CBSE Class 7 Mathematics Finding Common Ground Practice Paper Assignment Set 04 with Answers

School Assignments for Class 7 Mathematics: Chapter 03 Finding Common Ground

Explore structured practice materials through the CBSE Class 7 Mathematics Finding Common Ground Practice Paper Assignment Set 04 with Answers. Tailored for Class 7 learners, utilizing these Mathematics assignments ensures thorough preparation and strengthens foundational knowledge before final CBSE evaluations.

Practice Class 7 Mathematics Assignments: Chapter 03 Finding Common Ground

Navigate directly to the solved Mathematics assignments using the digital viewer below. Each practice set includes detailed step-by-step solutions, allowing students to instantly cross-check their work and identify areas requiring further revision.

Class 7 Maths Ganita Prakash Part 2 Chapter 3 Finding Common Ground Practice Paper 36 is a 40-mark paper of 20 questions in Sections A to E, given here with step-by-step solutions. It works on the HCF of 756 and 1134, the LCM of 540 and 810, numbers with exactly 15 factors, factors of 1296, what the LCM does when both numbers are divided by 5, and recovering two numbers from their HCF, LCM and difference. The case studies cover two factory machines finishing their cycles together and packing notebooks and pens into identical gift packs.

Section A: Multiple Choice Questions (1 Mark Each)

Question 1: Two machines operate every 16 minutes and 24 minutes. If they start together now, after how many minutes will they operate together again?
(a) 24
(b) 32
(c) 48
(d) 96
Show Answer & Explanation

Answer: (c) 48

Explanation:
1. We need the LCM of 16 and 24.
2. 16 = 2⁴ and 24 = 2³ × 3.
3. LCM = 2⁴ × 3 = 48.
4. So they operate together again after 48 minutes.

Teacher's Note:
1. 24 and 32 are multiples of only one of the two numbers.
2. 96 is a common multiple, but not the lowest.

Question 2: Which number has prime factorisation 2² × 3³ × 5?
(a) 540
(b) 360
(c) 720
(d) 900
Show Answer & Explanation

Answer: (a) 540

Explanation:
1. 2² = 4 and 3³ = 27.
2. 4 × 27 = 108.
3. 108 × 5 = 540.
4. So the number is 540.

Teacher's Note:
1. 3³ means 3 × 3 × 3 = 27, not 3 × 3 = 9.
2. 360 = 2³ × 3² × 5 has different powers.

Question 3: Find the HCF of 756 and 1260.
(a) 84
(b) 126
(c) 168
(d) 252
Show Answer & Explanation

Answer: (d) 252

Explanation:
1. 756 = 2² × 3³ × 7 and 1260 = 2² × 3² × 5 × 7.
2. Common primes with smaller powers: 2², 3² and 7.
3. HCF = 4 × 9 × 7 = 252.
4. So the HCF is 252.

Teacher's Note:
1. 84 and 126 are common factors, but smaller than the highest one.
2. Check: 756 ÷ 252 = 3 and 1260 ÷ 252 = 5.

Question 4: Which of the following pairs is co-prime?
(a) 49 and 64
(b) 21 and 35
(c) 18 and 24
(d) 27 and 81
Show Answer & Explanation

Answer: (a) 49 and 64

Explanation:
1. Co-prime numbers have HCF 1.
2. 49 = 7² and 64 = 2⁶ have no common prime, so HCF = 1.
3. 21 and 35 share 7; 18 and 24 share 6; 27 and 81 share 27.
4. So the co-prime pair is 49 and 64.

Teacher's Note:
1. Powers of different primes are always co-prime.
2. Neither 49 nor 64 is prime, yet they are co-prime.

Question 5: The LCM of two numbers is 900 and one of the numbers is 300. Which of the following could be the other number?
(a) 2
(b) 9
(c) 4
(d) 6
Show Answer & Explanation

Answer: (b) 9

Explanation:
1. 300 = 2² × 3 × 5² and 900 = 2² × 3² × 5².
2. The LCM has 3² but 300 has only one 3, so the other number must contain 3² = 9.
3. LCM(300, 9) = 2² × 3² × 5² = 900. ✓
4. With 2, 4 or 6 the LCM stays 300.
5. So the other number can be 9.

