CBSE Class 7 Mathematics Finding Common Ground Worksheet Set 01

Very Short Answer Questions

Question 1. Define the Highest Common Factor (HCF) of two numbers.
Answer:
The HCF of two numbers is the greatest of all their common factors, that is, the largest number that divides both of them exactly.

Teacher's Note:
1. Find the common factors first.
2. Then pick the largest one.

Question 2. Write the full form of GCD and state which of the HCF and the LCM it means.
Answer:
GCD stands for Greatest Common Divisor. It is another name for the HCF.

Teacher's Note:
1. Divisor means factor.
2. GCD never means the LCM.

Question 3. List all the common factors of 16 and 40.
Answer:
Factors of 16: 1, 2, 4, 8, 16. Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40.
Common factors: 1, 2, 4 and 8.

Teacher's Note:
1. List factors in pairs so none is missed.
2. 1 is always a common factor.

Question 4. What is a prime number? Write the smallest prime number.
Answer:
A prime is a number greater than 1 whose only factors are 1 and the number itself.
The smallest prime number is 2.

Teacher's Note:
1. 2 is the only even prime.
2. 1 is not a prime.

Question 5. Using the division method, write the prime factorisation of 84.
Answer:
84 ÷ 2 = 42, 42 ÷ 2 = 21, 21 ÷ 3 = 7, and 7 is a prime.
So 84 = 2 × 2 × 3 × 7.

Teacher's Note:
1. Keep dividing by the smallest prime that works.
2. Stop when the quotient is a prime.

Question 6. Define the Lowest Common Multiple (LCM) of two numbers.
Answer:
The LCM of two numbers is the smallest of all their common multiples, that is, the smallest number that is a multiple of both.

Teacher's Note:
1. Common multiples first, then the smallest.
2. Do not confuse with the HCF.

Question 7. Write the first three common multiples of 4 and 6.
Answer:
Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, …
Multiples of 6: 6, 12, 18, 24, 30, 36, …
First three common multiples: 12, 24 and 36.

Teacher's Note:
1. They are all multiples of the LCM, 12.
2. Leave out 0.

Question 8. Find the HCF of 20 and 28.
Answer:
20 = 2 × 2 × 5 and 28 = 2 × 2 × 7.
The common part is 2 × 2, so the HCF is 4.

Teacher's Note:
1. Take only the shared primes.
2. Check: 20 ÷ 4 = 5 and 28 ÷ 4 = 7.

Question 9. Find the LCM of 8 and 14.
Answer:
8 = 2 × 2 × 2 and 14 = 2 × 7.
Each prime the most times it occurs: LCM = 2 × 2 × 2 × 7 = 56.

Teacher's Note:
1. Do not count the shared 2 twice.
2. 56 ÷ 8 = 7 and 56 ÷ 14 = 4.

Question 10. What are co-prime numbers? Give one example of a pair of co-prime numbers.
Answer:
Co-prime numbers are two numbers whose HCF is 1, so their only common factor is 1.
Example: 8 and 9.

Teacher's Note:
1. They need not be primes.
2. Any two consecutive numbers are co-prime.

Question 11. What is the HCF of any two consecutive numbers?
Answer:
The HCF of two consecutive numbers is always 1, because a number bigger than 1 cannot divide two numbers that differ by just 1.

Teacher's Note:
1. A common factor divides the difference too.
2. The difference here is 1.

Question 12. Write the prime factorisation of 121 and state how many prime factors it has.
Answer:
121 = 11 × 11. It has two prime factors, both equal to 11.

Teacher's Note:
1. Count repeated primes each time they appear.
2. 121 is the square of 11.

Question 13. Two ropes of lengths 44 m and 66 m are to be cut into pieces of the same greatest possible whole-number length, with nothing left over. Find the length of each piece.
Answer:
The piece length must be a common factor of 44 and 66; the greatest one is the HCF.
44 = 2 × 2 × 11 and 66 = 2 × 3 × 11, so HCF = 2 × 11 = 22. Each piece is 22 m long.

Teacher's Note:
1. 'Greatest' and 'nothing left over' point to the HCF.
2. Check: 44 ÷ 22 = 2 and 66 ÷ 22 = 3.

Question 14. Using prime factorisation, find the HCF of 36 and 54.
Answer:
36 = 2 × 2 × 3 × 3 and 54 = 2 × 3 × 3 × 3.
Common primes, minimum times: HCF = 2 × 3 × 3 = 18.

Teacher's Note:
1. One 2 and two 3s are shared.
2. HCF uses the minimum count.

Question 15. Using prime factorisation, find the LCM of 36 and 54.
Answer:
36 = 2 × 2 × 3 × 3 and 54 = 2 × 3 × 3 × 3.
Each prime the maximum times: LCM = 2 × 2 × 3 × 3 × 3 = 108.

Teacher's Note:
1. Two 2s from 36, three 3s from 54.
2. Check: 18 × 108 = 1944 = 36 × 54.

Question 16. Why is there no greatest common multiple of two numbers?
Answer:
Multiples never stop, so the common multiples also go on for ever (for 6 and 8: 24, 48, 72, …).
A list that never ends has no greatest member.

Teacher's Note:
1. Add the LCM again to get a bigger one.
2. Only the lowest common multiple exists.

