ANSWER KEY - class 6 Chapter 5: PRIME TIME –question bank

 ANSWER KEY - class 6 Chapter 5: PRIME TIME  –question bank

Answer Key

Class: 6

Subject: Mathematics (Ganita Prakash)

Chapter 5: Prime Time

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Multiple Choice Questions (1 Mark Each)

1. c) 47

2. b) 15

3. a) 18 and 35

4. c) 2 x 2 x 3 x 7

5. c) The number formed by its last three digits is divisible by 8.

6. a) 12

7. c) 6 (Factors of 6: 1, 2, 3, 6; Sum = 12 = 2×6)

8. c) 97

9. b) Its last digit is 0 or 5.

10. b) 2

11. c) 28 (2+3+5+7+11 = 28)

12. c) Neither prime nor composite

13. c) 8460 (Divisible by 2 and 3)

14. b) 1

15. b) 2³ x 5³

16. c) The sum of its digits is divisible by 9.

17. d) 39 (Factors: 1, 3, 13, 39)

18. b) 1002 (1+0+0+2=3, divisible by 3)

19. b) 2

20. b) Prime numbers

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Assertion (A) and Reason (R) Type Questions (1 Mark Each)

1. a) Both A and R are true, and R is the correct explanation of A.

2. a) Both A and R are true, and R is the correct explanation of A. (HCF of 15 and 32 is 1)

3. d) A is false, but R is true. (2 is an even prime, so not all even numbers are composite)

4. a) Both A and R are true, and R is the correct explanation of A.

5. a) Both A and R are true, and R is the correct explanation of A.

6. d) A is false, but R is true. (91 = 7 × 13, so it is composite)

7. a) Both A and R are true, and R is the correct explanation of A.

8. a) Both A and R are true, and R is the correct explanation of A.

9. a) Both A and R are true, and R is the correct explanation of A.

10. a) Both A and R are true, and R is the correct explanation of A. (1001 ÷ 7 = 143)

11. a) Both A and R are true, and R is the correct explanation of A.

12. a) Both A and R are true, and R is the correct explanation of A.

13. c) A is true, but R is false. (The assertion is false, e.g., 3+5=8 (even) but 7+11=18 (even). The reason is true, but it does not correctly explain the false assertion. The sum of two odd primes is even, but the sum of 2 (even prime) and any other prime (odd) is odd.)

    • *Clarification: The Assertion is actually False. Counterexample: 2 (prime) + 3 (prime) = 5 (odd). The Reason is True. Therefore, the correct answer is d) A is false, but R is true.*

14. a) Both A and R are true, and R is the correct explanation of A. (864 ÷ 8 = 108)

15. b) Both A and R are true, but R is NOT the correct explanation of A. (Its uniqueness is due to its multiplicative identity property, which is the reason given. However, it is also unique because it is the only natural number with one factor.)

16. a) Both A and R are true, and R is the correct explanation of A.

17. a) Both A and R are true, and R is the correct explanation of A.

18. d) A is false, but R is true. (27 is composite, proving that not all numbers ending in 7 are prime)

19. a) Both A and R are true, and R is the correct explanation of A. (Any number can be written as a multiple of 100 plus its last two digits. Since 100 is divisible by 4, only the last two digits matter.)

20. d) A is false, but R is true.

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True or False (1 Mark Each)

1. False (57 ÷ 3 = 19, so it is composite)

2. True

3. False (2 is the smallest prime number)

4. True (Odd + Odd = Even)

5. True

6. False (2 is a prime number and it is even)

7. False (The Fundamental Theorem of Arithmetic states it is unique)

8. True

9. False (The sum of all digits must be divisible by 3)

10. False (e.g., 8 and 9 are co-prime but neither is prime)

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Short Answer Type Questions-I (2 Marks Each)

1. Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36.

2. Factors of 20: 1, 2, 4, 5, 10, 20. Factors of 28: 1, 2, 4, 7, 14, 28. Common factors: 1, 2, 4.

3. 98 = 2 × 7 × 7 or 2 × 7²

4. The smallest number divisible by both is their LCM.
   Multiples of 6: 6, 12, 18, 24, 30...
   Multiples of 8: 8, 16, 24, 32...
   LCM = 24

5. Yes. For divisibility by 4, check the last two digits. The last two digits are 56. Since 56 ÷ 4 = 14, 23456 is divisible by 4.