Teacher's Note:
1. 900 ÷ 300 = 3, but 3 does not work: LCM(300, 3) = 300. Look at prime powers instead of dividing.
2. Other numbers such as 36 or 225 would also give LCM 900.

Question 6: Which number has exactly 15 factors?
(a) 144
(b) 225
(c) 256
(d) 360
Show Answer & Explanation

Answer: (a) 144

Explanation:
1. Number of factors = product of (each power + 1).
2. 144 = 2⁴ × 3²: 5 × 3 = 15.
3. 225 = 3² × 5²: 3 × 3 = 9. 256 = 2⁸: 9. 360 = 2³ × 3² × 5: 4 × 3 × 2 = 24.
4. So 144 has exactly 15 factors.

Teacher's Note:
1. An odd number of factors happens only for perfect squares.
2. 15 = 5 × 3, so look for a number of the form p⁴ × q².

Question 7: Which of the following is not a factor of 1296?
(a) 36
(b) 54
(c) 72
(d) 125
Show Answer & Explanation

Answer: (d) 125

Explanation:
1. 1296 = 2⁴ × 3⁴.
2. 36 = 2² × 3², 54 = 2 × 3³ and 72 = 2³ × 3² all fit inside 2⁴ × 3⁴.
3. 125 = 5³ has the prime 5, which 1296 does not have.
4. So 125 is not a factor of 1296.

Teacher's Note:
1. 1296 = 36 × 36 = 6⁴.
2. An odd number ending in 5 cannot divide a number with no factor 5.

Question 8: If two numbers are in the ratio 7 : 11 and their HCF is 9, find the larger number.
(a) 63
(b) 77
(c) 81
(d) 99
Show Answer & Explanation

Answer: (d) 99

Explanation:
1. Write the numbers as 7k and 11k. As 7 and 11 are co-prime, HCF = k = 9.
2. The numbers are 7 × 9 = 63 and 11 × 9 = 99.
3. Check: HCF(63, 99) = 9. ✓
4. The larger number is 99.

Teacher's Note:
1. Multiply each ratio term by the HCF.
2. 77 = 7 × 11 does not use the HCF at all.

Question 9:
Assertion (A): The HCF of (18 × 20) and (18 × 35) is greater than 18.
Reason (R): Both numbers contain 90 as a common factor.
(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.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. 18 × 20 = 360 and 18 × 35 = 630. HCF = 18 × HCF(20, 35) = 18 × 5 = 90, which is greater than 18. So A is true.
2. 360 = 90 × 4 and 630 = 90 × 7, so 90 is a common factor. R is true.
3. Because 90 is a common factor, the HCF is at least 90 and so is greater than 18. R explains A.
4. The answer is (a).

Teacher's Note:
1. Any common factor is less than or equal to the HCF.
2. Here 18 × 5 = 90 because 20 and 35 share the factor 5.

Question 10:
Assertion (A): If two numbers that are both multiples of 5 are divided by 5, their LCM also gets divided by 5.
Reason (R): Dividing both numbers by 5 reduces the highest power of prime factor 5 in their LCM.
(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.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Example: LCM(50, 75) = 150; after dividing by 5, LCM(10, 15) = 30 = 150 ÷ 5. So A is true.
2. Dividing both numbers by 5 lowers the power of 5 in each by one; other primes stay the same.
3. So the highest power of 5 in the LCM drops by one, i.e. the LCM is divided by 5. R is true.
4. R explains A, so the answer is (a).

Teacher's Note:
1. Scaling both numbers down by 5 scales the LCM down by 5.
2. Test the statement with a simple example.

Section B: Short Answer Questions (2 Marks Each)

Question 11. (a) Write the prime factorisation of 756 and 1134.
(b) Find their HCF.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
756 = 2 × 2 × 3 × 3 × 3 × 7 = 2² × 3³ × 7.
1134 = 2 × 3 × 3 × 3 × 3 × 7 = 2 × 3⁴ × 7.
(b) Primes common to both: 2, 3, 7.
HCF = product of the common primes, each with its smallest power = 2 × 3³ × 7 = 378.
In simple words: 756 and 1134 share 2, 3³ and 7, so their HCF is 378.

