School Assignments for Class 7 Mathematics: Chapter 03 Finding Common Ground
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Here is Class 7 Maths Ganita Prakash Part 2 Chapter 3 Finding Common Ground Practice Paper 34: 40 marks, 20 questions in Sections A to E, each solved step by step. The questions focus on the HCF of 252 and 378, the LCM of 144 and 175 (two co-prime numbers), co-prime pairs, factors of 900, what happens to the LCM when both numbers are divided by 4, and finding two numbers when their HCF, LCM and difference are known. The case studies use LCM to time two village water pumps and HCF to pack emergency food kits.
Section A: Multiple Choice Questions (1 Mark Each)
(a) 90
(b) 180
(c) 225
(d) 360
Show Answer & Explanation
Answer: (b) 180
Explanation:
1. We need the LCM of 20 and 45.
2. 20 = 2² × 5 and 45 = 3² × 5.
3. LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180.
4. So they ring together again after 180 minutes.
Teacher's Note:
1. 90 is a multiple of 45 but not of 20 (90 ÷ 20 is not whole).
2. 360 is a common multiple too, but not the lowest.
(a) 360
(b) 720
(c) 1440
(d) 2880
Show Answer & Explanation
Answer: (b) 720
Explanation:
1. 2⁴ = 16 and 3² = 9.
2. 16 × 9 = 144.
3. 144 × 5 = 720.
4. So the number is 720.
Teacher's Note:
1. 360 = 2³ × 3² × 5 has one 2 fewer, so it is half of 720.
2. Group easy products first: 16 × 5 = 80, then 80 × 9 = 720.
(a) 42
(b) 63
(c) 84
(d) 126
Show Answer & Explanation
Answer: (d) 126
Explanation:
1. 252 = 2² × 3² × 7 and 378 = 2 × 3³ × 7.
2. Common primes with their smaller powers: 2, 3² and 7.
3. HCF = 2 × 9 × 7 = 126.
4. So the HCF is 126.
Teacher's Note:
1. Both numbers are even, so 2 must be part of the HCF; 63 is odd and misses it.
2. Check: 252 ÷ 126 = 2 and 378 ÷ 126 = 3.
(a) 16 and 25
(b) 21 and 35
(c) 18 and 24
(d) 14 and 49
Show Answer & Explanation
Answer: (a) 16 and 25
Explanation:
1. Co-prime numbers have no common prime factor (HCF = 1).
2. 16 = 2⁴ and 25 = 5²: nothing in common, HCF = 1.
3. 21 and 35 share 7; 18 and 24 share 6; 14 and 49 share 7.
4. So the co-prime pair is 16 and 25.
Teacher's Note:
1. Co-prime numbers can both be composite, like 16 and 25.
2. Find the prime factors of each number and look for any match.
(a) 4
(b) 6
(c) 8
(d) 12
Show Answer & Explanation
Answer: (c) 8
Explanation:
1. 150 = 2 × 3 × 5² and 600 = 2³ × 3 × 5².
2. The LCM has 2³, but 150 has only one 2, so the other number must contain 2³ = 8.
3. LCM(150, 8) = 2³ × 3 × 5² = 600. ✓
4. LCM(150, 4) = 300, LCM(150, 6) = 150 and LCM(150, 12) = 300.
5. So the other number can be 8.
Teacher's Note:
1. Find which prime power is missing from the given number.
2. Many numbers could be the other number (8, 24, 40, 200 …); among the options only 8 works.
(a) 60
(b) 48
(c) 80
(d) 100
Show Answer & Explanation
Answer: (a) 60
Explanation:
1. Number of factors = product of (each power + 1).
2. 60 = 2² × 3 × 5: 3 × 2 × 2 = 12.
3. 48 = 2⁴ × 3: 5 × 2 = 10. 80 = 2⁴ × 5: 5 × 2 = 10. 100 = 2² × 5²: 3 × 3 = 9.
4. So 60 has exactly 12 factors.
Teacher's Note:
1. Add 1 to each power, then multiply.
2. Check by listing factor pairs of 60: 1 × 60, 2 × 30, 3 × 20, 4 × 15, 5 × 12, 6 × 10.
(a) 36
(b) 45
(c) 75
(d) 128
Show Answer & Explanation
Answer: (d) 128
Explanation:
1. 900 = 2² × 3² × 5².
