Class 7 Mathematics Practice Sheet: CBSE Class 7 Mathematics Finding Common Ground Practice Paper Worksheet Set 03 with Answers
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This Class 7 Maths Ganita Prakash Part 2 Chapter 3 Finding Common Ground Practice Paper 35 has 20 questions worth 40 marks, split into Sections A to E, and every answer is worked out in steps. It covers the LCM of 18 and 27, the HCF of 432 and 540, numbers with exactly 16 factors, factors of 1000, how multiplying two numbers by 3 changes their LCM, and finding a pair of numbers from their HCF and LCM. In the case studies, students time two school buses leaving a depot and share shirts and trousers into identical uniform sets.
Section A: Multiple Choice Questions (1 Mark Each)
(a) 54
(b) 72
(c) 81
(d) 108
Show Answer & Explanation
Answer: (a) 54
Explanation:
1. We need the LCM of 18 and 27.
2. 18 = 2 × 3² and 27 = 3³.
3. LCM = 2 × 3³ = 2 × 27 = 54.
4. So they ring together again after 54 minutes.
Teacher's Note:
1. 72 is a multiple of 18 but not of 27, and 81 is a multiple of 27 but not of 18.
2. 108 is a common multiple, but twice the lowest one.
(a) 504
(b) 336
(c) 672
(d) 756
Show Answer & Explanation
Answer: (a) 504
Explanation:
1. 2³ = 8 and 3² = 9.
2. 8 × 9 = 72.
3. 72 × 7 = 504.
4. So the number is 504.
Teacher's Note:
1. 756 = 2² × 3³ × 7: the powers of 2 and 3 are swapped.
2. Work out the powers before multiplying.
(a) 54
(b) 72
(c) 108
(d) 216
Show Answer & Explanation
Answer: (c) 108
Explanation:
1. 432 = 2⁴ × 3³ and 540 = 2² × 3³ × 5.
2. Common primes with smaller powers: 2² and 3³.
3. HCF = 4 × 27 = 108.
4. So the HCF is 108.
Teacher's Note:
1. 5 is only in 540, so it is not part of the HCF.
2. Check: 432 ÷ 108 = 4 and 540 ÷ 108 = 5.
(a) 35 and 64
(b) 21 and 49
(c) 24 and 36
(d) 18 and 27
Show Answer & Explanation
Answer: (a) 35 and 64
Explanation:
1. Co-prime means HCF = 1.
2. 35 = 5 × 7 and 64 = 2⁶ share no prime, so HCF = 1.
3. 21 and 49 share 7; 24 and 36 share 12; 18 and 27 share 9.
4. So the co-prime pair is 35 and 64.
Teacher's Note:
1. An odd number and a power of 2 are always co-prime.
2. One common prime is enough to make a pair not co-prime.
(a) 2
(b) 8
(c) 6
(d) 16
Show Answer & Explanation
Answer: (b) 8
Explanation:
1. 210 = 2 × 3 × 5 × 7 and 840 = 2³ × 3 × 5 × 7.
2. The other number must supply 2³ = 8, but not more than 2³.
3. LCM(210, 8) = 840 ✓; LCM(210, 16) = 1680 (too many 2s); with 2 or 6 the LCM stays 210.
4. So the other number can be 8.
Teacher's Note:
1. The other number must have exactly the missing power of 2: not less and not more.
2. 16 = 2⁴ would push the LCM to 2⁴ × 3 × 5 × 7 = 1680.
(a) 120
(b) 144
(c) 180
(d) 240
Show Answer & Explanation
Answer: (a) 120
Explanation:
1. Number of factors = product of (each power + 1).
2. 120 = 2³ × 3 × 5: 4 × 2 × 2 = 16.
3. 144 = 2⁴ × 3²: 5 × 3 = 15. 180 = 2² × 3² × 5: 3 × 3 × 2 = 18. 240 = 2⁴ × 3 × 5: 5 × 2 × 2 = 20.
4. So 120 has exactly 16 factors.
Teacher's Note:
1. Write the prime factorisation first; the factor count comes straight from the powers.
2. 144 is a perfect square, so it has an odd number of factors.
