CBSE Class 6 Mathematics Prime Time Practice Paper Assignment Set 03 with Answers

Here is Class 6 Maths Ganita Prakash Chapter 5 Prime Time Practice Paper 15 (CBSE), with 20 questions worth 40 marks across Sections A to E and a worked answer for every question. The paper focuses on the divisibility tests for 4, 8 and 11, prime and composite numbers, the smallest common multiple of 1 to 10, the greatest common factor of 252 and 360, and case studies on planting rows of roses and jasmine and on two buses returning to a station.

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 16, 24, and 32.
Answer:
Prime factorise: 16 = 2⁴, 24 = 2³ × 3, 32 = 2⁵.
Take the largest power of each prime: 2⁵ and 3.
Smallest number = 2⁵ × 3 = 32 × 3 = 96.
In simple words: It must contain five 2s (for 32) and one 3 (for 24), so it is 96.

Teacher's Note:
1. Check: 96 ÷ 16 = 6, ÷ 24 = 4, ÷ 32 = 3.
2. Since 16 divides 32, you only need 32 and 24.

Question 12. Find the greatest common factor of 252 and 360.
Answer:
Prime factorise: 252 = 2² × 3² × 7, 360 = 2³ × 3² × 5.
Primes common to both: two 2s and two 3s.
Greatest common factor = 2² × 3² = 4 × 9 = 36.
In simple words: Multiply the primes shared by 252 and 360: 2 × 2 × 3 × 3 = 36.

Teacher's Note:
1. Check: 252 ÷ 36 = 7 and 360 ÷ 36 = 10.
2. 7 and 10 have no common factor, so 36 is the greatest.

Question 13. Find the prime factorisation of 735.
Answer:
735 ends in 5, so 735 = 5 × 147.
147 has digit sum 12, so 147 = 3 × 49 = 3 × 7 × 7.
So 735 = 3 × 5 × 7 × 7 = 3 × 5 × 7².
In simple words: Take out 5, then 3, and what is left is 7 × 7.

Teacher's Note:
1. Use the divisibility rules to find the first factor quickly.
2. Check: 15 × 49 = 735.

Question 14. Is the number 9135 divisible by 5 and 7? Explain your answer.
Answer:
Divisibility by 5: the last digit of 9135 is 5, so it is divisible by 5.
Divisibility by 7: divide by 7: 9135 ÷ 7 = 1305 exactly (7 × 1305 = 9135).
Another quick check: remove the last digit and subtract twice it: 913 − 10 = 903; again 90 − 6 = 84 = 7 × 12. Since 84 is a multiple of 7, so is 9135.
So 9135 is divisible by both 5 and 7 (9135 = 35 × 261).
In simple words: It ends in 5 and 7 × 1305 = 9135, so both 5 and 7 divide it.

Teacher's Note:
1. There is no simple digit rule for 7 in the textbook, so dividing is fine.
2. Divisible by 5 and 7 means divisible by 35.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Find a number between 200 and 300 that has a prime factor of 13 and is a multiple of 6.
Answer:
A multiple of 6 has the prime factors 2 and 3. It must also have the prime factor 13.
So the number is a multiple of 2 × 3 × 13 = 78.
Multiples of 78: 78 × 2 = 156 (too small), 78 × 3 = 234, 78 × 4 = 312 (too large).
So the number is 234.
In simple words: Multiply 6 by 13 to get 78; the only multiple of 78 between 200 and 300 is 234.

Teacher's Note:
1. Check: 234 = 2 × 3² × 13.
2. 6 and 13 are co-prime, so their product works.

Question 16. Using the divisibility test for 11, check if the numbers 121 and 1331 are divisible by 11. Explain your reasoning.
Answer:
Rule: a number is divisible by 11 if the difference between the sum of its digits in odd places and in even places (counted from the right) is 0 or a multiple of 11.
121: odd places 1 + 1 = 2, even place 2. Difference 2 − 2 = 0, so 121 is divisible by 11 (121 = 11 × 11).
1331: odd places 1 + 3 = 4, even places 3 + 1 = 4. Difference 4 − 4 = 0, so 1331 is divisible by 11 (1331 = 11 × 121).
In simple words: In both numbers the two alternate digit sums are equal, so 11 divides both.

Teacher's Note:
1. Count places from the units digit.
2. 121 = 11² and 1331 = 11³.