2. 36 = 2² × 3², 45 = 3² × 5 and 75 = 3 × 5² all fit inside 900.
3. 128 = 2⁷ needs seven 2s but 900 has only two.
4. So 128 is not a factor of 900.
Teacher's Note:
1. A factor cannot have a higher power of a prime than the number itself.
2. 128 is even, like 900, but that is not enough to be a factor.
(a) 56
(b) 77
(c) 88
(d) 99
Show Answer & Explanation
Answer: (b) 77
Explanation:
1. Write the numbers as 8k and 11k. As 8 and 11 are co-prime, HCF = k = 7.
2. The numbers are 8 × 7 = 56 and 11 × 7 = 77.
3. Check: HCF(56, 77) = 7. ✓
4. The larger number is 77.
Teacher's Note:
1. Multiply each ratio term by the HCF.
2. 88 = 8 × 11 mixes up the ratio and the HCF.
Assertion (A): The HCF of (20 × 15) and (20 × 25) is greater than 20.
Reason (R): Both numbers contain 100 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. 20 × 15 = 300 and 20 × 25 = 500. HCF = 20 × HCF(15, 25) = 20 × 5 = 100, which is greater than 20. So A is true.
2. 300 = 100 × 3 and 500 = 100 × 5, so 100 is a common factor. R is true.
3. Since 100 is a common factor, the HCF is at least 100, which is more than 20. So R explains A.
4. The answer is (a).
Teacher's Note:
1. A common factor bigger than 20 guarantees that the HCF is bigger than 20.
2. Compare with a reason like "both contain 20", which would not be enough.
Assertion (A): If two numbers that are both multiples of 4 are divided by 4, their LCM also gets divided by 4.
Reason (R): Dividing both numbers by 4 reduces the highest power of prime factors 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(8, 12) = 24; after dividing by 4, LCM(2, 3) = 6 = 24 ÷ 4. So A is true.
2. 4 = 2², so dividing both numbers by 4 lowers the power of 2 in each by 2.
3. The highest power of 2 in the LCM also drops by 2, so the LCM is divided by 2² = 4. R is true.
4. R explains A, so the answer is (a).
Teacher's Note:
1. Follow the prime 2 through the calculation, since 4 = 2².
2. Scaling both numbers by the same amount scales the LCM (and the HCF) by that amount.
Section B: Short Answer Questions (2 Marks Each)
Question 11. (a) Write the prime factorisation of 540 and 720.
(b) Find their HCF.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
540 = 2 × 2 × 3 × 3 × 3 × 5 = 2² × 3³ × 5.
720 = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 2⁴ × 3² × 5.
(b) Primes common to both: 2, 3, 5.
HCF = product of the common primes, each with its smallest power = 2² × 3² × 5 = 180.
In simple words: 540 and 720 share 2², 3² and 5, giving HCF 180.
Teacher's Note:
1. Take 2² (not 2⁴) and 3² (not 3³): the smaller powers.
2. Check: 540 ÷ 180 = 3 and 720 ÷ 180 = 4.
Question 12. (a) Write the prime factorisation of 196 and 294.
(b) Find their LCM.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
196 = 2 × 2 × 7 × 7 = 2² × 7².
294 = 2 × 3 × 7 × 7 = 2 × 3 × 7².
(b) Primes that appear in either number: 2, 3, 7.
LCM = product of every prime, each with its greatest power = 2² × 3 × 7² = 588.
In simple words: Combine 2², 3 and 7² to get LCM 588.
Teacher's Note:
1. 7² is common, so it appears only once in the LCM.
2. Check: 588 ÷ 196 = 3 and 588 ÷ 294 = 2.
Question 13. (a) Write the prime factorisation of 360 and 450.
(b) Find their HCF.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
360 = 2 × 2 × 2 × 3 × 3 × 5 = 2³ × 3² × 5.
450 = 2 × 3 × 3 × 5 × 5 = 2 × 3² × 5².
