Class 7 Mathematics Practice Sheet: CBSE Class 7 Mathematics Finding Common Ground Practice Paper Worksheet Set 02 with Answers
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Practice Paper 33 for Class 7 Maths Ganita Prakash Part 2 Chapter 3 Finding Common Ground carries 40 marks across 20 questions in five sections (A to E), with a clear, worked answer for every question. This paper tests prime factorisation of numbers like 504, 756 and 1008, HCF and LCM using powers of primes, which numbers are factors of a given number, numbers with exactly 12 factors, numbers in a given ratio with a known HCF, and finding two numbers from their HCF, LCM and difference. Its case studies are about two school buses arriving together and packing relief food kits.
Section A: Multiple Choice Questions (1 Mark Each)
(a) 48
(b) 72
(c) 96
(d) 108
Show Answer & Explanation
Answer: (b) 72
Explanation:
1. We need the smallest number that is a multiple of both 24 and 36, i.e. the LCM.
2. 24 = 2³ × 3 and 36 = 2² × 3².
3. LCM = 2³ × 3² = 8 × 9 = 72.
4. So they ring together again after 72 minutes.
Teacher's Note:
1. 48 is a multiple of 24 but not of 36, and 108 is a multiple of 36 but not of 24.
2. Take the greater power of each prime for the LCM.
(a) 168
(b) 252
(c) 336
(d) 504
Show Answer & Explanation
Answer: (a) 168
Explanation:
1. 2³ = 8.
2. 8 × 3 = 24.
3. 24 × 7 = 168.
4. So the number is 168.
Teacher's Note:
1. 336 = 2⁴ × 3 × 7 and 504 = 2³ × 3² × 7; one extra prime changes the number.
2. Multiply in any order, e.g. 8 × 21 = 168.
(a) 90
(b) 120
(c) 150
(d) 180
Show Answer & Explanation
Answer: (d) 180
Explanation:
1. 540 = 2² × 3³ × 5 and 900 = 2² × 3² × 5².
2. Common primes with smaller powers: 2², 3² and 5.
3. HCF = 4 × 9 × 5 = 180.
4. So the HCF is 180.
Teacher's Note:
1. 90 is a common factor, but not the highest one.
2. Check: 540 ÷ 180 = 3 and 900 ÷ 180 = 5, which are co-prime.
(a) 25 and 49
(b) 21 and 35
(c) 18 and 24
(d) 27 and 81
Show Answer & Explanation
Answer: (a) 25 and 49
Explanation:
1. Co-prime numbers have HCF 1.
2. 25 = 5² and 49 = 7² share no 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 25 and 49.
Teacher's Note:
1. Two squares of different primes are always co-prime.
2. When one number divides the other (27 and 81), the HCF is the smaller number.
(a) 2
(b) 3
(c) 8
(d) 6
Show Answer & Explanation
Answer: (c) 8
Explanation:
1. 210 = 2 × 3 × 5 × 7 and 840 = 2³ × 3 × 5 × 7.
2. The LCM has 2³ but 210 has only one 2, so the other number must contain 2³ = 8.
3. LCM(210, 8) = 2³ × 3 × 5 × 7 = 840. ✓
4. With 2, 3 or 6 the LCM stays 210.
5. So the other number can be 8.
Teacher's Note:
1. The missing power of a prime in the LCM must come from the other number.
2. 840 ÷ 210 = 4, but 4 does not work: LCM(210, 4) = 420.
(a) 60
(b) 48
(c) 100
(d) 120
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. 100 = 2² × 5²: 3 × 3 = 9. 120 = 2³ × 3 × 5: 4 × 2 × 2 = 16.
4. So 60 has exactly 12 factors.
Teacher's Note:
1. Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60, which is 12 factors.
2. Listing in pairs (1 × 60, 2 × 30, …) is a good check.
(a) 84
(b) 144
(c) 112
(d) 125
Show Answer & Explanation
Answer: (d) 125
Explanation:
1. 1008 = 2⁴ × 3² × 7.
2. 84 = 2² × 3 × 7, 144 = 2⁴ × 3² and 112 = 2⁴ × 7 all fit inside 2⁴ × 3² × 7.
3. 125 = 5³ needs the prime 5, which 1008 does not have.
4. So 125 is not a factor of 1008.
Teacher's Note:
1. Every prime power of a factor must be present in the number.
2. Check by division: 1008 ÷ 144 = 7 and 1008 ÷ 112 = 9.
(a) 35
(b) 56
(c) 63
(d) 70
Show Answer & Explanation
Answer: (b) 56
Explanation:
1. Write the numbers as 5k and 8k. Since 5 and 8 are co-prime, HCF = k = 7.
2. The numbers are 5 × 7 = 35 and 8 × 7 = 56.
3. Check: HCF(35, 56) = 7. ✓
4. The larger number is 56.
Teacher's Note:
1. The HCF is the common multiplier of the ratio terms.
2. Read carefully whether the larger or the smaller number is asked.
Assertion (A): The HCF of (15 × 12) and (15 × 20) is greater than 15.