Teacher's Note:
1. 1134 is 2 × 567, so it has just one 2.
2. Check: 756 ÷ 378 = 2 and 1134 ÷ 378 = 3.

 

Question 12. (a) Write the prime factorisation of 360 and 540.
(b) Find their LCM.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
360 = 2 × 2 × 2 × 3 × 3 × 5 = 2³ × 3² × 5.
540 = 2 × 2 × 3 × 3 × 3 × 5 = 2² × 3³ × 5.
(b) Primes that appear in either number: 2, 3, 5.
LCM = product of every prime, each with its greatest power = 2³ × 3³ × 5 = 1080.
In simple words: LCM = 2³ × 3³ × 5 = 1080.

Teacher's Note:
1. 2³ comes from 360 and 3³ from 540.
2. Check: 1080 ÷ 360 = 3 and 1080 ÷ 540 = 2.

 

Question 13. (a) Write the prime factorisation of 196 and 245.
(b) Find their HCF.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
196 = 2 × 2 × 7 × 7 = 2² × 7².
245 = 5 × 7 × 7 = 5 × 7².
(b) Primes common to both: 7.
HCF = 7² = 49.
In simple words: Only 7² is common, so the HCF is 49.

Teacher's Note:
1. 196 is even and 245 is odd, so 2 cannot be in the HCF.
2. Check: 196 ÷ 49 = 4 and 245 ÷ 49 = 5.

 

Question 14. (a) Write the prime factorisation of 84 and 210.
(b) Find their LCM.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
84 = 2 × 2 × 3 × 7 = 2² × 3 × 7.
210 = 2 × 3 × 5 × 7.
(b) Primes that appear in either number: 2, 3, 5, 7.
LCM = product of every prime, each with its greatest power = 2² × 3 × 5 × 7 = 420.
In simple words: Taking 2², 3, 5 and 7 gives LCM 420.

Teacher's Note:
1. Each prime is written once, with its greatest power.
2. Check: 420 ÷ 84 = 5 and 420 ÷ 210 = 2.

 

Section C: Short Answer Questions (3 Marks Each)

Question 15. (a) Write the prime factorisation of 540 and 810.
(b) Find their HCF.
(c) Find their LCM.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
540 = 2 × 2 × 3 × 3 × 3 × 5 = 2² × 3³ × 5.
810 = 2 × 3 × 3 × 3 × 3 × 5 = 2 × 3⁴ × 5.
(b) Common primes with smallest powers: 2, 3³, 5.
HCF = product of the common primes, each with its smallest power = 2 × 3³ × 5 = 270.
(c) All primes with greatest powers: 2², 3⁴, 5.
LCM = product of every prime, each with its greatest power = 2² × 3⁴ × 5 = 1620.
Check: HCF × LCM = 270 × 1620 = 437400 and 540 × 810 = 437400. They match.
In simple words: 540 and 810 have HCF 270 and LCM 1620.

Teacher's Note:
1. 810 = 2 × 3⁴ × 5, so the power of 3 in the LCM is 3⁴.
2. HCF × LCM = 270 × 1620 = 437400 = 540 × 810.

 

Question 16. (a) Write the prime factorisation of 90, 120 and 180.
(b) Identify the highest powers of all primes involved.
(c) Find their LCM.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
90 = 2 × 3 × 3 × 5 = 2 × 3² × 5.
120 = 2 × 2 × 2 × 3 × 5 = 2³ × 3 × 5.
180 = 2 × 2 × 3 × 3 × 5 = 2² × 3² × 5.
(b) Compare the power of each prime in the three numbers:
Prime 2: 2¹ in 90, 2³ in 120, 2² in 180; highest power 2³.
Prime 3: 3² in 90, 3¹ in 120, 3² in 180; highest power 3².
Prime 5: 5¹ in 90, 5¹ in 120, 5¹ in 180; highest power 5.
(c) LCM = 2³ × 3² × 5 = 360.
In simple words: Highest powers 2³, 3² and 5 give LCM 360.