6. Yes. For divisibility by 3, check the sum of digits: 1+2+3+4+5+6 = 21. Since 21 is divisible by 3 (21÷3=7), 123,456 is divisible by 3.

7. Any number like 25, 75, 125, 175, etc. (Must be a multiple of 25 but not even).

8. Factors of 18: 1, 2, 3, 6, 9, 18. Factors of 35: 1, 5, 7, 35. The only common factor is 1. Therefore, HCF = 1.

9. Smallest: 11. Largest: 97.

10. Yes. Since 9 is a multiple of 3, any number divisible by 9 will automatically be divisible by 3. Example: 18 is divisible by 9 and also by 3.

11. Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24...
    Multiples of 4: 4, 8, 12, 16, 20, 24...
    Common multiples: 12, 24, 36...

12. Factors of 31: 1, 31. Factors of 44: 1, 2, 4, 11, 22, 44. The only common factor is 1. Therefore, 31 and 44 are co-prime.

13. 2³ × 3 × 5 = 8 × 3 × 5 = 120

14. The first three composite numbers are 4, 6, 8. Their sum is 4+6+8=18.

15. No. The HCF must always be a factor of the LCM. Here, 18 is not a factor of 380 (380 ÷ 18 is not a whole number).

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Short Answer Type Questions-II (3 Marks Each)

1. First, find the LCM of 15, 20, and 25.
   15 = 3 × 5
   20 = 2² × 5
   25 = 5²
   LCM = 2² × 3 × 5² = 4 × 3 × 25 = 300
   The required number is LCM + Remainder = 300 + 5 = 305.

2. The minimum distance each should cover is the LCM of their step sizes.
   LCM of 45 cm, 50 cm, and 60 cm.
   45 = 3² × 5
   50 = 2 × 5²
   60 = 2² × 3 × 5
   LCM = 2² × 3² × 5² = 4 × 9 × 25 = 900 cm or 9 m.

3. 1728 ÷ 2 = 864
   864 ÷ 2 = 432
   432 ÷ 2 = 216
   216 ÷ 2 = 108
   108 ÷ 2 = 54
   54 ÷ 2 = 27
   27 ÷ 3 = 9
   9 ÷ 3 = 3
   3 ÷ 3 = 1
   Therefore, 1728 = 2⁶ × 3³

4. A number less than 100 with factors 3 and 5 must be a multiple of 15 (LCM of 3 and 5). Multiples of 15 under 100: 15, 30, 45, 60, 75, 90. From these, find numbers where one digit is one more than the other: 45 (5 is 1 more than 4).

5. The greatest number that will divide 33, 87, and 129 leaving the same remainder is the HCF of the differences between the numbers.
   First, find the differences:
   87 - 33 = 54
   129 - 87 = 42
   129 - 33 = 96
   Now, find the HCF of 54, 42, and 96.
   54 = 2 × 3³
   42 = 2 × 3 × 7
   96 = 2⁵ × 3
   HCF = 2 × 3 = 6

6. The Sieve of Eratosthenes is a method to find prime numbers by iteratively marking multiples of primes starting from 2.
   Prime numbers between 1 and 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

7. The smallest 4-digit number is 1000. We need the LCM of 6, 8, and 9 to find the smallest number divisible by all.
   6 = 2 × 3
   8 = 2³
   9 = 3²
   LCM = 2³ × 3² = 8 × 9 = 72.
   Now, find the smallest 4-digit multiple of 72.
   1000 ÷ 72 ≈ 13.88, so the next whole number is 14.
   72 × 14 = 1008.

8. For a number to end with digit 0, its prime factorisation must include 2 and 5. The prime factorisation of 12ⁿ is (2² × 3)ⁿ = 2²ⁿ × 3ⁿ. This does not contain 5 as a prime factor. Therefore, 12ⁿ can never end with the digit 0 for any natural number n.

9. 144 and 180:
   144 = 12² = (2²×3)² = 2⁴ × 3²
   180 = 18 × 10 = (2×3²)×(2×5) = 2² × 3² × 5
   HCF = product of smallest powers of common primes = 2² × 3² = 4 × 9 = 36
   LCM = product of highest powers of all primes = 2⁴ × 3² × 5 = 16 × 9 × 5 = 720

10. Number: 1 0 9 3 4
    Digits at odd places (from right): 4, 9, 1 → Sum = 4+9+1 = 14
    Digits at even places: 3, 0 → Sum = 3+0 = 3
    Difference = 14 - 3 = 11, which is divisible by 11.
    Therefore, 10934 is divisible by 11.