Reason (R): Both numbers contain 15 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: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. HCF(180, 300) = 15 × HCF(12, 20) = 15 × 4 = 60, which is greater than 15. So A is true.
2. Both numbers are multiples of 15, so R is true.
3. R only shows that the HCF is at least 15. It is greater than 15 because 12 and 20 also share the factor 4.
4. So the answer is (b) Both A and R are true but R is not the correct explanation of A.
Teacher's Note:
1. 15 × 2 and 15 × 3 also share 15, but their HCF is exactly 15.
2. R must be enough on its own to make A true for option (a).
Assertion (A): If two numbers that are both multiples of 6 are divided by 6, their LCM also gets divided by 6.
Reason (R): Dividing both numbers by 6 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(36, 60) = 180; after dividing by 6, LCM(6, 10) = 30 = 180 ÷ 6. So A is true.
2. 6 = 2 × 3, so dividing both numbers by 6 lowers the power of 2 and of 3 in each number by one.
3. The LCM uses the highest powers, so its powers of 2 and 3 also drop by one, i.e. the LCM is divided by 2 × 3 = 6. R is true.
4. R explains A, so the answer is (a).
Teacher's Note:
1. Think of 6 as 2 × 3 and follow each prime separately.
2. The numbers must be multiples of 6, otherwise dividing by 6 does not give whole numbers.
Section B: Short Answer Questions (2 Marks Each)
Question 11. (a) Write the prime factorisation of 360 and 540.
(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.
540 = 2 × 2 × 3 × 3 × 3 × 5 = 2² × 3³ × 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: 360 and 540 share 2², 3² and 5, so the HCF is 4 × 9 × 5 = 180.
Teacher's Note:
1. 360 has 2³ and 540 has 2², so the HCF takes 2².
2. Check: 360 ÷ 180 = 2 and 540 ÷ 180 = 3.
Question 12. (a) Write the prime factorisation of 168 and 210.
(b) Find their LCM.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
168 = 2 × 2 × 2 × 3 × 7 = 2³ × 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 = 2³ × 3 × 5 × 7 = 840.
In simple words: The LCM uses 2³, 3, 5 and 7, which gives 840.
Teacher's Note:
1. 5 appears only in 210 but must still be included in the LCM.
2. Check: 840 ÷ 168 = 5 and 840 ÷ 210 = 4.
Question 13. (a) Write the prime factorisation of 196 and 294.
(b) Find their HCF.
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 common to both: 2, 7.
HCF = product of the common primes, each with its smallest power = 2 × 7² = 98.
In simple words: 196 and 294 share one 2 and 7², so the HCF is 2 × 49 = 98.
Teacher's Note:
1. Both numbers contain 7², so the HCF keeps the full 7².
2. Check: 196 ÷ 98 = 2 and 294 ÷ 98 = 3.
Question 14. (a) Write the prime factorisation of 150 and 240.
(b) Find their LCM.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
150 = 2 × 3 × 5 × 5 = 2 × 3 × 5².
240 = 2 × 2 × 2 × 2 × 3 × 5 = 2⁴ × 3 × 5.
(b) Primes that appear in either number: 2, 3, 5.
LCM = product of every prime, each with its greatest power = 2⁴ × 3 × 5² = 1200.
In simple words: Take 2⁴, 3 and 5²: 16 × 3 × 25 = 1200.
Teacher's Note:
1. The power of 2 comes from 240 and the power of 5 comes from 150.
2. Check: 1200 ÷ 150 = 8 and 1200 ÷ 240 = 5.
Section C: Short Answer Questions (3 Marks Each)
Question 15. (a) Write the prime factorisation of 420 and 630.
(b) Find their HCF.
(c) Find their LCM.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
420 = 2 × 2 × 3 × 5 × 7 = 2² × 3 × 5 × 7.
630 = 2 × 3 × 3 × 5 × 7 = 2 × 3² × 5 × 7.
(b) Common primes with smallest powers: 2, 3, 5, 7.
HCF = product of the common primes, each with its smallest power = 2 × 3 × 5 × 7 = 210.
(c) All primes with greatest powers: 2², 3², 5, 7.
LCM = product of every prime, each with its greatest power = 2² × 3² × 5 × 7 = 1260.
Check: HCF × LCM = 210 × 1260 = 264600 and 420 × 630 = 264600. They match.
In simple words: For 420 and 630 the HCF is 210 and the LCM is 1260.
Teacher's Note:
1. Write both factorisations one below the other so the powers are easy to compare.
2. The product check HCF × LCM = 420 × 630 confirms the work.
Question 16. (a) Write the prime factorisation of 72, 90 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:
72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3².
90 = 2 × 3 × 3 × 5 = 2 × 3² × 5.
180 = 2 × 2 × 3 × 3 × 5 = 2² × 3² × 5.
