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

Practice Paper 14 for CBSE Class 6 Mathematics (Ganita Prakash) Chapter 5 Prime Time is a full 40-mark paper of 20 questions in Sections A to E, solved here step by step. Its questions cover prime factorisation of 108, 490, 999, 288 and 432, the divisibility rules for 3, 4, 5, 8 and 9, remainders on division, finding common multiples within a range, and two case studies on sharing students equally among buses and on alarms that ring together.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Find the smallest number that is divisible by 9, 15, and 24.
Answer:
Prime factorise: 9 = 3², 15 = 3 × 5, 24 = 2³ × 3.
Take the largest power of each prime: 2³, 3² and 5.
Smallest number = 2³ × 3² × 5 = 8 × 9 × 5 = 360.
In simple words: Collect each prime the greatest number of times it appears in any one number: 8 × 9 × 5 = 360.

Teacher's Note:
1. Check: 360 ÷ 9 = 40, ÷ 15 = 24, ÷ 24 = 15.
2. Do not just multiply 9 × 15 × 24; that is a common multiple but not the smallest.

Question 12. Using the divisibility test for 4, determine if 984 is divisible by 4. Explain your answer.
Answer:
Test for 4: a number is divisible by 4 if the number formed by its last two digits is divisible by 4.
The last two digits of 984 form 84, and 84 ÷ 4 = 21 exactly.
So 984 is divisible by 4 (984 ÷ 4 = 246).
In simple words: Look only at 84; since 4 divides 84, it divides 984.

Teacher's Note:
1. This works because 100 is a multiple of 4.
2. Write the rule first, then apply it.

Question 13. Write the prime factorisation of 999.
Answer:
Digit sum = 9 + 9 + 9 = 27, so 999 is divisible by 9: 999 = 9 × 111.
111 has digit sum 3, so 111 = 3 × 37, and 37 is prime.
So 999 = 3 × 3 × 3 × 37 = 3³ × 37.
In simple words: 999 = 9 × 111 = 9 × 3 × 37.

Teacher's Note:
1. 37 is prime: 2, 3 and 5 do not divide it, and 7 × 7 = 49 > 37.
2. Check: 27 × 37 = 999.

Question 14. State the divisibility rule for 9 and check if the number 729 is divisible by 9.
Answer:
Rule: A number is divisible by 9 if the sum of its digits is divisible by 9.
For 729: 7 + 2 + 9 = 18, and 18 ÷ 9 = 2.
So 729 is divisible by 9 (729 ÷ 9 = 81).
In simple words: The digits of 729 add up to 18, a multiple of 9.

Teacher's Note:
1. State the rule in words before using it.
2. 729 = 9 × 81 = 3 × 3 × 3 × 3 × 3 × 3.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Find a number between 400 and 500 that is a multiple of 6, 9, and 12.
Answer:
Prime factorise: 6 = 2 × 3, 9 = 3², 12 = 2² × 3.
Smallest common multiple = 2² × 3² = 36, so the number must be a multiple of 36.
Multiples of 36: 36 × 11 = 396 (too small), 36 × 12 = 432, 36 × 13 = 468, 36 × 14 = 504 (too large).
So a required number is 432 (468 is also correct).
In simple words: Every common multiple of 6, 9 and 12 is a multiple of 36, and 432 = 36 × 12 lies between 400 and 500.

Teacher's Note:
1. The question asks for 'a number', so either 432 or 468 is fine.
2. Check: 432 ÷ 6 = 72, ÷ 9 = 48, ÷ 12 = 36.

Question 16. Which of the numbers 1234, 5678, and 9012 is divisible by 4 and 8? Explain your reasoning using divisibility rules.
Answer:
Rule for 4: the last two digits must form a number divisible by 4. Rule for 8: the last three digits must form a number divisible by 8.
1234: 34 ÷ 4 leaves remainder 2, so 1234 is not divisible by 4 (and so not by 8).
5678: 78 ÷ 4 leaves remainder 2, so 5678 is not divisible by 4 (and so not by 8).
9012: 12 ÷ 4 = 3, so 9012 is divisible by 4. But its last three digits form 012 = 12, and 12 ÷ 8 leaves remainder 4, so it is not divisible by 8.
So none of the three numbers is divisible by both 4 and 8; only 9012 is divisible by 4.
In simple words: Only 9012 passes the test for 4, and it fails the test for 8.

Teacher's Note:
1. A number not divisible by 4 cannot be divisible by 8.
2. 'None' can be a correct answer if you explain it.