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Long Answer Type Questions (5 Marks Each)

1. a) Definitions:

   • Prime Number: A number greater than 1 with exactly two distinct factors, 1 and itself. Example: 5.

   • Composite Number: A number greater than 1 with more than two factors. Example: 4.
     b) Primes between 50 and 80: 53, 59, 61, 67, 71, 73, 79.
     c) 0 has an infinite number of divisors (every number is a factor of 0). 1 has only one divisor (itself). Since a prime number must have exactly two distinct positive divisors, neither 0 nor 1 qualifies.

2. a) The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
   b)

   • 1000 = 10³ = (2×5)³ = 2³ × 5³

   • 1728 = 2⁶ × 3³ (from SAQ-II, Q3)
     c)

   • HCF = product of smallest powers of common primes. Common primes: 2. HCF = 2³ = 8 (3 is not common, 5 is not common).

   • LCM = product of highest powers of all primes = 2⁶ × 3³ × 5³ = 64 × 27 × 125 = 216,000

3. a) Rules:

   • 2: Last digit is 0, 2, 4, 6, or 8.

   • 3: Sum of digits is divisible by 3.

   • 4: Number formed by last two digits is divisible by 4.

   • 5: Last digit is 0 or 5.

   • 8: Number formed by last three digits is divisible by 8.

   • 9: Sum of digits is divisible by 9.

   • 10: Last digit is 0.
     b)

   • 12,345: Last digit is 5, so divisible by 5. Sum of digits=1+2+3+4+5=15, divisible by 3. Last digit is not 0, so not divisible by 10.

   • 10,248: Last two digits are 48. 48÷4=12, so divisible by 4. Last three digits are 248. 248÷8=31, so divisible by 8.

4. a) Co-prime numbers are numbers whose Highest Common Factor (HCF) is 1. Example: 8 and 9 (HCF=1).
   b)

   • 81 = 3⁴

   • 16 = 2⁴

   • There are no common prime factors. Therefore, HCF=1. So, yes, 81 and 16 are co-prime.
     c) The prime factorisation of 217 is 7 × 31. Since the numbers are co-prime, they must be 7 and 31. (7 × 31 = 217).

5. a) We need the greatest number that divides (70-5)=65 and (125-8)=117 exactly. This is the HCF of 65 and 117.
   65 = 5 × 13
   117 = 3² × 13
   HCF = 13. The greatest number is 13.
   b) The two bells will ring together again after the LCM of 30 and 40 minutes.
   30 = 2 × 3 × 5
   40 = 2³ × 5
   LCM = 2³ × 3 × 5 = 8 × 3 × 5 = 120 minutes or 2 hours.
   Therefore, they will ring together again at 9:00 + 2:00 = 11:00 a.m..

6. a) Twin primes are pairs of prime numbers that differ by 2. Pairs under 50: (3,5), (5,7), (11,13), (17,19), (29,31), (41,43).
   b) A perfect number is a number equal to the sum of its proper divisors (factors excluding itself). Factors of 28: 1, 2, 4, 7, 14, 28. Sum of proper divisors: 1+2+4+7+14 = 28. Therefore, 28 is a perfect number.
   c) The smallest perfect number is 6 (1+2+3=6).

7. a) The required number will be 5 less than a common multiple of 28, 36, and 45. (Number + 5 is divisible by 28, 36, 45). Find the LCM of 28, 36, 45.
   28 = 2² × 7
   36 = 2² × 3²
   45 = 3² × 5
   LCM = 2² × 3² × 5 × 7 = 4 × 9 × 5 × 7 = 1260
   The required number is LCM - 5 = 1260 - 5 = 1255.
   b) The maximum capacity of the container is the HCF of 403, 434, and 465.
   403 = 13 × 31
   434 = 2 × 7 × 31
   465 = 3 × 5 × 31
   HCF = 31. The maximum capacity is 31 litres.