Teacher's Note:
1. 120 supplies 2³; 90 and 180 supply 3².
2. Check: 360 ÷ 90 = 4, 360 ÷ 120 = 3 and 360 ÷ 180 = 2.

 

Question 17. (a) Write the prime factorisation of 630 and 1050.
(b) Identify the common prime factors.
(c) Find their HCF.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
630 = 2 × 3 × 3 × 5 × 7 = 2 × 3² × 5 × 7.
1050 = 2 × 3 × 5 × 5 × 7 = 2 × 3 × 5² × 7.
(b) The primes 2, 3, 5, 7 divide both numbers. Taking the smaller power of each, the common prime factors are 2 × 3 × 5 × 7.
(c) HCF = 2 × 3 × 5 × 7 = 210.
In simple words: 630 and 1050 share 2, 3, 5 and 7 once each, so the HCF is 210.

Teacher's Note:
1. 630 has 3² and 1050 has 5², but only the single powers are common.
2. Check: 630 ÷ 210 = 3 and 1050 ÷ 210 = 5.

 

Section D: Long Answer Question (5 Marks)

Question 18. The difference of two numbers is 180, their HCF is 36, and their LCM is 1296.
(a) Find the product of the two numbers.
(b) Let the numbers be 36x and 36y. Find x and y.
(c) Hence, find the two numbers.
(d) Verify your result using the HCF–LCM relation.
(e) Check whether the difference matches the given condition.
Answer:
(a) For two numbers, Product = HCF × LCM = 36 × 1296 = 46656.
(b) Both numbers are multiples of the HCF, so write them as 36x and 36y, where x and y are co-prime (otherwise the HCF would be bigger than 36).
Then LCM = 36 × x × y = 1296, so x × y = 1296 ÷ 36 = 36.
Difference: 36y − 36x = 180, so y − x = 180 ÷ 36 = 5.
Co-prime pairs with product 36 (smaller first): (1, 36), (4, 9).
The pair whose difference is 5 is (4, 9). So x = 4 and y = 9.
(c) The numbers are 36 × 4 = 144 and 36 × 9 = 324. So the numbers are 144 and 324.
(d) 144 = 2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3² and 324 = 2 × 2 × 3 × 3 × 3 × 3 = 2² × 3⁴.
HCF = 2² × 3² = 36 and LCM = 2⁴ × 3⁴ = 1296.
HCF × LCM = 36 × 1296 = 46656 and 144 × 324 = 46656. Both are equal, so the result is verified.
(e) 324 − 144 = 180, which is the given difference. ✓
In simple words: The numbers are 144 and 324: HCF 36, LCM 1296 and difference 180.

Teacher's Note:
1. x = 4, y = 6 (numbers 144 and 216) is a trap: 4 and 6 are not co-prime, and HCF(144, 216) is 72, not 36.
2. Always check that x and y have no common factor.

 

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

Question 19. Case Study 1: Scheduling Machine Inspections
In a manufacturing plant, Machine A completes a production cycle every 30 minutes, while Machine B completes a cycle every 45 minutes. Both machines start together at 8:00 a.m. The manager wants to schedule inspections when both machines complete their cycles at the same time again. Use mathematical reasoning to find the next common completion time.
Answer the following questions:
(a) Which concept is used?
(b) Find the required value.
(c) After how many minutes will both machines complete cycles together again?
(d) At what exact time will this occur?
Answer:
(a) We need the first time that is a multiple of both 30 and 45 minutes, i.e. the smallest common multiple. So the concept is LCM (Lowest Common Multiple).
(b) 30 = 2 × 3 × 5 and 45 = 3 × 3 × 5 = 3² × 5.
LCM = 2 × 3² × 5 = 90.
(c) Both machines complete their cycles together again after 90 minutes (1 hour 30 minutes).
(d) 8:00 a.m. + 1 hour 30 minutes = 9:30 a.m.
In simple words: The machines finish together every 90 minutes, so next at 9:30 a.m.

