CBSE Class 7 Mathematics Finding Common Ground Worksheet Set 01

Long Answer Questions

Question 1. Study the number 84 carefully and answer the following.
(i) Find the prime factorisation of 84 using the division method, showing every step.
(ii) Using subparts of this prime factorisation, list all the factors of 84 in increasing order.
(iii) Is 2 × 2 × 2 a factor of 84? Give a reason for your answer using the prime factorisation.
Answer:
(i) 84 ÷ 2 = 42, 42 ÷ 2 = 21, 21 ÷ 3 = 7, and 7 is a prime, so we stop.
So 84 = 2 × 2 × 3 × 7.
(ii) Every factor of 84 is a subpart of this factorisation, together with 1.
Single primes: 2, 3, 7
Two primes: 2 × 2 = 4, 2 × 3 = 6, 2 × 7 = 14, 3 × 7 = 21
Three primes: 2 × 2 × 3 = 12, 2 × 2 × 7 = 28, 2 × 3 × 7 = 42
All four primes: 2 × 2 × 3 × 7 = 84
Factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84 (12 factors).
(iii) No. 84 has the prime 2 only twice, so 2 × 2 × 2 = 8 cannot be a subpart of 2 × 2 × 3 × 7. Indeed, 84 ÷ 8 is not a whole number.
In simple words: 84 = 2 × 2 × 3 × 7 has 12 factors, and 8 is not one of them because 84 has only two 2s.

Teacher's Note:
1. Group the subparts by how many primes they use.
2. A factor cannot use a prime more often than the number does.
3. Count the factors to check nothing is missed.

Question 2. Consider the numbers 72 and 120.
(i) Write the prime factorisation of each number.
(ii) Find their HCF by taking the common primes the minimum number of times.
(iii) List all the common factors of 72 and 120, and state what you notice about these common factors and the HCF.
Answer:
(i) 72 = 2 × 2 × 2 × 3 × 3 and 120 = 2 × 2 × 2 × 3 × 5.
(ii) 2 occurs three times in each, so take three 2s.
3 occurs twice in 72 but once in 120, so take one 3. 5 is only in 120, so leave it out.
HCF = 2 × 2 × 2 × 3 = 24.
(iii) A common factor must be a subpart of 2 × 2 × 2 × 3. With 1, these are 1, 2, 3, 4, 6, 8, 12, 24.
This is exactly the list of factors of 24. So every common factor is a factor of the HCF, and the HCF is the largest of them.
In simple words: The HCF of 72 and 120 is 24, and the common factors are just the factors of 24.

Teacher's Note:
1. Compare the count of each prime one by one.
2. The common factors are always the factors of the HCF.
3. This gives a quick way to list common factors.

Question 3. Consider the numbers 24 and 36.
(i) Write the first six multiples of 24 and the first six multiples of 36, and mark the common ones.
(ii) Find the LCM of 24 and 36 using prime factorisation.
(iii) Is there a largest common multiple of 24 and 36? Explain your answer.
Answer:
(i) Multiples of 24: 24, 48, 72, 96, 120, 144
Multiples of 36: 36, 72, 108, 144, 180, 216
The common ones are 72 and 144.
(ii) 24 = 2 × 2 × 2 × 3 and 36 = 2 × 2 × 3 × 3.
Take three 2s (from 24) and two 3s (from 36): LCM = 2 × 2 × 2 × 3 × 3 = 72. This agrees with the lists.
(iii) No. Every multiple of 72 (72, 144, 216, 288, …) is a common multiple, and this list never ends.
Adding 72 to any common multiple gives a bigger one, so there is a lowest common multiple but no largest.
In simple words: The LCM of 24 and 36 is 72, and its multiples go on for ever, so there is no largest.

Teacher's Note:
1. The listing method and the prime method must agree.
2. Common multiples are the multiples of the LCM.
3. A never-ending list has no largest member.

Question 4. Take the numbers 90 and 126.
(i) Carry out the common-division procedure, dividing both numbers by a common prime factor at each step until the two quotients have no common prime factor left.
(ii) Write the HCF and the LCM of 90 and 126 from your working.
(iii) Verify that HCF × LCM = the product of the two numbers.
Answer:
(i) 2: 90, 126 → 45, 63
3: 45, 63 → 15, 21
3: 15, 21 → 5, 7
5 and 7 have no common prime factor, so we stop.
(ii) HCF = product of the divisors = 2 × 3 × 3 = 18.
LCM = divisors × last quotients = 2 × 3 × 3 × 5 × 7 = 630.
This works because 90 = 18 × 5 and 126 = 18 × 7, so the smallest number containing both is 18 × 5 × 7.
(iii) HCF × LCM = 18 × 630 = 11340.
Product = 90 × 126 = 11340.
Both are equal, so HCF × LCM = product holds here.
In simple words: Dividing by 2, 3, 3 leaves 5 and 7, so HCF = 18, LCM = 630, and 18 × 630 = 90 × 126.

