CBSE Class 7 Mathematics Finding Common Ground Practice Paper Assignment Set 02 with Answers

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

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Practice Class 7 Mathematics Assignments: Chapter 03 Finding Common Ground

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Class 7 Maths Ganita Prakash Part 2 Chapter 3 Finding Common Ground Practice Paper 32 is a 40-mark test of 20 questions in Sections A to E, and each question below is followed by a step-by-step answer. It practises prime factorisation of numbers such as 924, 1540 and 1260, finding the HCF and LCM by comparing powers of primes, co-prime pairs, counting the factors of a number and using HCF × LCM = product of two numbers. The case studies apply LCM to two school bells and HCF to packing identical sports kits.

Section A: Multiple Choice Questions (1 Mark Each)

Question 1: Two lights blink every 28 minutes and 42 minutes. If they blink together now, after how many minutes will they blink together again?
(a) 84
(b) 168
(c) 196
(d) 336
Show Answer & Explanation

Answer: (a) 84

Explanation:
1. The lights blink together at a time that is a multiple of both 28 and 42, so we need the LCM.
2. 28 = 2² × 7 and 42 = 2 × 3 × 7.
3. LCM = 2² × 3 × 7 = 84.
4. So they blink together again after 84 minutes.

Teacher's Note:
1. "Together again" questions always need the LCM, the smallest common multiple.
2. 168 and 336 are also common multiples, but not the first one.

Question 2: Which number has prime factorisation 2⁴ × 3 × 7?
(a) 336
(b) 672
(c) 1008
(d) 1344
Show Answer & Explanation

Answer: (a) 336

Explanation:
1. 2⁴ = 2 × 2 × 2 × 2 = 16.
2. 16 × 3 = 48.
3. 48 × 7 = 336.
4. So the number is 336.

Teacher's Note:
1. Find the value of the power first, then multiply.
2. 672 is 2⁵ × 3 × 7; one extra 2 doubles the number.

Question 3: Find the HCF of 924 and 1540.
(a) 344
(b) 308
(c) 288
(d) 154
Show Answer & Explanation

Answer: (b) 308

Explanation:
1. 924 = 2² × 3 × 7 × 11.
2. 1540 = 2² × 5 × 7 × 11.
3. Common primes with their smaller powers: 2², 7 and 11.
4. HCF = 2² × 7 × 11 = 4 × 77 = 308.

Teacher's Note:
1. Take only the primes that appear in both numbers.
2. Check: 924 ÷ 308 = 3 and 1540 ÷ 308 = 5, and 3 and 5 are co-prime.

Question 4: Which of the following pairs is co-prime?
(a) 49 and 81
(b) 21 and 28
(c) 18 and 27
(d) 35 and 70
Show Answer & Explanation

Answer: (a) 49 and 81

Explanation:
1. Two numbers are co-prime if their HCF is 1, that is, they share no prime factor.
2. 49 = 7² and 81 = 3⁴: no common prime, HCF = 1.
3. 21 and 28 share 7; 18 and 27 share 9; 35 and 70 share 35.
4. So the co-prime pair is 49 and 81.

Teacher's Note:
1. Co-prime numbers need not be prime themselves; 49 and 81 are both composite.
2. Look for any common prime factor; one is enough to rule out a pair.

Question 5: The LCM of two numbers is 1320 and one of the numbers is 330. Which of the following could be the other number?
(a) 2
(b) 3
(c) 8
(d) 6
Show Answer & Explanation

Answer: (c) 8

Explanation:
1. 330 = 2 × 3 × 5 × 11 and 1320 = 2³ × 3 × 5 × 11.
2. The LCM has 2³ but 330 has only 2¹, so the other number must contain 2³ = 8.
3. LCM(330, 8) = 2³ × 3 × 5 × 11 = 1320. ✓
4. LCM(330, 2), LCM(330, 3) and LCM(330, 6) are all 330, not 1320.
5. So the other number can be 8.

Teacher's Note:
1. The LCM takes the highest power of each prime, so the missing power of 2 must come from the other number.
2. The other number is not fixed (24, 40 or 120 would also work); that is why the question asks which option could be the number.

