic, HCF x LCM = product of two numbers, and decimal expansion rules.

Chapter target: 10 MCQs in Module 1

Practice style: concept MCQ, calculation MCQ and assertion-reason

Weightage signal: High for basic number theory and decimal expansion

Attempt rule: solve first, then check solution direction

Class 10Amit Shrivastava

80/80 Module

Page 2

Real Numbers MCQ Set 1

Q1. For positive integers $a$ and $b$ , Euclid's division lemma states that $a = b q + r$ . Which condition is correct for $r$ ?

0 < r < b

0 \leq r < b

0 \leq b < r

r = b

Answer: B

Solution: In Euclid's division lemma, the remainder may be zero, but it must always be smaller than the divisor. Therefore,

0 \leq r < b

Q2. The HCF of $96$ and $404$ is:

2

4

8

12

Answer: B

Solution:

404 = 96 \times 4 + 20

96 = 20 \times 4 + 16

20 = 16 \times 1 + 4

16 = 4 \times 4 + 0

. The last non-zero remainder is

4

, so HCF =

4

Q3. If $H C F (26, 91) = 13$ , then $L C M (26, 91)$ is:

182

169

117

91

Answer: A

Solution: For two numbers,

H C F \times L C M = P r o d u c t o f n u m b e r s

. So

L C M = (26 \times 91) / 13 = 182

Accuracy checkpoint: If Q2 is wrong, revise Euclid's division algorithm.

Speed checkpoint: Q3 should take less than 45 seconds.

Class 10Amit Shrivastava

80/80 Module

Page 3

Real Numbers MCQ Set 2

Q4. The prime factorisation of $156$ is:

2 \times 2 \times 3 \times 13

2 \times 3 \times 3 \times 13

2 \times 2 \times 3 \times 7

3 \times 4 \times 13

Answer: A

Solution:

156 = 12 \times 13 = 2 \times 2 \times 3 \times 13

. Option D is not prime factorisation because

4

is not prime.

Q5. If $p$ is a prime number and $p$ divides $a 2$ , then $p$ divides:

a

2 a

p 2

a + p

Answer: A

Solution: This result is used in irrationality proofs. If a prime number divides the square of a number, it must divide the number itself.

Q6. Which of the following is irrational?

0.75

227

2

0.333 . . .

Answer: C

Solution: Terminating decimals and recurring decimals are rational.

227

is also rational because it is in

p q

form.

2

is irrational.

PYQ pattern: Irrationality questions often test $2$ , $3$ , $5$ or their simple combinations.

Common trap: $227$ is rational because it is in $p q$ form.

Class 10Amit Shrivastava

80/80 Module

Page 4

Real Numbers Decimal Expansion

Q7. The decimal expansion of $13125$ is:

ATerminating

BNon-terminating repeating

CNon-terminating non-repeating

Answer: A

Solution:

125 = 53

. The denominator has only the prime factor

5

, so the decimal expansion is terminating.

Q8. The decimal expansion of $712$ is:

ATerminating

BNon-terminating repeating

CNon-terminating non-repeating

DInteger

Answer: B

Solution:

12 = 22 \times 3

. Since the denominator has factor

3

apart from

2

and

5

, the decimal expansion is non-terminating repeating.

Q9. If a rational number has a terminating decimal expansion, its denominator in simplest form must be of the form:

2 m 5 n

3 m 5 n

2 m 7 n

3 m 7 n

Answer: A

Solution: A rational number has terminating decimal expansion only when the denominator has no prime factors other than

2

and

5

Formula memory: Terminating decimal denominator = $2 m \times 5 n$ after simplification.

Mistake alert: Always reduce the fraction to simplest form first.

Class 10Amit Shrivastava

80/80 Module

Page 5

Real Numbers Assertion Reason

Q10. Assertion: The number $6 n$ can never end with digit $0$ for any natural number $n$ . Reason: Any number ending with $0$ must have $5$ as a prime factor.

Answer: A

Solution:

6 n = 2 n \times 3 n

, so its prime factors are only

2

and

3

. A number ending in

0

must be divisible by

10 = 2 \times 5

. Since factor

5

is absent,

6 n

cannot end with

0

Q11. Assertion: HCF of two different prime numbers is always $1$ . Reason: Prime numbers have only two factors, $1$ and the number itself.

Answer: A

Solution: Two different prime numbers cannot have any common factor except

1

. The reason correctly explains why the HCF is

1

Assertion-reason rule: Check Assertion independently, then Reason independently, then connection.

Exam habit: Do not select option A unless the reason actually explains the assertion.

Class 10Amit Shrivastava

80/80 Module

Page 6

Chapter-End Quiz And Score

Mini Quiz: Attempt these without seeing the solution. Give yourself 8 minutes.

Q12. $H C F (65, 117)$ is:

9

13

17

26

Q13. LCM of $12$ and $18$ is:

24

36

48

72

Q14. Decimal expansion of $29 / 40$ is:

ATerminating

BRepeating

CIrrational

DInteger

Q15. Which number is irrational?

58

0.121212 . . .

7

3.5

Quiz Answer Key: Q12-B, Q13-B, Q14-A, Q15-C.

Score check: 4/4 means strong start. 3/4 means revise one weak subtopic. 2/4 or below means repeat Pages 1 to 5 before moving to Polynomials.

Next chapter after this: Polynomials

Target before next chapter: zero confusion in HCF-LCM and decimal expansion

Class 10Amit Shrivastava

Continue reading after purchase

You have completed the free preview. Add this book to cart and purchase to unlock the remaining pages.