Teacher's Note:
1. 90 minutes = 1 hour 30 minutes.
2. Check: 90 = 3 × 30 = 2 × 45.

 

Question 20. Case Study 2: Preparing Gift Packs
A charity organization has 360 notebooks and 540 pens to distribute. The organizers want to prepare the maximum possible number of identical gift packs, without any leftover items. Use mathematical reasoning to determine the number of gift packs and the contents of each pack.
Answer the following questions:
(a) Which concept is used?
(b) Find the required value.
(c) How many gift packs can be prepared?
(d) How many notebooks and pens will each pack contain?
Answer:
(a) The number of identical gift packs must divide both 360 and 540 exactly, and we want the largest such number. So the concept is HCF (Highest Common Factor).
(b) 360 = 2 × 2 × 2 × 3 × 3 × 5 = 2³ × 3² × 5 and 540 = 2 × 2 × 3 × 3 × 3 × 5 = 2² × 3³ × 5.
HCF = 2² × 3² × 5 = 180.
(c) Number of gift packs = 180.
(d) Each gift pack gets 360 ÷ 180 = 2 notebooks and 540 ÷ 180 = 3 pens. So each gift pack contains 2 notebooks and 3 pens.
Check: 180 × 2 = 360 and 180 × 3 = 540, so nothing is left over.
In simple words: 180 gift packs can be made, each with 2 notebooks and 3 pens.

Teacher's Note:
1. The number of packs must divide 360 and 540 exactly; the largest such number is the HCF.
2. 2 : 3 is 360 : 540 in its simplest form.

 

Free study material for Mathematics

Chapter 03 Finding Common Ground Printable Assignments & Solutions for Class 7 Mathematics

Revision Assignment: Chapter 03 Finding Common Ground (CBSE)

Explore reliable practice questions for Chapter 03 Finding Common Ground tailored for Class 7 learners. Use these structured worksheets to evaluate preparedness and strengthen core problem-solving skills.

Maximize Exam Scores with Chapter Practice Sets

  • Syllabus Compliance: Sets reflect current CBSE evaluation criteria and official marking frameworks.
  • Multi-Format Practice: Includes varied problem types designed to deepen comprehension across all sub-topics.
  • Time Management: Routine practice optimizes pacing to finish school examinations comfortably within schedule.

Steps to Complete Chapter 03 Finding Common Ground Assignments Successfully

  1. Initial Reading: Begin by reading the NCERT book for Class 7 Mathematics to build a baseline understanding.
  2. Independent Testing: Attempt assignment questions unassisted, then verify work using provided answer keys.
  3. Supplementary Aids: Leverage revision notes and worksheets whenever you encounter difficult topics.

FAQs

Where can I download the latest CBSE Class 7 Mathematics Chapter 03 Finding Common Ground assignments?

You can download free PDF assignments for Class 7 Mathematics Chapter 03 Finding Common Ground from StudiesToday.com. These practice sheets have been updated for the 2026-27 session covering all concepts from latest NCERT textbook.

Do these Mathematics Chapter 03 Finding Common Ground assignments include solved questions?

Yes, our teachers have given solutions for all questions in the Class 7 Mathematics Chapter 03 Finding Common Ground assignments. This will help you to understand step-by-step methodology to get full marks in school tests and exams.

Are the assignments for Class 7 Mathematics Chapter 03 Finding Common Ground based on the 2026 exam pattern?

Yes. These assignments are designed as per the latest CBSE syllabus for 2026. We have included huge variety of question formats such as MCQs, Case-study based questions and important diagram-based problems found in Chapter 03 Finding Common Ground.

How can practicing Chapter 03 Finding Common Ground assignments help in Mathematics preparation?

Practicing topicw wise assignments will help Class 7 students understand every sub-topic of Chapter 03 Finding Common Ground. Daily practice will improve speed, accuracy and answering competency-based questions.

Can I download Mathematics Chapter 03 Finding Common Ground assignments for free on mobile?

Yes, all printable assignments for Class 7 Mathematics Chapter 03 Finding Common Ground are available for free download in mobile-friendly PDF format.