CBSE Class 6 Mathematics Prime Time Practice Paper Assignment Set 02 with Answers

This is CBSE Class 6 Maths Ganita Prakash Chapter 5 Prime Time Practice Paper 13, a 40-mark test of 20 questions arranged in Sections A to E, given here with step-by-step answers. It checks prime factorisation of numbers such as 84, 168 and 840, divisibility tests for 6, 8, 9 and 11, co-prime numbers, listing primes between two numbers, and real-life problems on common factors (library shelves) and common multiples (traffic lights).

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Find the prime factorisation of 168.
Answer:
Keep dividing by the smallest prime that divides the number:
168 ÷ 2 = 84, 84 ÷ 2 = 42, 42 ÷ 2 = 21, 21 ÷ 3 = 7, and 7 is prime.
So 168 = 2 × 2 × 2 × 3 × 7 = 2³ × 3 × 7.
In simple words: Halve 168 three times to reach 21, then split 21 into 3 × 7.

Teacher's Note:
1. Check: 8 × 3 × 7 = 168.
2. Stop only when every factor is prime.

Question 12. Using the divisibility test for 8, determine if 3456 is divisible by 8. Explain your answer.
Answer:
Test for 8: a number is divisible by 8 if the number formed by its last three digits is divisible by 8.
The last three digits of 3456 form 456.
456 ÷ 8 = 57 exactly (8 × 57 = 456).
So 3456 is divisible by 8.
In simple words: Only the last three digits matter for 8, and 456 = 8 × 57.

Teacher's Note:
1. Why it works: 1000 = 8 × 125, so thousands are always divisible by 8.
2. 3456 ÷ 8 = 432 confirms it.

Question 13. Find the prime factorisation of 840 using the factor tree method.
Answer:
Split 840 step by step, always taking out a prime factor:
840 = 2 × 420, 420 = 2 × 210, 210 = 2 × 105, 105 = 3 × 35, 35 = 5 × 7.

The green circles (the leaves of the tree) are the prime factors.
So 840 = 2 × 2 × 2 × 3 × 5 × 7 = 2³ × 3 × 5 × 7.
In simple words: Keep breaking 840 into two factors until every branch ends in a prime.

Teacher's Note:
1. You may start with 840 = 84 × 10; the final primes are the same.
2. Check: 8 × 3 × 5 × 7 = 840.

Question 14. Check if the number 840 is divisible by 8 and 11. Explain your reasoning.
Answer:
Divisibility by 8: 840 has only three digits, so we test 840 itself: 840 ÷ 8 = 105 exactly. So 840 is divisible by 8.
Divisibility by 11: find the alternating sum of digits from the right: 0 − 4 + 8 = 4.
4 is neither 0 nor a multiple of 11, so 840 is not divisible by 11.
Hence 840 is divisible by 8 but not by 11.
In simple words: 840 = 8 × 105, but its alternating digit sum is 4, so 11 does not divide it.

Teacher's Note:
1. For 11, add and subtract the digits alternately.
2. Check: 11 × 76 = 836, which leaves remainder 4.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Find all the prime numbers between 80 and 110.
Answer:
Strike out even numbers and multiples of 3, 5 and 7 from 81 to 109 (since 11 × 11 = 121 > 110, these are enough).
Removed examples: 81 = 9 × 9, 87 = 3 × 29, 91 = 7 × 13, 93 = 3 × 31, 99 = 9 × 11, 105 = 3 × 35.
The numbers left are 83, 89, 97, 101, 103, 107 and 109.
So the primes between 80 and 110 are 83, 89, 97, 101, 103, 107 and 109 (7 primes).
In simple words: Remove every number with a small factor; the seven survivors are prime.

Teacher's Note:
1. 91 = 7 × 13 looks prime but is not.
2. Only primes up to 10 need checking, because 11 × 11 = 121.

Question 16. What is the smallest 4-digit number that is divisible by both 4 and 10? Explain your method.
Answer:
A number divisible by both 4 and 10 must be a multiple of their smallest common multiple.
Multiples of 10: 10, 20, 30, …; the first one divisible by 4 is 20. So we need multiples of 20.
The smallest 4-digit number is 1000, and 1000 ÷ 20 = 50 exactly.
So the answer is 1000.
In simple words: 1000 ends in 0 (so 10 divides it) and its last two digits 00 are divisible by 4.

Teacher's Note:
1. Always check the smallest 4-digit number itself first.
2. 4 × 10 = 40 is a common multiple, but 20 is the smallest.