(a) 40
(b) 50
(c) 125
(d) 64
Show Answer & Explanation
Answer: (d) 64
Explanation:
1. 1000 = 2³ × 5³.
2. 40 = 2³ × 5, 50 = 2 × 5² and 125 = 5³ all fit inside 2³ × 5³.
3. 64 = 2⁶ needs six 2s, but 1000 has only three.
4. So 64 is not a factor of 1000.
Teacher's Note:
1. 1000 = 10³ = 2³ × 5³.
2. 1000 ÷ 64 = 15.625, which is not a whole number.
(a) 30
(b) 54
(c) 60
(d) 72
Show Answer & Explanation
Answer: (b) 54
Explanation:
1. Write the numbers as 5k and 9k. As 5 and 9 are co-prime, HCF = k = 6.
2. The numbers are 5 × 6 = 30 and 9 × 6 = 54.
3. Check: HCF(30, 54) = 6. ✓
4. The larger number is 54.
Teacher's Note:
1. The ratio terms must be co-prime for k to be the HCF.
2. 30 is the smaller number.
Assertion (A): The HCF of (15 × 12) and (15 × 18) is greater than 15.
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. 15 × 12 = 180 and 15 × 18 = 270. HCF = 15 × HCF(12, 18) = 15 × 6 = 90, which is greater than 15. So A is true.
2. 180 = 90 × 2 and 270 = 90 × 3, so 90 is a common factor. R is true.
3. A common factor of 90 means the HCF is at least 90, so it must be greater than 15. R explains A.
4. The answer is (a).
Teacher's Note:
1. The HCF is the biggest common factor, so it is at least as big as any common factor.
2. Here 90 is in fact the HCF itself.
Assertion (A): If two numbers are multiplied by 3, their LCM also gets multiplied by 3.
Reason (R): Multiplying both numbers by 3 increases the highest power of prime factor 3 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(12, 18) = 36 and LCM(36, 54) = 108 = 3 × 36. So A is true.
2. Multiplying both numbers by 3 raises the power of 3 in each number by one; other primes do not change.
3. So the highest power of 3 in the LCM rises by one, which multiplies the LCM by 3. R is true.
4. R explains A, so the answer is (a).
Teacher's Note:
1. Multiplying both numbers by the same number multiplies the LCM by that number.
2. Check with small numbers like 2 and 3: LCM 6, and for 6 and 9 the LCM is 18.
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: The common part of 540 and 720 is 2² × 3² × 5 = 180.
Teacher's Note:
1. Write the smaller power for each shared prime: 2² from 540 and 3² from 720.
2. 180 × 3 = 540 and 180 × 4 = 720.
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: LCM = 2² × 3 × 7² = 588.
Teacher's Note:
1. Use 2² from 196 and 3 from 294; 7² is in both.
2. Check: 588 is 3 × 196 and 2 × 294.
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: 360 and 450 have HCF 2 × 9 × 5 = 90.
Teacher's Note:
1. Only one 2 is common because 450 has a single 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: No prime is shared, so LCM = 144 × 175 = 25200.
Teacher's Note:
1. Co-prime numbers: LCM = product, HCF = 1.
2. Multiply carefully: 144 × 175 = 144 × 100 + 144 × 75 = 14400 + 10800 = 25200.
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: 420 and 630: HCF 210, LCM 1260.
Teacher's Note:
1. HCF divides both numbers; LCM is divisible by both.
2. Check the HCF × LCM rule to catch mistakes.
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 LCM of 72, 90 and 180 is 2³ × 3² × 5 = 360.
Teacher's Note:
1. Only 72 has 2³; that power must be kept.
2. 360 ÷ 72 = 5, 360 ÷ 90 = 4 and 360 ÷ 180 = 2.
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: The shared primes 2², 3² and 7 give HCF 252.
Teacher's Note:
1. Smaller power of 2 is 2² (from 756); smaller power of 3 is 3² (from 504).
2. 252 × 2 = 504 and 252 × 3 = 756.
