CBSE Class 7 Mathematics Finding Common Ground Assignment Set 01

Short Answer Questions

Question 1. Write in your own words what the Highest Common Factor (HCF) of two numbers means. Then list all the common factors of 18 and 24 and state their HCF.
Answer:
The HCF of two numbers is the largest number that divides both of them; it is the greatest of their common factors.
Factors of 18: 1, 2, 3, 6, 9, 18.
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24.
Common factors: 1, 2, 3, 6. The largest is 6, so HCF = 6.
In simple words: The common factors of 18 and 24 are 1, 2, 3, 6, and the biggest, 6, is the HCF.

Teacher's Note:
1. Write the factors in order to compare easily.
2. The HCF is the last number in the common list.

Question 2. Find the prime factorisation of 126 using the division method. Show each division step and write the final answer as a product of primes.
Answer:
We keep dividing by a prime until a prime is left.
126 ÷ 2 = 63
63 ÷ 3 = 21
21 ÷ 3 = 7, and 7 is a prime, so we stop.
So 126 = 2 × 3 × 3 × 7.
In simple words: Divide by 2, then 3, then 3, and the 7 left over is prime.

Teacher's Note:
1. Try 2 first, then 3, 5, 7, … .
2. Multiply back to check: 2 × 3 × 3 × 7 = 126.

Question 3. Write what the Lowest Common Multiple (LCM) of two numbers means. By listing multiples, find the LCM of 8 and 12.
Answer:
The LCM of two numbers is the smallest number that is a multiple of both.
Multiples of 8: 8, 16, 24, 32, 40, 48, …
Multiples of 12: 12, 24, 36, 48, …
Common multiples: 24, 48, … The lowest is 24, so LCM = 24.
In simple words: 24 is the first number in both lists of multiples.

Teacher's Note:
1. List the multiples of the bigger number and test each one.
2. Stop at the first common one.

Question 4. List the factors of 36 and the factors of 60. From these lists, write all their common factors and state their HCF.
Answer:
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36.
Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.
Common factors: 1, 2, 3, 4, 6, 12.
The greatest is 12, so HCF = 12.
In simple words: Six numbers divide both 36 and 60, and the biggest of them is 12.

Teacher's Note:
1. Find factors in pairs: 1 × 36, 2 × 18, and so on.
2. Every common factor divides the HCF 12.

Question 5. Using prime factorisation, find the LCM of 15 and 20. Write the prime factorisation of each number first.
Answer:
15 = 3 × 5
20 = 2 × 2 × 5
For the LCM take each prime the greatest number of times it appears: two 2s, one 3 and one 5.
LCM = 2 × 2 × 3 × 5 = 60.
In simple words: Put together two 2s, one 3 and one 5 to get 60.

Teacher's Note:
1. The 5 is shared, so take it only once.
2. Check: 60 ÷ 15 = 4 and 60 ÷ 20 = 3.

Question 6. The prime factorisation of 98 is 2 × 7 × 7.
(i) Using subparts of this factorisation, list all the factors of 98.
(ii) How many factors does 98 have in all?
Answer:
(i) The subparts of 2 × 7 × 7 are 2, 7, 7 × 7 = 49, 2 × 7 = 14 and 2 × 7 × 7 = 98.
Adding 1, the factors of 98 are 1, 2, 7, 14, 49, 98.
(ii) 98 has 6 factors in all.
In simple words: Multiply some of the primes 2, 7, 7 together, and add 1, to get every factor.

Teacher's Note:
1. Remember to include 1 and the number itself.
2. Do not list 7 twice.

Question 7. Find the HCF of 21 and 40 by writing their prime factorisations. What name is given to a pair of numbers like this?
Answer:
21 = 3 × 7 and 40 = 2 × 2 × 2 × 5.
No prime is common, so the only common factor is 1 and HCF = 1.
Numbers whose HCF is 1 are called co-prime numbers. So 21 and 40 are co-prime.
In simple words: 21 and 40 share no prime, so their HCF is 1 and they are co-prime.

Teacher's Note:
1. Neither 21 nor 40 is prime, yet they are co-prime.
2. No common prime means HCF 1.