8. a) 7×11×13 + 13 = 13 × (7×11 + 1) = 13 × (77 + 1) = 13 × 78. Since it has factors other than 1 and itself, it is a composite number.
   b) For 15ⁿ to end with a zero, it must be divisible by 10, meaning its prime factors must include 2 and 5. 15ⁿ = (3×5)ⁿ = 3ⁿ × 5ⁿ. This does not contain the prime factor 2. Therefore, 15ⁿ can never end with the digit zero.
   c) For 37a5 to be divisible by 3, the sum of its digits must be divisible by 3. Sum = 3 + 7 + a + 5 = 15 + a.
   For 15+a to be divisible by 3, 'a' can be 0, 3, 6, or 9. (15, 18, 21, 24 are all divisible by 3).

9. a) & b) Factor Tree and Prime Factorisation:
   One possible factor tree:
   240
   /
   24 10
   / \ /
   4 6 2 5
   / \ /
   2 2 2 3
   Prime Factorisation: 240 = 2 × 2 × 2 × 2 × 3 × 5 = 2⁴ × 3¹ × 5¹
   c) Number of factors = (4+1) × (1+1) × (1+1) = 5 × 2 × 2 = 20 factors.

10. a) The greatest possible volume of the tin is the HCF of 120, 180, and 240.
    120 = 2³ × 3 × 5
    180 = 2² × 3² × 5
    240 = 2⁴ × 3 × 5
    HCF = 2² × 3 × 5 = 4 × 3 × 5 = 60 litres.
    b) The traffic lights will change simultaneously after the LCM of 48, 72, and 108 seconds.
    48 = 2⁴ × 3
    72 = 2³ × 3²
    108 = 2² × 3³
    LCM = 2⁴ × 3³ = 16 × 27 = 432 seconds.
    432 seconds = 432 ÷ 60 = 7 minutes 12 seconds.
    They will change together again at 7:07:12 a.m..

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Case-Based Questions (4 MCQs each)

Case Study 1: The Idli-Vada Game

1. c) 15 (LCM of 3 and 5)

2. b) 10 (Multiples of 3 up to 30: 3,6,9,12,15,18,21,24,27,30 → 10 times. This includes 'Idli-Vada' numbers.)

3. b) 4 (Multiples of 5 up to 30: 5,10,15,20,25,30. Subtract the 'Idli-Vada' numbers (15,30). So, 6 - 2 = 4 times.)

4. a) 12 (LCM of 4 and 6)

Case Study 2: Jumpy's Treasure Hunt

1. a) 3 (3 is a common factor of 18 and 24)

2. c) 6 (HCF of 18 and 24 is 6)

3. b) 15 and 28 (15 and 28 are co-prime. Their HCF is 1, so no jump size >1 will land on both.)

4. b) multiple

Case Study 3: Packing Figs

1. b) 3 (Factor pairs of 12: 1×12, 2×6, 3×4)

2. b) 1 (7 is prime, its only factor pair is 1×7)

3. a) Prime

4. c) 36 (36 has the most factors: 1,2,3,4,6,9,12,18,36 → 9 factors)

Case Study 4: The Sieve of Eratosthenes

1. b) 2

2. a) 3 (The next uncrossed number after 2)

3. b) Composite numbers

4. b) 15 (Primes: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47)

Case Study 5: Leap Years

1. c) 2000 (2000÷400=5, so it is a leap year. 1900÷400=4.75, not whole, so it was not.)

2. d) 2028 (2024 + 4 = 2028)

3. b) 24 (Number of multiples of 4 from 2004 to 2096: (2096-2004)/4 + 1 = 92/4 + 1 = 23 + 1 = 24. Remember, 2100 is not included in this range and is not a leap year.)

4. d) 400

class 6 Chapter 5: PRIME TIME –question bank

 

class 6 Chapter 5: PRIME TIME  –question bank

Class: 6

Subject: Mathematics (Ganita Prakash)

Chapter 5: Prime Time

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Multiple Choice Questions (1 Mark Each)

1. Which of the following is a prime number?
a) 21
b) 33
c) 47
d) 51
(Competency: Identifying prime numbers)

2. The first number for which a player says 'idli-vada' in the game is:
a) 10
b) 15
c) 20
d) 30
(Competency: Finding common multiples)

3. Which pair of numbers is co-prime?
a) 18 and 35
b) 15 and 39
c) 30 and 415
d) 81 and 18
(Competency: Identifying co-prime numbers)

4. The prime factorisation of 84 is:
a) 2 x 2 x 21
b) 2 x 3 x 14
c) 2 x 2 x 3 x 7
d) 3 x 4 x 7
(Competency: Prime factorisation)