(b) Primes common to both: 2, 3, 5.
HCF = product of the common primes, each with its smallest power = 2 × 3² × 5 = 90.
In simple words: The shared part of 360 and 450 is 2 × 3² × 5 = 90.
Teacher's Note:
1. 450 is 2 × 225, so it has only one 2.
2. Check: 360 ÷ 90 = 4 and 450 ÷ 90 = 5.
Question 14. (a) Write the prime factorisation of 144 and 175.
(b) Find their LCM.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
144 = 2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3².
175 = 5 × 5 × 7 = 5² × 7.
(b) Primes that appear in either number: 2, 3, 5, 7.
LCM = product of every prime, each with its greatest power = 2⁴ × 3² × 5² × 7 = 25200.
In simple words: 144 and 175 have no common prime, so the LCM is their product, 25200.
Teacher's Note:
1. For co-prime numbers the LCM is simply the product.
2. Check: 144 × 175 = 25200 and HCF(144, 175) = 1.
Section C: Short Answer Questions (3 Marks Each)
Question 15. (a) Write the prime factorisation of 450 and 600.
(b) Find their HCF.
(c) Find their LCM.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
450 = 2 × 3 × 3 × 5 × 5 = 2 × 3² × 5².
600 = 2 × 2 × 2 × 3 × 5 × 5 = 2³ × 3 × 5².
(b) Common primes with smallest powers: 2, 3, 5².
HCF = product of the common primes, each with its smallest power = 2 × 3 × 5² = 150.
(c) All primes with greatest powers: 2³, 3², 5².
LCM = product of every prime, each with its greatest power = 2³ × 3² × 5² = 1800.
Check: HCF × LCM = 150 × 1800 = 270000 and 450 × 600 = 270000. They match.
In simple words: 450 and 600 have HCF 150 and LCM 1800.
Teacher's Note:
1. Line up the powers of 2, 3 and 5 for both numbers before choosing smaller or greater.
2. HCF × LCM = 150 × 1800 = 270000 = 450 × 600.
Question 16. (a) Write the prime factorisation of 48, 75 and 150.
(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:
48 = 2 × 2 × 2 × 2 × 3 = 2⁴ × 3.
75 = 3 × 5 × 5 = 3 × 5².
150 = 2 × 3 × 5 × 5 = 2 × 3 × 5².
(b) Compare the power of each prime in the three numbers:
Prime 2: 2⁴ in 48, none in 75, 2¹ in 150; highest power 2⁴.
Prime 3: 3¹ in 48, 3¹ in 75, 3¹ in 150; highest power 3.
Prime 5: none in 48, 5² in 75, 5² in 150; highest power 5².
(c) LCM = 2⁴ × 3 × 5² = 1200.
In simple words: Greatest powers 2⁴, 3 and 5² give LCM 1200.
Teacher's Note:
1. 48 gives the power of 2, while 75 and 150 give 5².
2. Check: 1200 ÷ 48 = 25, 1200 ÷ 75 = 16 and 1200 ÷ 150 = 8.
Question 17. (a) Write the prime factorisation of 420 and 840.
(b) Identify the common prime factors.
(c) Find their HCF.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
420 = 2 × 2 × 3 × 5 × 7 = 2² × 3 × 5 × 7.
840 = 2 × 2 × 2 × 3 × 5 × 7 = 2³ × 3 × 5 × 7.
(b) The primes 2, 3, 5, 7 divide both numbers. Taking the smaller power of each, the common prime factors are 2² × 3 × 5 × 7.
(c) HCF = 2² × 3 × 5 × 7 = 420.
In simple words: 420 divides 840 exactly, so the HCF is 420 itself.
Teacher's Note:
1. When one number is a factor of the other, the HCF is the smaller number.
2. 840 = 2 × 420, so every prime factor of 420 is in 840.
Section D: Long Answer Question (5 Marks)
Question 18. The HCF of two numbers is 24 and their LCM is 864. The difference of the two numbers is 120.
(a) Find the product of the two numbers.
(b) Let the numbers be 24x and 24y. Find x and y.
(c) Hence, find the two numbers.