Section D: Long Answer Question (5 Marks)
Question 18. The LCM of two numbers is 1512 and their HCF is 36. The difference of the two numbers is 36.
(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.
Answer:
(a) For two numbers, Product = HCF × LCM = 36 × 1512 = 54432.
(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 = 1512, so x × y = 1512 ÷ 36 = 42.
Difference: 36y − 36x = 36, so y − x = 36 ÷ 36 = 1.
Co-prime pairs with product 42 (smaller first): (1, 42), (2, 21), (3, 14), (6, 7).
The pair whose difference is 1 is (6, 7). So x = 6 and y = 7.
(c) The numbers are 36 × 6 = 216 and 36 × 7 = 252. So the numbers are 216 and 252.
(d) 216 = 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 3³ and 252 = 2 × 2 × 3 × 3 × 7 = 2² × 3² × 7.
HCF = 2² × 3² = 36 and LCM = 2³ × 3³ × 7 = 1512.
HCF × LCM = 36 × 1512 = 54432 and 216 × 252 = 54432. Both are equal, so the result is verified. Also 252 − 216 = 36, as given.
In simple words: The two numbers are 216 and 252, with HCF 36, LCM 1512 and difference 36.
Teacher's Note:
1. xy = 42 has four co-prime pairs, so an extra condition is needed to get a single answer.
2. Check every condition: HCF, LCM and difference.
Section E: Case Study Based Questions (4 Marks Each)
Question 19. Case Study 1: Synchronizing School Bus Timings
Two school buses leave the depot at regular intervals. Bus A departs every 36 minutes, while Bus B departs every 48 minutes. Both buses depart together at 7:00 a.m. The transport coordinator wants to know when the buses will depart together again to plan route adjustments efficiently.
Answer the following questions:
(a) Which concept is used?
(b) Find the required value.
(c) After how many minutes will both buses depart together again?
(d) At what exact time will this occur?
Answer:
(a) We need the first time that is a multiple of both 36 and 48 minutes, i.e. the smallest common multiple. So the concept is LCM (Lowest Common Multiple).
(b) 36 = 2 × 2 × 3 × 3 = 2² × 3² and 48 = 2 × 2 × 2 × 2 × 3 = 2⁴ × 3.
LCM = 2⁴ × 3² = 144.
(c) Both buses depart together again after 144 minutes (2 hours 24 minutes).
(d) 7:00 a.m. + 2 hours 24 minutes = 9:24 a.m.
In simple words: The buses leave together every 144 minutes, so next at 9:24 a.m.
Teacher's Note:
1. 144 minutes = 2 hours 24 minutes (120 + 24).
2. LCM of 36 and 48 takes 2⁴ from 48 and 3² from 36.
Question 20. Case Study 2: Preparing Uniform Sets
A school receives 480 shirts and 720 trousers for student uniforms. The administration wants to create the maximum possible number of identical uniform sets, without any leftover items. Use mathematical reasoning to determine the number of sets and the contents of each set.
Answer the following questions:
(a) Which concept is used?
(b) Find the required value.
(c) How many uniform sets can be prepared?
(d) How many shirts and trousers will each set contain?
Answer:
(a) The number of identical sets must divide both 480 and 720 exactly, and we want the largest such number. So the concept is HCF (Highest Common Factor).
(b) 480 = 2 × 2 × 2 × 2 × 2 × 3 × 5 = 2⁵ × 3 × 5 and 720 = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 2⁴ × 3² × 5.
HCF = 2⁴ × 3 × 5 = 240.
(c) Number of sets = 240.
(d) Each set gets 480 ÷ 240 = 2 shirts and 720 ÷ 240 = 3 trousers. So each set contains 2 shirts and 3 trousers.
Check: 240 × 2 = 480 and 240 × 3 = 720, so nothing is left over.
In simple words: 240 identical sets can be made, each with 2 shirts and 3 trousers.
Teacher's Note:
1. The HCF gives the largest number of identical sets.
2. Each set's contents are the totals divided by the HCF.
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