5. A number is divisible by 8 if:
a) Its last digit is even.
b) The number formed by its last two digits is divisible by 4.
c) The number formed by its last three digits is divisible by 8.
d) It is divisible by both 2 and 4.
(Competency: Applying divisibility rules)

6. The smallest number that is a multiple of both 4 and 6 is:
a) 12
b) 24
c) 36
d) 48
(Competency: Finding LCM)

7. Which of the following is a perfect number?
a) 10
b) 12
c) 6
d) 8
(Competency: Understanding properties of numbers)

8. The largest 2-digit prime number is:
a) 91
b) 93
c) 97
d) 99
(Competency: Identifying prime numbers)

9. A number is divisible by 5 if:
a) It is an odd number.
b) Its last digit is 0 or 5.
c) The sum of its digits is 5.
d) It ends with 0.
(Competency: Applying divisibility rules)

10. Twin primes are pairs of primes that differ by:
a) 1
b) 2
c) 3
d) 4
(Competency: Understanding properties of primes)

11. The sum of the first five prime numbers is:
a) 18
b) 26
c) 28
d) 30
(Competency: Calculation and properties of primes)

12. The number 1 is:
a) A prime number
b) A composite number
c) Neither prime nor composite
d) An even number
(Competency: Classifying numbers)

13. Which of the following numbers is divisible by 6?
a) 2341
b) 5732
c) 8460
d) 2953
(Competency: Applying divisibility rules)

14. The HCF of two co-prime numbers is always:
a) 0
b) 1
c) The smaller number
d) The larger number
(Competency: Relating HCF and co-prime numbers)

15. The prime factorisation of 1000 is:
a) 10³
b) 2³ x 5³
c) 2² x 5³
d) 2³ x 5²
(Competency: Prime factorisation)

16. A number is divisible by 9 if:
a) It is divisible by 3.
b) Its last digit is 9.
c) The sum of its digits is divisible by 9.
d) It is an odd number.
(Competency: Applying divisibility rules)

17. Which of the following is a composite number?
a) 17
b) 23
c) 31
d) 39
(Competency: Identifying composite numbers)

18. The smallest 4-digit number divisible by 3 is:
a) 1000
b) 1002
c) 1005
d) 1010
(Competency: Applying divisibility rules)

19. The number of factors of a prime number is:
a) 1
b) 2
c) 3
d) Infinite
(Competency: Understanding factors of primes)

20. The Sieve of Eratosthenes is used to find:
a) Multiples of a number
b) Prime numbers
c) Composite numbers
d) Common factors
(Competency: Understanding mathematical methods)

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Assertion (A) and Reason (R) Type Questions (1 Mark Each)

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
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.

1. Assertion (A): 1 is neither a prime nor a composite number.
Reason (R): 1 has only one factor.
(Competency: Classifying numbers)

2. Assertion (A): 15 and 32 are co-prime numbers.
Reason (R): Two numbers are co-prime if their HCF is 1.
(Competency: Understanding co-prime numbers)

3. Assertion (A): All even numbers are composite numbers.
Reason (R): 2 is a prime number.
(Competency: Properties of even numbers and primes)

4. Assertion (A): The number 123456 is divisible by 3.
Reason (R): A number is divisible by 3 if the sum of its digits is divisible by 3. The sum of the digits of 123456 is 21, which is divisible by 3.
(Competency: Applying divisibility rules)

5. Assertion (A): The prime factorisation of a number is unique.
Reason (R): The order of factors does not matter in multiplication.
(Competency: Fundamental Theorem of Arithmetic)

6. Assertion (A): 91 is a prime number.
Reason (R): It has no factors other than 1 and itself.
(Competency: Identifying prime numbers)

7. Assertion (A): If a number is divisible by both 2 and 3, it is divisible by 6.
Reason (R): 6 is the LCM of 2 and 3.
(Competency: Applying divisibility rules)

8. Assertion (A): The numbers 0 and 1 are not considered prime numbers.
Reason (R): A prime number must have exactly two distinct factors.
(Competency: Definition of prime numbers)

9. Assertion (A): The HCF of two consecutive numbers is always 1.
Reason (R): Two consecutive numbers are always co-prime.
(Competency: Properties of consecutive numbers)