Question 17. A number when divided by 7 leaves a remainder of 1. What is the smallest such 3-digit number that is a multiple of 4?
Answer:
List 3-digit multiples of 4 from the smallest and find the remainder on dividing by 7:
100 → remainder 2, 104 → 6, 108 → 3, 112 → 0, 116 → 4, 120 → 1.
120 = 7 × 17 + 1 and 120 = 4 × 30.
So the smallest such number is 120.
In simple words: Go through 100, 104, 108, … until one leaves remainder 1 on dividing by 7; the first is 120.

Teacher's Note:
1. Start from the smallest 3-digit multiple of 4, which is 100.
2. Check both conditions for the final answer.

Section D: Long Answer Question (5 Marks)

Question 18. Find the largest 4-digit number that is a common multiple of 24, 32, and 40.
Answer:
Step 1: Prime factorise: 24 = 2³ × 3, 32 = 2⁵, 40 = 2³ × 5.
Step 2: Smallest common multiple = 2⁵ × 3 × 5 = 32 × 15 = 480.
Step 3: Every common multiple of 24, 32 and 40 is a multiple of 480.
Step 4: The largest 4-digit number is 9999. 9999 ÷ 480 gives quotient 20 and remainder 399.
Step 5: So the largest 4-digit multiple of 480 is 480 × 20 = 9600 (480 × 21 = 10080 has five digits).
The required number is 9600.
In simple words: Find the smallest common multiple (480), then the biggest multiple of 480 that still has four digits.

Teacher's Note:
1. Check: 9600 ÷ 24 = 400, ÷ 32 = 300, ÷ 40 = 240.
2. Equivalently, 9999 − 399 = 9600.

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

Question 19. Case Study 1: The School Field Trip
A school is planning a field trip for its students. There are 288 students from class 6 and 432 students from class 7. The school wants to hire the largest possible buses, with each bus carrying the same number of students. Each bus must only have students from a single class to keep the groups separate. To minimize the number of buses needed, the school must find the greatest number of students that can fit on each bus. The solution is the greatest common factor of the two student numbers.
(a) Find the prime factorisation of 288.
(b) Find the prime factorisation of 432.
(c) Determine the largest number that is a factor of both 288 and 432.
(d) Use the divisibility rule for 9 to show that 432 is divisible by 9.
Answer:
(a) 288 = 2 × 144 = 2 × 2 × 2 × 2 × 2 × 9 = 2⁵ × 3².
(b) 432 = 16 × 27 = 2⁴ × 3³.
(c) Common primes: four 2s and two 3s. Greatest common factor = 2⁴ × 3² = 16 × 9 = 144.
(d) Rule for 9: the digit sum must be divisible by 9. For 432: 4 + 3 + 2 = 9, which is divisible by 9, so 432 is divisible by 9 (432 ÷ 9 = 48).
So each bus can carry 144 students: 2 buses for class 6 and 3 buses for class 7.
In simple words: Take the 2s and 3s that 288 and 432 share: 2⁴ × 3² = 144.

Teacher's Note:
1. For a common factor, take the smaller power of each shared prime.
2. Check: 288 ÷ 144 = 2 and 432 ÷ 144 = 3.

Question 20. Case Study 2: The Clock Alarms
A person sets three different alarms to wake up. The first alarm rings every 15 minutes, the second every 20 minutes, and the third every 30 minutes. They all just went off at 6:00 AM. The person wants to know the exact time when all three alarms will ring together again. To solve this, they must find the least common multiple of the three time intervals. The result will provide the number of minutes until the next synchronized alarm.
(a) Find the prime factorisation of 15.
(b) Find the prime factorisation of 20.
(c) Find the smallest number that is a multiple of 15, 20, and 30.
(d) At what time will all three alarms ring together again?
Answer:
(a) 15 = 3 × 5.
(b) 20 = 2 × 2 × 5 = 2² × 5.
(c) Also 30 = 2 × 3 × 5. Take the largest power of each prime: 2² × 3 × 5 = 60.
(d) The alarms ring together every 60 minutes = 1 hour. One hour after 6:00 AM is 7:00 AM.
In simple words: All three alarms meet every 60 minutes, so the next time is 7:00 AM.

Teacher's Note:
1. Check: 60 ÷ 15 = 4, 60 ÷ 20 = 3, 60 ÷ 30 = 2.
2. Part (d) asks for a clock time, not just a number of minutes.