Question 17. A number when divided by 8 leaves a remainder of 3. What is the largest such 2-digit number that is also a multiple of 5?
Answer:
A multiple of 5 ends in 0 or 5.
A number of the form (multiple of 8) + 3 is odd, so it cannot end in 0. It must end in 5.
Try 2-digit numbers ending in 5 from the top: 95 = 8 × 11 + 7, 85 = 8 × 10 + 5, 75 = 8 × 9 + 3.
So the largest such number is 75.
In simple words: Check 95, 85, 75, …; the first one that leaves remainder 3 on dividing by 8 is 75.

Teacher's Note:
1. Start from the largest candidate and move down.
2. Check: 72 + 3 = 75 and 75 = 5 × 15.

Section D: Long Answer Question (5 Marks)

Question 18. Find the smallest number that is a common multiple of all numbers from 1 to 10.
Answer:
Step 1: The primes up to 10 are 2, 3, 5 and 7.
Step 2: Find the largest power of each prime among 1 to 10: 8 = 2³, 9 = 3², 5 = 5 and 7 = 7.
Step 3: Every other number is built from these: 4 = 2², 6 = 2 × 3, 10 = 2 × 5.
Step 4: Smallest common multiple = 2³ × 3² × 5 × 7 = 8 × 9 × 5 × 7.
Step 5: 8 × 9 = 72, 72 × 5 = 360, 360 × 7 = 2520.
So the smallest number divisible by every number from 1 to 10 is 2520.
In simple words: Use 8, 9, 5 and 7 (the largest prime powers up to 10) and multiply them: 2520.

Teacher's Note:
1. Check: 2520 ÷ 8 = 315, ÷ 9 = 280, ÷ 7 = 360.
2. Multiplying 1 × 2 × … × 10 gives a much larger common multiple, not the smallest.

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

Question 19. Case Study 1: The Gardener's Plants
A gardener, Mr. Kumar, has 180 rose plants and 270 jasmine plants. He wants to arrange them in rows with the same number of plants in each row. Each row must contain only one type of plant. To plant the maximum number of plants in a single row, he must determine the greatest common factor of the number of rose and jasmine plants. This will help him create the most efficient layout and ensure no plants are left out.
(a) Find the prime factorisation of 180.
(b) Find the prime factorisation of 270.
(c) What is the greatest common factor of 180 and 270?
(d) Use the divisibility rule for 10 to show that the largest common number of plants per row (the answer to part c) is divisible by 10.
Answer:
(a) 180 = 2 × 2 × 3 × 3 × 5 = 2² × 3² × 5.
(b) 270 = 2 × 3 × 3 × 3 × 5 = 2 × 3³ × 5.
(c) Common primes: one 2, two 3s and one 5. Greatest common factor = 2 × 3² × 5 = 90.
(d) Rule for 10: a number is divisible by 10 if its last digit is 0. The last digit of 90 is 0, so 90 is divisible by 10.
So Mr. Kumar can plant 90 plants in each row: 2 rows of roses and 3 rows of jasmine.
In simple words: The largest number that divides both 180 and 270 is 90.

Teacher's Note:
1. Take the smaller power of each prime that appears in both numbers.
2. Check: 180 ÷ 90 = 2 and 270 ÷ 90 = 3.

Question 20. Case Study 2: The Bus Schedule
Two buses, Bus A and Bus B, depart from the same station at the same time. Bus A completes its route and returns to the station every 25 minutes, while Bus B returns every 30 minutes. A passenger wants to know when both buses will next be at the station together. To figure this out, the passenger needs to find the least common multiple of their departure intervals. The result will tell them the number of minutes until the next time the two buses will be at the station at the exact same moment.
(a) What are the prime factors of 25?
(b) What are the prime factors of 30?
(c) Find the smallest number that is a multiple of 25 and 30.
(d) Based on your answer from (c), after how many minutes will both buses be at the station together again?
Answer:
(a) 25 = 5 × 5 = 5², so its only prime factor is 5.
(b) 30 = 2 × 3 × 5, so its prime factors are 2, 3 and 5.
(c) Take the largest power of each prime: 2 × 3 × 5² = 6 × 25 = 150.
(d) Both buses will be at the station together again after 150 minutes (2 hours 30 minutes).
In simple words: Bus A returns at 25, 50, 75, … and Bus B at 30, 60, 90, …; the first common time is 150 minutes.

Teacher's Note:
1. Check: 150 ÷ 25 = 6 and 150 ÷ 30 = 5.
2. 25 × 30 = 750 is a common multiple, but not the smallest.