Teacher's Note:
1. Write each step as divisor: numbers → quotients.
2. Stop when the quotients are co-prime.
3. The verification is a check on the working.

Question 5. Consider the numbers 180 and 210.
(i) Write the prime factorisation of each number.
(ii) Find their HCF and their LCM using the prime factorisations.
(iii) Verify the relation HCF × LCM = product of the two numbers, and explain why the LCM must always be a multiple of the HCF.
Answer:
(i) 180 = 2 × 2 × 3 × 3 × 5 and 210 = 2 × 3 × 5 × 7.
(ii) HCF: common primes 2, 3, 5, each the minimum number of times (one each): HCF = 2 × 3 × 5 = 30.
LCM: every prime the maximum number of times (two 2s, two 3s, one 5, one 7): LCM = 2 × 2 × 3 × 3 × 5 × 7 = 1260.
(iii) HCF × LCM = 30 × 1260 = 37800 and 180 × 210 = 37800. The relation holds.
The HCF uses the common primes the fewest times, while the LCM uses each of those primes the most times.
So every prime of the HCF appears in the LCM at least as often, which makes the HCF a factor of the LCM. Here 1260 = 30 × 42.
In simple words: HCF = 30 and LCM = 1260; their product equals 180 × 210, and 1260 is 42 times 30.

Teacher's Note:
1. Minimum counts for the HCF, maximum counts for the LCM.
2. Max is never less than min, so the HCF divides the LCM.
3. Always multiply out to verify.

Question 6. Answer the following about co-prime numbers.
(i) What are co-prime numbers? Give two examples of co-prime pairs in which neither number is prime.
(ii) Find the HCF and the LCM of 16 and 21, and say whether they are co-prime.
(iii) Show, with two examples and a reason, that any two consecutive whole numbers are always co-prime.
Answer:
(i) Two numbers are co-prime if their HCF is 1, that is, their only common factor is 1. They need not be primes.
Examples: 8 and 9 (8 = 2 × 2 × 2, 9 = 3 × 3) and 15 and 16 (15 = 3 × 5, 16 = 2 × 2 × 2 × 2).
(ii) 16 = 2 × 2 × 2 × 2 and 21 = 3 × 7. No common prime, so HCF = 1 and they are co-prime.
LCM = 2 × 2 × 2 × 2 × 3 × 7 = 336, which is 16 × 21.
(iii) Examples: HCF(14, 15) = 1 and HCF(20, 21) = 1.
Reason: if a number d divides both n and n + 1, it must also divide their difference, 1. Only 1 divides 1, so d = 1.
Hence two consecutive whole numbers always have HCF 1.
In simple words: Co-prime numbers share only the factor 1; 16 and 21 are co-prime, and so are any two consecutive numbers.

Teacher's Note:
1. Co-prime is about the pair, not each number.
2. For co-prime numbers LCM = product.
3. A common factor of two numbers divides their difference.

Question 7. Make and explain general statements about HCF and LCM.
(i) Take the pairs 18 and 20, and 34 and 36. Find the HCF of each pair and write a general statement about the HCF of two consecutive even numbers.
(ii) Find the LCM of each of these two pairs and write a general statement connecting the LCM of two consecutive even numbers with their product.
(iii) When is the HCF of two numbers equal to one of the numbers themselves? Describe such pairs using a letter-number.
Answer:
(i) 18 = 2 × 3 × 3, 20 = 2 × 2 × 5, so HCF = 2.
34 = 2 × 17, 36 = 2 × 2 × 3 × 3, so HCF = 2.
General statement: the HCF of two consecutive even numbers is always 2. Both are even, so 2 is common; removing one 2 from each leaves two consecutive numbers, which are co-prime.
(ii) LCM(18, 20) = 180, while 18 × 20 = 360.
LCM(34, 36) = 612, while 34 × 36 = 1224.
General statement: the LCM of two consecutive even numbers is half their product. This follows from HCF × LCM = product with HCF = 2.
(iii) This happens exactly when one number is a factor of the other.
Using a letter-number, the pair is n and kn for a counting number k, and then HCF = n. Example: HCF(7, 35) = 7.
In simple words: Consecutive even numbers have HCF 2 and LCM half their product; HCF(n, kn) = n.