(d) Verify your result using the HCF–LCM relation.
Answer:
(a) For two numbers, Product = HCF × LCM = 24 × 864 = 20736.
(b) Both numbers are multiples of the HCF, so write them as 24x and 24y, where x and y are co-prime (otherwise the HCF would be bigger than 24).
Then LCM = 24 × x × y = 864, so x × y = 864 ÷ 24 = 36.
Difference: 24y − 24x = 120, so y − x = 120 ÷ 24 = 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 24 × 4 = 96 and 24 × 9 = 216. So the numbers are 96 and 216.
(d) 96 = 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3 and 216 = 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 3³.
HCF = 2³ × 3 = 24 and LCM = 2⁵ × 3³ = 864.
HCF × LCM = 24 × 864 = 20736 and 96 × 216 = 20736. Both are equal, so the result is verified. Also 216 − 96 = 120, as given.
In simple words: The numbers are 96 and 216: HCF 24, LCM 864 and difference 120.
Teacher's Note:
1. Without an extra condition there are two answers (24 and 864, or 96 and 216); the difference fixes one.
2. x and y must be co-prime, so pairs like (2, 18) or (3, 12) are not allowed.
Section E: Case Study Based Questions (4 Marks Each)
Question 19. Case Study 1: Synchronizing Community Water Supply
Two water pumps in a village operate at regular intervals. Pump A runs every 24 minutes, while Pump B runs every 36 minutes. Both pumps start together at 6:00 a.m. The village committee wants to schedule maintenance when both pumps operate together again. Use mathematical reasoning to determine the next time of simultaneous operation.
Answer the following questions:
(a) Which concept is used?
(b) Find the required value.
(c) After how many minutes will both pumps operate together again?
(d) At what exact time will this occur?
Answer:
(a) We need the first time that is a multiple of both 24 and 36 minutes, i.e. the smallest common multiple. So the concept is LCM (Lowest Common Multiple).
(b) 24 = 2 × 2 × 2 × 3 = 2³ × 3 and 36 = 2 × 2 × 3 × 3 = 2² × 3².
LCM = 2³ × 3² = 72.
(c) Both pumps operate together again after 72 minutes (1 hour 12 minutes).
(d) 6:00 a.m. + 1 hour 12 minutes = 7:12 a.m.
In simple words: The pumps run together every 72 minutes, so next at 7:12 a.m.
Teacher's Note:
1. 72 minutes = 1 hour 12 minutes.
2. Always state the concept first; it earns a mark in case studies.
Question 20. Case Study 2: Packing Emergency Food Kits
A disaster-relief team has 360 packets of biscuits and 480 packets of juice. The team wants to prepare the maximum possible number of identical food kits, without any leftover items. Use mathematical reasoning to determine the number of kits and the contents of each kit.
Answer the following questions:
(a) Which concept is used?
(b) Find the required value.
(c) How many kits can be prepared?
(d) How many biscuit packets and juice packets will each kit contain?
Answer:
(a) The number of identical kits must divide both 360 and 480 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 480 = 2 × 2 × 2 × 2 × 2 × 3 × 5 = 2⁵ × 3 × 5.
HCF = 2³ × 3 × 5 = 120.
(c) Number of kits = 120.
(d) Each kit gets 360 ÷ 120 = 3 biscuit packets and 480 ÷ 120 = 4 juice packets. So each kit contains 3 biscuit packets and 4 juice packets.
Check: 120 × 3 = 360 and 120 × 4 = 480, so nothing is left over.
In simple words: 120 kits can be made, each with 3 biscuit packets and 4 juice packets.
Teacher's Note:
1. The number of kits is the HCF; the contents come from dividing.
2. 3 : 4 is the ratio 360 : 480 in lowest terms, a quick check.
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CBSE Class 7 Mathematics Assignments for Chapter 03 Finding Common Ground
Revision Assignment: Chapter 03 Finding Common Ground (CBSE)
Review targeted chapter assignments for Class 7 Mathematics Chapter 03 Finding Common Ground. Built according to official CBSE guidelines, these downloadable problem sets help students build accuracy and prepare effectively for school tests.
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