Question 6: Which number has exactly 18 factors?
(a) 180
(b) 240
(c) 360
(d) 420
Show Answer & Explanation

Answer: (a) 180

Explanation:
1. If n = pa × qb × …, the number of factors is (a + 1)(b + 1)…
2. 180 = 2² × 3² × 5: (2 + 1)(2 + 1)(1 + 1) = 18.
3. 240 = 2⁴ × 3 × 5: 5 × 2 × 2 = 20. 360 = 2³ × 3² × 5: 4 × 3 × 2 = 24. 420 = 2² × 3 × 5 × 7: 3 × 2 × 2 × 2 = 24.
4. So the number with exactly 18 factors is 180.

Teacher's Note:
1. Each factor picks a power of every prime, from 0 up to the power in the number; that is why we add 1 to each power.
2. Bigger numbers do not always have more factors.

Question 7: Which of the following is not a factor of 1728?
(a) 72
(b) 96
(c) 108
(d) 125
Show Answer & Explanation

Answer: (d) 125

Explanation:
1. 1728 = 2⁶ × 3³. Its factors can contain only the primes 2 and 3.
2. 72 = 2³ × 3², 96 = 2⁵ × 3 and 108 = 2² × 3³ all fit inside 2⁶ × 3³.
3. 125 = 5³ contains the prime 5, which 1728 does not have.
4. So 125 is not a factor of 1728.

Teacher's Note:
1. A number is a factor only if its primes and powers are all present in the bigger number.
2. 1728 = 12 × 12 × 12 is a useful cube to remember.

Question 8: If two numbers are in the ratio 4 : 7 and their HCF is 9, find the larger number.
(a) 28
(b) 36
(c) 49
(d) 63
Show Answer & Explanation

Answer: (d) 63

Explanation:
1. Numbers in the ratio 4 : 7 can be written as 4k and 7k. As 4 and 7 are co-prime, their HCF is k.
2. So k = 9.
3. The numbers are 4 × 9 = 36 and 7 × 9 = 63.
4. The larger number is 63.

Teacher's Note:
1. The HCF is the common multiplier only when the ratio is in its lowest terms.
2. 36 is the smaller number, a common trap.

Question 9:
Assertion (A): The HCF of (24 × 18) and (24 × 30) is greater than 24.
Reason (R): Both numbers contain 24 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(24 × 18, 24 × 30) = 24 × HCF(18, 30) = 24 × 6 = 144, which is greater than 24. So A is true.
2. Both numbers are multiples of 24, so R is true.
3. But R only tells us the HCF is at least 24. It is greater than 24 because 18 and 30 also share the factor 6.
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. A common factor 24 alone would allow the HCF to be exactly 24 (for example 24 × 2 and 24 × 3).
2. Ask whether R by itself is enough to make A true.

Question 10:
Assertion (A): If two numbers are multiplied by 4, their LCM also gets multiplied by 4.
Reason (R): Multiplying both numbers by 4 increases the highest power of prime factor 2 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(6, 10) = 30 and LCM(24, 40) = 120 = 4 × 30. So A is true.
2. 4 = 2², so multiplying both numbers by 4 raises the power of 2 in each number by 2, and the other primes stay the same.
3. The LCM uses the highest power of each prime, so its power of 2 also goes up by 2, i.e. the LCM is multiplied by 2² = 4. R is true.
4. R explains why A happens, so the answer is (a).

Teacher's Note:
1. Work with prime powers to see what happens to the LCM.
2. Try a quick example to check an assertion.

Section B: Short Answer Questions (2 Marks Each)

Question 11. (a) Write the prime factorisation of 924 and 1540.
(b) Find their HCF.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
924 = 2 × 2 × 3 × 7 × 11 = 2² × 3 × 7 × 11.
1540 = 2 × 2 × 5 × 7 × 11 = 2² × 5 × 7 × 11.
(b) Primes common to both: 2, 7, 11.
HCF = product of the common primes, each with its smallest power = 2² × 7 × 11 = 308.
In simple words: Both numbers share 2², 7 and 11, so their HCF is 4 × 7 × 11 = 308.