Question 17. Ravi visits the village library every 10 days and Sara visits it every 12 days. If both visited it today, after how many days will they next visit it on the same day?
Answer:
They meet again after the first common multiple of 10 and 12, the LCM.
10 = 2 × 5, 12 = 2 × 2 × 3, so LCM = 2 × 2 × 3 × 5 = 60. They meet again after 60 days.

Teacher's Note:
1. 'Next time together' is an LCM question.
2. 60 ÷ 10 = 6 and 60 ÷ 12 = 5.

Question 18. The HCF of two numbers is 9 and their LCM is 108. Find the product of the two numbers.
Answer:
Product = HCF × LCM = 9 × 108 = 972.

Teacher's Note:
1. HCF × LCM = product of the two numbers.
2. We do not need the numbers themselves.

Question 19. What happens to the HCF of two numbers when both the numbers are doubled? Give one example.
Answer:
The HCF also gets doubled, because each number gains one extra factor 2.
Example: HCF(6, 8) = 2, while HCF(12, 16) = 4.

Teacher's Note:
1. The extra 2 is common to both.
2. So it joins the HCF.

Question 20. Give a counterexample to the claim that the larger a number is, the longer its prime factorisation will be.
Answer:
Take 96 and 121. 96 = 2 × 2 × 2 × 2 × 2 × 3 has six prime factors.
The larger number 121 = 11 × 11 has only two. So the claim is false.

Teacher's Note:
1. One counterexample is enough to disprove a claim.
2. Big primes give short factorisations.

Question 21. Is 2 × 3 × 3 a factor of 2 × 2 × 3 × 5 × 7? Give a reason for your answer.
Answer:
No. A factor must be a subpart of the prime factorisation.
2 × 2 × 3 × 5 × 7 has only one 3, but 2 × 3 × 3 needs two 3s.

Teacher's Note:
1. Compare the count of each prime.
2. Check: 420 ÷ 18 is not a whole number.

Question 22. Find the HCF of 7 and 42, and state why it is one of the two given numbers.
Answer:
HCF(7, 42) = 7. It is one of the numbers because 7 is a factor of 42 (42 = 7 × 6).
When one number divides the other, the smaller number is their HCF.

Teacher's Note:
1. 7 divides itself and 42.
2. No bigger number divides 7.

Question 23. While finding the HCF of 84 and 132 by repeated division, the divisors used on the left were 2, 2 and 3, and the last pair of quotients was 7 and 11. Write the HCF and the LCM of 84 and 132.
Answer:
HCF = product of the divisors = 2 × 2 × 3 = 12.
LCM = 12 × 7 × 11 = 924.

Teacher's Note:
1. Left column gives the HCF.
2. Multiply by the last quotients for the LCM.

Question 24. Find the smallest number of chocolates that can be shared equally among 4 children, among 6 children, or among 9 children, with none left over.
Answer:
The number must be a common multiple of 4, 6 and 9; the smallest is the LCM.
4 = 2 × 2, 6 = 2 × 3, 9 = 3 × 3, so LCM = 2 × 2 × 3 × 3 = 36. 36 chocolates are needed.

Teacher's Note:
1. Shared equally with none left means a multiple.
2. Smallest means LCM.

Question 25. The HCF of two numbers is 6 and their LCM is 72. If one of the numbers is 24, find the other number.
Answer:
Product = HCF × LCM = 6 × 72 = 432.
Other number = 432 ÷ 24 = 18. Check: HCF(24, 18) = 6 and LCM(24, 18) = 72.

Teacher's Note:
1. Find the product first.
2. Always check both HCF and LCM.

Question 26. What is the HCF of two consecutive even numbers? Explain why it cannot be 4.
Answer:
The HCF is always 2.
Any common factor must divide their difference, which is 2, so it cannot be 4. Example: HCF(14, 16) = 2.

Teacher's Note:
1. Both are even, so 2 is common.
2. A common factor divides the difference.

Question 27. Can two numbers have HCF 15 and LCM 175? Give a reason for your answer.
Answer:
No. The HCF must be a factor of the LCM, but 175 ÷ 15 is not a whole number.

Teacher's Note:
1. The LCM is always a multiple of the HCF.
2. 175 = 5 × 5 × 7 has no 3.

Question 28. Find the smallest number greater than 3 that leaves a remainder of 3 when divided by 8 and also when divided by 12.
Answer:
Such a number is 3 more than a common multiple of 8 and 12.
LCM(8, 12) = 24, so the number is 24 + 3 = 27.

Teacher's Note:
1. Remove the remainder first.
2. Check: 27 = 8 × 3 + 3 = 12 × 2 + 3.

Question 29. Both 42 and 70 are multiples of 7. Is 7 their HCF? Justify your answer.
Answer:
No. 7 is only a common factor.
42 = 2 × 3 × 7 and 70 = 2 × 5 × 7, so the common part is 2 × 7 and the HCF is 14.

Teacher's Note:
1. A common factor need not be the highest.
2. Look for every shared prime.

Question 30. A courtyard is 12 m long and 18 m wide. Find the least number of identical square slabs of whole-number side (in metres) needed to pave it completely.
Answer:
Largest slab side = HCF(12, 18) = 6 m.
Slabs along the length = 12 ÷ 6 = 2, along the width = 18 ÷ 6 = 3, so the least number is 2 × 3 = 6 slabs.

Teacher's Note:
1. Largest slab gives fewest slabs.
2. Multiply the counts along the two sides.