Question 8. Two numbers have the prime factorisations 2 × 2 × 3 × 5 × 7 and 2 × 3 × 3 × 5. Find their HCF without first multiplying out the two numbers, and explain the rule you used.
Answer:
Rule: take only the primes common to both, each the smaller number of times it occurs.
Common primes are 2, 3 and 5 (7 is only in the first).
2: twice and once, take one 2. 3: once and twice, take one 3. 5: once in each, take one 5.
HCF = 2 × 3 × 5 = 30.
In simple words: Only one 2, one 3 and one 5 are shared, so the HCF is 30.

Teacher's Note:
1. Compare prime by prime.
2. A prime in only one number is left out of the HCF.

Question 9. A shopkeeper has 60 pens and 48 pencils. He wants to make identical gift packs, each pack holding the same number of pens and the same number of pencils, with nothing left over. Find the greatest number of packs he can make and what goes into each pack.
Answer:
The number of packs must divide 60 and 48 exactly, and be as large as possible, so we need the HCF.
60 = 2 × 2 × 3 × 5 and 48 = 2 × 2 × 2 × 2 × 3.
HCF = 2 × 2 × 3 = 12.
He can make 12 packs, each with 60 ÷ 12 = 5 pens and 48 ÷ 12 = 4 pencils.
In simple words: 12 packs share 60 pens and 48 pencils equally: 5 pens and 4 pencils each.

Teacher's Note:
1. Identical packs with nothing left over: HCF.
2. Divide each total by the number of packs.

Question 10. Two runners start together from the starting line of a circular track. One completes a round in 36 s and the other in 48 s. After how many seconds will they be at the starting line together again? Give the answer in minutes and seconds also.
Answer:
The first runner is at the start after multiples of 36 s and the second after multiples of 48 s, so we need the LCM.
36 = 2 × 2 × 3 × 3 and 48 = 2 × 2 × 2 × 2 × 3.
LCM = 2 × 2 × 2 × 2 × 3 × 3 = 144.
They meet at the starting line after 144 s, that is, 2 minutes 24 seconds.
In simple words: 144 s is the first time that is a multiple of both 36 s and 48 s.

Teacher's Note:
1. 'Together again' means LCM.
2. 144 = 2 × 60 + 24.

Question 11. Find the HCF of 48, 72 and 180 using prime factorisation. State clearly how many times each common prime is taken.
Answer:
48 = 2 × 2 × 2 × 2 × 3
72 = 2 × 2 × 2 × 3 × 3
180 = 2 × 2 × 3 × 3 × 5
Primes common to all three: 2 and 3.
The fewest 2s in any one number is two (in 180), and the fewest 3s is one (in 48).
HCF = 2 × 2 × 3 = 12.
In simple words: All three numbers share at least two 2s and one 3, giving 12.

Teacher's Note:
1. With three numbers, the prime must be in all three.
2. Take the smallest count among all three.

Question 12. Find the LCM of 12, 15 and 20 using prime factorisation. State clearly how many times each prime is taken.
Answer:
12 = 2 × 2 × 3
15 = 3 × 5
20 = 2 × 2 × 5
Each prime is taken the greatest number of times it occurs anywhere: two 2s (from 12 or 20), one 3 and one 5.
LCM = 2 × 2 × 3 × 5 = 60.
In simple words: Two 2s, one 3 and one 5 cover all three numbers, giving 60.

Teacher's Note:
1. Every prime that appears anywhere goes into the LCM.
2. Check: 60 is divisible by 12, 15 and 20.

Question 13. Use the common-division procedure (dividing both numbers by a common prime at each step) on 60 and 90.
(i) Write their HCF.
(ii) Write their LCM.
Answer:
Divide both numbers by a common prime at each step:
2: 60, 90 → 30, 45
3: 30, 45 → 10, 15
5: 10, 15 → 2, 3 (no common prime is left)
(i) HCF = product of the divisors = 2 × 3 × 5 = 30.
(ii) LCM = divisors × last two quotients = 2 × 3 × 5 × 2 × 3 = 180.
In simple words: The divisors 2, 3, 5 give the HCF 30; including the last quotients 2 and 3 gives the LCM 180.

Teacher's Note:
1. Stop when the two quotients are co-prime.
2. Check: 30 × 180 = 5400 = 60 × 90.

Question 14. Find the HCF of 15 × 8 and 15 × 14 without multiplying the numbers out. Is the HCF the same as the common multiplier 15? Give a reason.
Answer:
15 × 8 = 3 × 5 × 2 × 2 × 2
15 × 14 = 3 × 5 × 2 × 7
Common part: 3 × 5 × 2 = 30, so HCF = 30.
No, the HCF is not 15. It is bigger because 8 and 14 also share the factor 2, which adds to the HCF.
In simple words: Besides 15, the numbers 8 and 14 share a 2, so the HCF is 15 × 2 = 30.