Teacher's Note:
1. Use the smaller power of each common prime for the HCF.
2. A common slip is to leave out 2² and get 77; multiply every common prime power.

 

Question 12. (a) Write the prime factorisation of 288 and 360.
(b) Find their LCM.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
288 = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 2⁵ × 3².
360 = 2 × 2 × 2 × 3 × 3 × 5 = 2³ × 3² × 5.
(b) Primes that appear in either number: 2, 3, 5.
LCM = product of every prime, each with its greatest power = 2⁵ × 3² × 5 = 1440.
In simple words: Take 2⁵, 3² and 5 together: 32 × 9 × 5 = 1440.

Teacher's Note:
1. For LCM take every prime, with the greatest power found in any number.
2. Check: 1440 ÷ 288 = 5 and 1440 ÷ 360 = 4.

 

Question 13. (a) Write the prime factorisation of 420 and 630.
(b) 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.
630 = 2 × 3 × 3 × 5 × 7 = 2 × 3² × 5 × 7.
(b) Primes common to both: 2, 3, 5, 7.
HCF = product of the common primes, each with its smallest power = 2 × 3 × 5 × 7 = 210.
In simple words: 420 and 630 share 2, 3, 5 and 7 once each, so the HCF is 210.

Teacher's Note:
1. 420 has 2² but 630 has only 2, so the HCF gets just one 2.
2. Check: 420 ÷ 210 = 2 and 630 ÷ 210 = 3.

 

Question 14. (a) Write the prime factorisation of 175 and 280.
(b) Find their LCM.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
175 = 5 × 5 × 7 = 5² × 7.
280 = 2 × 2 × 2 × 5 × 7 = 2³ × 5 × 7.
(b) Primes that appear in either number: 2, 5, 7.
LCM = product of every prime, each with its greatest power = 2³ × 5² × 7 = 1400.
In simple words: Multiply 2³, 5² and 7: 8 × 25 × 7 = 1400.

Teacher's Note:
1. The prime 2 is only in 280, but it still goes into the LCM.
2. Check: 1400 ÷ 175 = 8 and 1400 ÷ 280 = 5.

 

Section C: Short Answer Questions (3 Marks Each)

Question 15. (a) Write the prime factorisation of 600 and 900.
(b) Find their HCF.
(c) Find their LCM.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
600 = 2 × 2 × 2 × 3 × 5 × 5 = 2³ × 3 × 5².
900 = 2 × 2 × 3 × 3 × 5 × 5 = 2² × 3² × 5².
(b) Common primes with smallest powers: 2², 3, 5².
HCF = product of the common primes, each with its smallest power = 2² × 3 × 5² = 300.
(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 = 300 × 1800 = 540000 and 600 × 900 = 540000. They match.
In simple words: 600 and 900 have HCF 300 and LCM 1800.

Teacher's Note:
1. HCF uses smaller powers, LCM uses greater powers of the same primes.
2. HCF × LCM = product of the two numbers is a quick check.

 

Question 16. (a) Write the prime factorisation of 72, 108 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.
108 = 2 × 2 × 3 × 3 × 3 = 2² × 3³.
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, 2² in 108, 2² in 180; highest power 2³.
Prime 3: 3² in 72, 3³ in 108, 3² in 180; highest power 3³.
Prime 5: none in 72, none in 108, 5¹ in 180; highest power 5.
(c) LCM = 2³ × 3³ × 5 = 1080.
In simple words: The highest powers are 2³, 3³ and 5, so the LCM is 8 × 27 × 5 = 1080.

Teacher's Note:
1. With three numbers, compare the power of each prime in all three.
2. A prime present in only one number (here 5) is still part of the LCM.

 

Question 17. (a) Write the prime factorisation of 840 and 1260.
(b) Identify the common prime factors.
(c) Find their HCF.
Answer:
(a) Divide repeatedly by the smallest prime that divides the number:
840 = 2 × 2 × 2 × 3 × 5 × 7 = 2³ × 3 × 5 × 7.
1260 = 2 × 2 × 3 × 3 × 5 × 7 = 2² × 3² × 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: 840 and 1260 share 2², 3, 5 and 7, so their HCF is 420.