10. Assertion (A): The number 1001 is divisible by 7.
Reason (R): The difference between the number formed by the last three digits and the remaining part (1 - 1 = 0) is divisible by 7.
(Competency: Applying divisibility rules)

11. Assertion (A): Every composite number can be expressed as a product of primes.
Reason (R): This is known as the Fundamental Theorem of Arithmetic.
(Competency: Fundamental Theorem of Arithmetic)

12. Assertion (A): 45, 60, 75, and 90 are all multiples of 15.
Reason (R): 15 is a common factor of all these numbers.
(Competency: Understanding multiples and factors)

13. Assertion (A): The sum of two prime numbers is always even.
Reason (R): 2 is the only even prime number.
(Competency: Properties of prime numbers)

14. Assertion (A): 864 is divisible by 8.
Reason (R): The number formed by its last three digits, 864, is divisible by 8.
(Competency: Applying divisibility rules)

15. Assertion (A): The number 1 is a unique number.
Reason (R): It is the multiplicative identity.
(Competency: Properties of 1)

16. Assertion (A): The LCM of two co-prime numbers is their product.
Reason (R): Co-prime numbers have no common prime factors.
(Competency: Relating LCM and co-prime numbers)

17. Assertion (A): 0 is a multiple of every number.
Reason (R): 0 divided by any number gives 0.
(Competency: Understanding multiples)

18. Assertion (A): All numbers ending with 7 are prime numbers.
Reason (R): 27 is a composite number.
(Competency: Identifying prime and composite numbers)

19. Assertion (A): The divisibility rule for 4 involves the last two digits.
Reason (R): 100 is divisible by 4.
(Competency: Understanding divisibility rules)

20. Assertion (A): The number of prime numbers is finite.
Reason (R): Euclid proved that there are infinitely many primes.
(Competency: Knowledge about primes)

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True or False (1 Mark Each)

1. The number 57 is prime. (True/False)
(Competency: Identifying prime numbers)

2. Every multiple of a number is greater than or equal to that number. (True/False)
(Competency: Understanding multiples)

3. 1 is the smallest prime number. (True/False)
(Competency: Identifying prime numbers)

4. The sum of two odd numbers is always even. (True/False)
(Competency: Properties of odd and even numbers)

5. If a number is divisible by 10, it must be divisible by 5. (True/False)
(Competency: Relating divisibility rules)

6. All prime numbers are odd. (True/False)
(Competency: Properties of prime numbers)

7. Two different numbers can have the same prime factorisation. (True/False)
(Competency: Fundamental Theorem of Arithmetic)

8. The HCF of two numbers is always a factor of their LCM. (True/False)
(Competency: Relationship between HCF and LCM)

9. A number is divisible by 3 if its last digit is divisible by 3. (True/False)
(Competency: Applying divisibility rules)

10. Co-prime numbers must be prime numbers. (True/False)
(Competency: Understanding co-prime numbers)

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Short Answer Type Questions-I (2 Marks Each)

1. List all the factors of 36.
(Competency: Finding factors)

2. Find the common factors of 20 and 28.
(Competency: Finding common factors)

3. Write the prime factorisation of 98.
(Competency: Prime factorisation)

4. Find the smallest number that is divisible by both 6 and 8.
(Competency: Finding LCM)

5. Check whether 23456 is divisible by 4. Give a reason for your answer.
(Competency: Applying divisibility rules)

6. Is 1,23,456 divisible by 3? Justify your answer.
(Competency: Applying divisibility rules)

7. Find any two numbers which are multiples of 25 but not multiples of 50.
(Competency: Understanding multiples)

8. Find the HCF of 18 and 35.
(Competency: Finding HCF)

9. Write the smallest and largest 2-digit prime numbers.
(Competency: Identifying prime numbers)

10. If a number is divisible by 9, will it always be divisible by 3? Justify with an example.
(Competency: Relating divisibility rules)

11. Find the first three common multiples of 3 and 4.
(Competency: Finding common multiples)

12. State whether 31 and 44 are co-prime. Give reasons.
(Competency: Identifying co-prime numbers)

13. Find the missing number: The prime factorisation of ______ is 2³ × 3 × 5.
(Competency: Prime factorisation)

14. Find the sum of the first three composite numbers.
(Competency: Identifying composite numbers)

15. Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons.
(Competency: Relationship between HCF and LCM)

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Short Answer Type Questions-II (3 Marks Each)