Teacher's Note:
1. A common multiplier is a common factor, not always the highest.
2. Look for anything else shared.

Question 15. Guna says that while finding the HCF and LCM of 150 and 225, he first divides both numbers by 25 instead of one prime at a time. Carry out his method and write the HCF and the LCM of 150 and 225.
Answer:
Guna divides by any common factor, not only a prime.
25: 150, 225 → 6, 9
3: 6, 9 → 2, 3 (no common factor left)
HCF = 25 × 3 = 75.
LCM = 25 × 3 × 2 × 3 = 450.
Check: 75 × 450 = 33750 = 150 × 225.
In simple words: Dividing by 25 and then 3 leaves 2 and 3, so HCF = 75 and LCM = 450.

Teacher's Note:
1. Any common factor may be used at a step.
2. The answer is the same as with primes.

Question 16. Find the smallest number that is a multiple of 6, 9 and 15. Then write the next two numbers that are multiples of all three.
Answer:
The smallest common multiple is the LCM.
6 = 2 × 3, 9 = 3 × 3, 15 = 3 × 5
LCM = 2 × 3 × 3 × 5 = 90.
Every common multiple is a multiple of the LCM, so the next two are 180 and 270.
In simple words: 90 is the smallest; then add 90 each time to get 180 and 270.

Teacher's Note:
1. Common multiples are multiples of the LCM.
2. Two 3s are needed because of 9.

Question 17. The product of two numbers is 1620 and their HCF is 9. Find their LCM, and write one possible pair of such numbers.
Answer:
HCF × LCM = product of the numbers.
So LCM = \( \frac{\text{product}}{\text{HCF}} \) = \( \frac{1620}{9} \) = 180.
One possible pair is 36 and 45: 36 × 45 = 1620, and HCF = 9 since 36 = 9 × 4, 45 = 9 × 5 and 4, 5 are co-prime.
The pair 9 and 180 also works.
In simple words: LCM = 1620 ÷ 9 = 180, and 36, 45 is one pair that fits.

Teacher's Note:
1. Divide the product by the HCF to get the LCM.
2. Both numbers must be multiples of 9.

Question 18. Find two numbers whose HCF is 1 and whose LCM is 84. Write all the possible pairs of such numbers and say what these pairs are called.
Answer:
If the HCF is 1, the numbers share no prime, so their LCM equals their product.
So we need two co-prime numbers whose product is 84 = 2 × 2 × 3 × 7. The two 2s must stay together.
Possible pairs: 1 and 84, 3 and 28, 4 and 21, 7 and 12.
These are the only pairs. Pairs with HCF 1 are called co-prime pairs.
In simple words: Split 84 into two parts that share nothing: 1 × 84, 3 × 28, 4 × 21, 7 × 12.

Teacher's Note:
1. 2 and 42 fail because both are even.
2. Co-prime means LCM = product.

Question 19. m and n are two different prime numbers. Write their HCF and their LCM in terms of m and n, and explain why the LCM must be greater than both m and n.
Answer:
Different primes have only 1 as a common factor, so HCF = 1.
They share no prime, so the LCM must contain both: LCM = m × n.
Both m and n are greater than 1, so mn is greater than m and also greater than n.
Hence the LCM of two different primes is bigger than both of them.
In simple words: Two different primes share nothing, so HCF = 1 and LCM = m × n, which is bigger than each.

Teacher's Note:
1. Try 3 and 5: HCF 1, LCM 15.
2. Multiplying by a number bigger than 1 makes it bigger.

Question 20. Explain why the LCM of two numbers can never be greater than the product of the two numbers. Illustrate your reasoning using the numbers 9 and 12.
Answer:
The product of two numbers is always a common multiple: it is the first number times the second, and the second times the first.
The LCM is the lowest of all common multiples.
So the LCM can never be more than the product; at most it is equal to it (when the numbers are co-prime).
For 9 and 12: product = 108, while LCM = 36, which is indeed less than 108.
In simple words: The product is one common multiple, so the lowest one cannot be bigger.

Teacher's Note:
1. LCM = product only for co-prime numbers.
2. 9 and 12 share a 3, so the LCM is smaller.