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