(b) Compare the power of each prime in the three numbers:
Prime 2: 2³ in 72, 2¹ in 90, 2² in 180; highest power 2³.
Prime 3: 3² in 72, 3² in 90, 3² in 180; highest power 3².
Prime 5: none in 72, 5¹ in 90, 5¹ in 180; highest power 5.
(c) LCM = 2³ × 3² × 5 = 360.
In simple words: The greatest powers 2³, 3² and 5 give LCM = 360.
Teacher's Note:
1. All three numbers share 3², so the LCM gets just 3².
2. Check: 360 is divisible by 72, 90 and 180.
Question 17. (a) Write the prime factorisation of 504 and 756.
(b) Identify the common prime factors.
(c) Find their HCF.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
504 = 2 × 2 × 2 × 3 × 3 × 7 = 2³ × 3² × 7.
756 = 2 × 2 × 3 × 3 × 3 × 7 = 2² × 3³ × 7.
(b) The primes 2, 3, 7 divide both numbers. Taking the smaller power of each, the common prime factors are 2² × 3² × 7.
(c) HCF = 2² × 3² × 7 = 252.
In simple words: 504 and 756 share 2², 3² and 7, so their HCF is 252.
Teacher's Note:
1. 504 has 2³ and 756 has 3³; the HCF takes only 2² and 3².
2. Check: 504 ÷ 252 = 2 and 756 ÷ 252 = 3.
Section D: Long Answer Question (5 Marks)
Question 18. The LCM of two numbers is 1680 and their HCF is 120. The difference of the two numbers is 600.
(a) Find the product of the two numbers.
(b) Let the numbers be 120x and 120y. 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 = 120 × 1680 = 201600.
(b) Both numbers are multiples of the HCF, so write them as 120x and 120y, where x and y are co-prime (otherwise the HCF would be bigger than 120).
Then LCM = 120 × x × y = 1680, so x × y = 1680 ÷ 120 = 14.
Difference: 120y − 120x = 600, so y − x = 600 ÷ 120 = 5.
Co-prime pairs with product 14 (smaller first): (1, 14), (2, 7).
The pair whose difference is 5 is (2, 7). So x = 2 and y = 7.
(c) The numbers are 120 × 2 = 240 and 120 × 7 = 840. So the numbers are 240 and 840.
(d) 240 = 2 × 2 × 2 × 2 × 3 × 5 = 2⁴ × 3 × 5 and 840 = 2 × 2 × 2 × 3 × 5 × 7 = 2³ × 3 × 5 × 7.
HCF = 2³ × 3 × 5 = 120 and LCM = 2⁴ × 3 × 5 × 7 = 1680.
HCF × LCM = 120 × 1680 = 201600 and 240 × 840 = 201600. Both are equal, so the result is verified.
(e) 840 − 240 = 600, which is the given difference. ✓
In simple words: The numbers are 240 and 840: HCF 120, LCM 1680 and difference 600.
Teacher's Note:
1. x and y must be co-prime; that keeps the HCF equal to the given value.
2. The difference condition picks one pair from all the possible pairs.
Section E: Case Study Based Questions (4 Marks Each)
Question 19. Case Study 1: Coordinating School Bus Arrivals
Two school buses arrive at the school gate at regular intervals. Bus A arrives every 30 minutes, while Bus B arrives every 45 minutes. Both buses arrive together at 7:30 a.m. The transport coordinator wants to schedule inspection when both buses arrive together again. Use mathematical reasoning to find the next common arrival time.
Answer the following questions:
(a) Which concept is used?
(b) Find the required value.
(c) After how many minutes will both buses arrive 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 buses arrive together again after 90 minutes (1 hour 30 minutes).
(d) 7:30 a.m. + 1 hour 30 minutes = 9:00 a.m.
In simple words: The buses meet at the gate every 90 minutes, so next at 9:00 a.m.
Teacher's Note:
1. 90 minutes is 1 hour 30 minutes; adding it to 7:30 a.m. gives 9:00 a.m.
2. LCM gives the first common time; later ones are its multiples (180, 270 minutes …).
Question 20. Case Study 2: Packing Relief Food Kits
A relief organization has 360 kg of rice and 540 kg of wheat. The team wants to pack these into the maximum possible number of identical food kits, with no food wasted. 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 much rice and wheat will each kit contain?
Answer:
(a) The number of identical kits 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 kits = 180.
(d) Each kit gets 360 ÷ 180 = 2 kg rice and 540 ÷ 180 = 3 kg wheat. So each kit contains 2 kg rice and 3 kg wheat.
Check: 180 × 2 = 360 and 180 × 3 = 540, so nothing is left over.
In simple words: 180 kits can be packed, each with 2 kg rice and 3 kg wheat.
Teacher's Note:
1. Identical kits with nothing wasted means the number of kits divides both amounts.
2. Do not write 180 kg in each kit; 180 is the number of kits.
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