Question 17. A number when divided by 4 leaves a remainder of 3. What is the largest such 3-digit number that is a multiple of 5?
Answer:
A multiple of 5 ends in 0 or 5.
If it ends in 0, it is even; but an even number divided by 4 leaves remainder 0 or 2, never 3. So the number ends in 5.
Try 3-digit numbers ending in 5 from the top: 995 = 4 × 248 + 3, remainder 3.
So the largest such number is 995.
In simple words: The number must end in 5, and 995 already leaves remainder 3 when divided by 4.

Teacher's Note:
1. Start from the largest candidate and work down.
2. Check: 4 × 248 = 992 and 992 + 3 = 995.

Section D: Long Answer Question (5 Marks)

Question 18. Find the smallest 5-digit number that is a multiple of 15, 20, and 25.
Answer:
Step 1: Prime factorise: 15 = 3 × 5, 20 = 2² × 5, 25 = 5².
Step 2: The smallest common multiple uses the largest power of each prime: 2² × 3 × 5² = 4 × 3 × 25 = 300.
Step 3: Every common multiple of 15, 20 and 25 is a multiple of 300.
Step 4: The smallest 5-digit number is 10000. 10000 ÷ 300 gives quotient 33 and remainder 100, so 10000 is not a multiple.
Step 5: The next multiple is 300 × 34 = 10200 (or 10000 + (300 − 100) = 10200).
So the smallest 5-digit number that is a multiple of 15, 20 and 25 is 10200.
In simple words: Find the smallest common multiple (300), then the first multiple of 300 that has five digits.

Teacher's Note:
1. Check: 10200 ÷ 15 = 680, ÷ 20 = 510, ÷ 25 = 408.
2. 300 × 33 = 9900 has only four digits.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Case Study 1: The School Library
A librarian, Mrs. Sharma, is organizing a new collection of books. She has 210 adventure books and 350 mystery books. She wants to arrange the books on shelves, with each shelf containing the same number of books of a single genre. To use the minimum number of shelves and organize the books most efficiently, she must find the greatest possible number of books that can be placed on each shelf. The solution requires finding the greatest common factor of the two book quantities.
(a) Find the prime factorisation of 210.
(b) Find the prime factorisation of 350.
(c) Determine the largest number that is a factor of both 210 and 350.
(d) Use the divisibility rule for 10 to show that the largest common number of books per shelf (the answer to part c) is divisible by 10.
Answer:
(a) 210 = 2 × 105 = 2 × 3 × 35 = 2 × 3 × 5 × 7.
(b) 350 = 2 × 175 = 2 × 5 × 35 = 2 × 5 × 5 × 7 = 2 × 5² × 7.
(c) The primes common to both are one 2, one 5 and one 7. Greatest common factor = 2 × 5 × 7 = 70.
(d) A number is divisible by 10 if its last digit is 0. The last digit of 70 is 0, so 70 is divisible by 10.
So Mrs. Sharma can keep 70 books on each shelf: 3 shelves of adventure books and 5 shelves of mystery books.
In simple words: Take the primes that 210 and 350 share (2, 5, 7) and multiply them to get 70.

Teacher's Note:
1. Only primes common to both numbers go into the common factor.
2. Check: 210 ÷ 70 = 3 and 350 ÷ 70 = 5.

Question 20. Case Study 2: The Traffic Lights
Three traffic lights at different intersections turn green at different intervals. The first light turns green every 40 seconds, the second every 60 seconds, and the third every 75 seconds. They all just turned green at the same time. A driver wonders when all three lights will turn green simultaneously again. To find this, the driver must calculate the least common multiple of their respective timing intervals. This calculation will reveal the total number of seconds that must pass for the event to repeat.
(a) Find the prime factorisation of 40.
(b) Find the prime factorisation of 60.
(c) Find the smallest number that is a multiple of 40, 60, and 75.
(d) Based on your answer from (c), after how many minutes will all three lights be green again at the same time?
Answer:
(a) 40 = 2 × 2 × 2 × 5 = 2³ × 5.
(b) 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5.
(c) Also 75 = 3 × 5². Take the largest power of each prime: 2³ × 3 × 5² = 8 × 3 × 25 = 600.
(d) 600 seconds ÷ 60 = 10 minutes. All three lights will be green together again after 10 minutes.
In simple words: The lights meet again after the smallest common multiple of 40, 60 and 75, which is 600 seconds or 10 minutes.

Teacher's Note:
1. Check: 600 ÷ 40 = 15, 600 ÷ 60 = 10, 600 ÷ 75 = 8.
2. Remember to change seconds into minutes in part (d).