Teacher's Note:
1. List the primes of each number first, then pick the smaller power of the shared ones.
2. Check: 840 ÷ 420 = 2 and 1260 ÷ 420 = 3.

 

Section D: Long Answer Question (5 Marks)

Question 18. The LCM of two numbers is 1800 and their HCF is 60. The difference of the two numbers is 780.
(a) Find the product of the two numbers.
(b) Let the numbers be 60x and 60y. 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 = 60 × 1800 = 108000.
(b) Both numbers are multiples of the HCF, so write them as 60x and 60y, where x and y are co-prime (otherwise the HCF would be bigger than 60).
Then LCM = 60 × x × y = 1800, so x × y = 1800 ÷ 60 = 30.
Difference: 60y − 60x = 780, so y − x = 780 ÷ 60 = 13.
Co-prime pairs with product 30 (smaller first): (1, 30), (2, 15), (3, 10), (5, 6).
The pair whose difference is 13 is (2, 15). So x = 2 and y = 15.
(c) The numbers are 60 × 2 = 120 and 60 × 15 = 900. So the numbers are 120 and 900.
(d) 120 = 2 × 2 × 2 × 3 × 5 = 2³ × 3 × 5 and 900 = 2 × 2 × 3 × 3 × 5 × 5 = 2² × 3² × 5².
HCF = 2² × 3 × 5 = 60 and LCM = 2³ × 3² × 5² = 1800.
HCF × LCM = 60 × 1800 = 108000 and 120 × 900 = 108000. Both are equal, so the result is verified.
(e) 900 − 120 = 780, which is the given difference. ✓
In simple words: The two numbers are 120 and 900: HCF 60, LCM 1800 and difference 780.

Teacher's Note:
1. Writing the numbers as HCF × x and HCF × y with co-prime x, y is the key idea.
2. Always check all the given conditions at the end.

 

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

Question 19. Case Study 1: Synchronizing Classroom Periods
Two school bells ring at regular intervals. Bell A rings every 40 minutes, while Bell B rings every 60 minutes. Both bells ring together at 9:00 a.m. The school administration wants to schedule a special announcement when both bells ring together again. Use mathematical reasoning to determine the time interval and the exact time of their next simultaneous ringing.
Answer the following questions:
(a) Which mathematical concept should be used?
(b) Find the required value using that concept.
(c) After how many minutes will both bells ring together again?
(d) At what exact time will this occur?
Answer:
(a) We need the first time that is a multiple of both 40 and 60 minutes, i.e. the smallest common multiple. So the concept is LCM (Lowest Common Multiple).
(b) 40 = 2 × 2 × 2 × 5 = 2³ × 5 and 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5.
LCM = 2³ × 3 × 5 = 120.
(c) Both bells ring together again after 120 minutes (2 hours).
(d) 9:00 a.m. + 2 hours = 11:00 a.m.
In simple words: The bells ring together every 120 minutes, so next at 11:00 a.m.

Teacher's Note:
1. Repeating events that must happen together point to LCM.
2. Convert minutes into hours and minutes before adding to a clock time.

 

Question 20. Case Study 2: Preparing Sports Kits
A sports department has 480 balls and 720 bats to distribute to various schools. The organizers want to prepare the maximum possible number of identical sports kits, using all the balls and bats without any items left unused. This ensures equal distribution and efficient use of resources. Use mathematical reasoning to determine the number of kits and the contents of each kit.
Answer the following questions:
(a) Which mathematical concept should be applied?
(b) Find the required value using that concept.
(c) How many sports kits can be prepared?
(d) How many balls and bats will each kit contain?
Answer:
(a) The number of identical kits 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 kits = 240.
(d) Each kit gets 480 ÷ 240 = 2 balls and 720 ÷ 240 = 3 bats. So each kit contains 2 balls and 3 bats.
Check: 240 × 2 = 480 and 240 × 3 = 720, so nothing is left over.
In simple words: 240 kits can be made, each with 2 balls and 3 bats.

Teacher's Note:
1. Equal sharing with nothing left over, as many groups as possible, points to HCF.
2. The HCF is the number of kits; divide each total by it to get the contents of one kit.

 

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