1. Find the smallest number which when divided by 15, 20, and 25 leaves a remainder of 5 in each case.
(Competency: Finding LCM in context)

2. Three boys step off together from the same spot. Their steps measure 45 cm, 50 cm, and 60 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?
(Competency: Real-life application of LCM)

3. Find the prime factorisation of 1728.
(Competency: Prime factorisation)

4. I am a number less than 100. Two of my factors are 3 and 5. One of my digits is 1 more than the other. Who am I? (Find all possible numbers)
(Competency: Problem-solving based on factors)

5. Find the greatest number that will divide 33, 87, and 129 leaving the same remainder in each case.
(Competency: Finding HCF in context)

6. Explain the Sieve of Eratosthenes. Use it to find all prime numbers between 1 and 30.
(Competency: Understanding and applying the Sieve)

7. Find the smallest 4-digit number which is divisible by 6, 8, and 9.
(Competency: Finding LCM)

8. Check whether 12ⁿ can end with the digit 0 for any natural number n. Justify your answer.
(Competency: Reasoning with prime factorisation)

9. Find the HCF and LCM of 144 and 180 by prime factorisation method.
(Competency: Finding HCF and LCM using prime factors)

10. A number is divisible by 11 if the difference between the sum of the digits at odd places and the sum of the digits at even places is either 0 or divisible by 11. Check the divisibility of 10934 by 11.
(Competency: Applying divisibility rule for 11)

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Long Answer Type Questions (5 Marks Each)

1. a) Define prime and composite numbers with examples.
b) Find all prime numbers between 50 and 80 using the Sieve of Eratosthenes.
c) Explain why 0 and 1 are not considered prime numbers.
(Competency: Understanding, applying, and explaining concepts)

2. a) State the Fundamental Theorem of Arithmetic.
b) Find the prime factorisation of 1000 and 1728.
c) Using prime factorisation, find the HCF and LCM of 1000 and 1728.
(Competency: Stating theorem and application)

3. a) Explain the rules for divisibility by 2, 3, 4, 5, 8, 9, and 10 with examples.
b) Using these rules, check the divisibility of 12,345 by 3, 5, and 10. Check the divisibility of 10,248 by 4 and 8.
(Competency: Understanding and applying multiple divisibility rules)

4. a) What are co-prime numbers? Give examples.
b) Check whether 81 and 16 are co-prime.
c) Two numbers are co-prime and their product is 217. Find the numbers.
(Competency: Understanding and problem-solving with co-prime numbers)

5. a) Find the greatest number that divides 70 and 125 leaving remainders 5 and 8 respectively.
b) In a school, the duration of a period in junior section is 30 minutes and in senior section is 40 minutes. If the first bell for both sections rings at 9:00 AM, when will the two bells ring together again?
(Competency: Application of HCF and LCM in problems)

6. a) Define twin primes. List all twin prime pairs under 50.
b) Define perfect numbers. Verify that 28 is a perfect number.
c) Find the smallest perfect number.
(Competency: Understanding special types of numbers)

7. a) Find the smallest number which when increased by 17 is exactly divisible by 28, 36, and 45.
b) Three tankers contain 403 litres, 434 litres, and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three tankers an exact number of times.
(Competency: Complex application of LCM and HCF)

8. a) Is 7 x 11 x 13 + 13 a composite number? Justify your answer.
b) Check whether 15ⁿ can end with the digit zero for any natural number n.
c) Find the digit ‘a’ if 37a5 is divisible by 3.
(Competency: Reasoning and justification using number theory)

9. a) Draw a factor tree for 240.
b) Write its prime factorisation.
c) Using this, find the number of factors of 240. (Hint: If N = a^p x b^q x c^r, number of factors = (p+1)(q+1)(r+1))
(Competency: Factor tree, prime factorisation, and finding number of factors)

10. a) A merchant has 120 litres, 180 litres, and 240 litres of three kinds of oil. He wants to sell the oil by filling the three kinds of oil in tins of equal volume. What is the greatest possible volume of such a tin?
b) Traffic lights at three different road crossings change after every 48 seconds, 72 seconds, and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?
(Competency: Real-life application of HCF and LCM)

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Case-Based Questions (4 MCQs each)

Case Study 1: The Idli-Vada Game
In a game, children sit in a circle. They call out numbers from 1. For multiples of 3, they say 'Idli'. For multiples of 5, they say 'Vada'. For numbers which are multiples of both 3 and 5, they say 'Idli-Vada'.

1. On which number will a child first say 'Idli-Vada'?
a) 5
b) 10
c) 15
d) 30
(Competency: Finding common multiples)

2. Up to 30, how many times will 'Idli' be said?
a) 9
b) 10
c) 11
d) 12
(Competency: Counting multiples)

3. Up to 30, how many times will only 'Vada' be said (and not 'Idli-Vada')?
a) 3
b) 4
c) 5
d) 6
(Competency: Analyzing multiples)

4. If the game is played with rules for 4 ('Idli') and 6 ('Vada'), what is the first number for which a child will say 'Idli-Vada'?
a) 12
b) 18
c) 24
d) 36
(Competency: Finding LCM)

Case Study 2: Jumpy's Treasure Hunt
Jumpy can only jump in steps of equal size, starting from 0. He wins a treasure if he lands exactly on the number where it is placed. Grumpy places treasures on two numbers: 18 and 24.

1. Which of the following jump sizes will land Jumpy on both 18 and 24?
a) 3
b) 4
c) 5
d) 7
(Competency: Finding common factors)

2. What is the largest jump size that can land him on both treasures?
a) 2
b) 3
c) 6
d) 8
(Competency: Finding HCF)

3. If Grumpy wants to make it impossible for Jumpy to win both treasures with any jump size greater than 1, which pair of numbers should he choose?
a) 12 and 18
b) 15 and 28
c) 16 and 24
d) 20 and 30
(Competency: Identifying co-prime numbers)

4. If a jump size lands Jumpy on number A, then A must be a ______ of the jump size.
a) factor
b) multiple
c) divisor
d) prime
(Competency: Understanding multiples and factors)

Case Study 3: Packing Figs
Guna and Anshu are packing figs. Guna wants to pack 12 figs per box, and Anshu wants to pack 7 figs per box. They are exploring the different rectangular arrangements possible for each number.

1. How many different rectangular arrangements can Guna make for 12 figs?
a) 2
b) 3
c) 4
d) 6
(Competency: Finding number of factor pairs)

2. How many different rectangular arrangements can Anshu make for 7 figs?
a) 0
b) 1
c) 2
d) 7
(Competency: Properties of prime numbers)

3. Based on this, 7 is a ______ number.
a) Prime
b) Composite
c) Co-prime
d) Even
(Competency: Classifying numbers)

4. Which of the following numbers will have the most number of rectangular arrangements?
a) 17
b) 23
c) 36
d) 41
(Competency: Relating to number of factors)

Case Study 4: The Sieve of Eratosthenes
The Sieve of Eratosthenes is an ancient algorithm used to find all prime numbers up to any given limit. It works by iteratively marking the multiples of each prime number starting from 2.

1. The first step in the Sieve is to circle the number __ and cross out all its multiples.
a) 1
b) 2
c) 3
d) 4
(Competency: Understanding the Sieve method)

2. After circling 2 and crossing out its multiples, the next number to be circled is:
a) 3
b) 4
c) 5
d) 6
(Competency: Applying the Sieve method)

3. The numbers crossed out as multiples of 2 and 3 (like 6, 12, 18) are:
a) Prime numbers
b) Composite numbers
c) Co-prime numbers
d) Odd numbers
(Competency: Identifying composite numbers)

4. How many prime numbers are there between 1 and 50?
a) 14
b) 15
c) 16
d) 17
(Competency: Applying the Sieve and counting)

Case Study 5: Leap Years
A leap year has 366 days, with February having 29 days. The rule for leap years is: If a year is divisible by 4, it is a leap year, except for end-of-century years, which must be divisible by 400 to be leap years.

1. Which of the following was a leap year?
a) 1900
b) 1978
c) 2000
d) 2022
(Competency: Applying divisibility rules in context)

2. The year 2024 is a leap year. What will be the next leap year?
a) 2025
b) 2026
c) 2027
d) 2028
(Competency: Finding multiples)

3. How many leap years are there between the years 2001 and 2100?
a) 23
b) 24
c) 25
d) 26
(Competency: Problem-solving with divisibility)

4. A century year (like 2100, 2200) will be a leap year only if it is divisible by:
a) 4
b) 40
c) 100
d) 400
(Competency: Remembering the exception rule)