Teacher's Note:
1. Test a general statement on more than one example.
2. HCF × LCM = product links parts (i) and (ii).
3. A letter-number covers every case at once.

Question 8. Explore what happens when both numbers are multiplied by the same number.
(i) Find the HCF and the LCM of 20 and 30.
(ii) Multiply both numbers by 3 to get 60 and 90. Find their HCF and LCM.
(iii) Compare your answers in (i) and (ii), and explain, using prime factorisation, what happens to the HCF and the LCM when both numbers are multiplied by 3.
Answer:
(i) 20 = 2 × 2 × 5, 30 = 2 × 3 × 5.
HCF = 2 × 5 = 10; LCM = 2 × 2 × 3 × 5 = 60.
(ii) 60 = 2 × 2 × 3 × 5, 90 = 2 × 3 × 3 × 5.
HCF = 2 × 3 × 5 = 30; LCM = 2 × 2 × 3 × 3 × 5 = 180.
(iii) The HCF went from 10 to 30 and the LCM from 60 to 180, so both became 3 times as large.
Multiplying each number by 3 adds one extra 3 to both factorisations. This 3 is common, so it joins the HCF.
For the LCM, the greatest count of the prime 3 also goes up by one, so the LCM is multiplied by 3 too.
Check: 30 × 180 = 5400 = 60 × 90.
In simple words: Multiplying both numbers by 3 multiplies both the HCF and the LCM by 3.

Teacher's Note:
1. The extra 3 appears in both numbers.
2. The same idea works for doubling.
3. Use HCF × LCM = product to check.

Question 9. The HCF of two numbers is 12 and their LCM is 180.
(i) Find the product of the two numbers.
(ii) If one of the numbers is 36, find the other number and check your answer by finding the HCF and LCM of the pair.
(iii) Can two numbers have HCF 8 and LCM 180? Give a clear reason.
Answer:
(i) Product = HCF × LCM = 12 × 180 = 2160.
(ii) Other number = 2160 ÷ 36 = 60.
Check: 36 = 2 × 2 × 3 × 3 and 60 = 2 × 2 × 3 × 5.
HCF = 2 × 2 × 3 = 12 (correct); LCM = 2 × 2 × 3 × 3 × 5 = 180 (correct).
(iii) No. The HCF is always a factor of the LCM, because every prime of the HCF occurs in the LCM at least as often.
Here 180 ÷ 8 = 22.5, which is not whole, so 8 is not a factor of 180. No such pair exists.
(Another way: both numbers would be multiples of 8, so their product 8 × 180 = 1440 would have to be divisible by 8 × 8 = 64, and it is not.)
In simple words: The product is 2160, the other number is 60, and HCF 8 with LCM 180 is impossible since 8 does not divide 180.

Teacher's Note:
1. Find the product before the missing number.
2. Always check that the HCF divides the LCM.
3. Verify both HCF and LCM of your answer.

Question 10. Answer the following about numbers divisible by many small numbers.
(i) Find the smallest number that is a multiple of 2, 3, 4, 5, 6, 8, 9 and 10.
(ii) Now find the smallest number that is a multiple of every number from 1 to 10.
(iii) Explain why the answer to (ii) is exactly 7 times the answer to (i), using prime factorisation in your explanation.
Answer:
(i) We need the LCM, taking each prime the greatest number of times.
2, 3, 4 = 2 × 2, 5, 6 = 2 × 3, 8 = 2 × 2 × 2, 9 = 3 × 3, 10 = 2 × 5
Greatest counts: three 2s (from 8), two 3s (from 9), one 5.
LCM = 2 × 2 × 2 × 3 × 3 × 5 = 360.
(ii) The new numbers are 1 and 7. 1 adds nothing; 7 is a prime, so one 7 must be added.
LCM = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520.
(iii) The numbers in (i) are made only of the primes 2, 3 and 5, so none of them contains 7.
Adding 7 to the list does not change the counts of 2, 3 and 5; it only adds one factor 7.
Hence the new LCM is 360 × 7 = 2520.
In simple words: The LCM of 2 to 10 without 7 is 360; bringing in the new prime 7 makes it 360 × 7 = 2520.

Teacher's Note:
1. Only the largest power of each prime matters.
2. 8 already covers 2 and 4; 9 covers 3.
3. A new prime multiplies the LCM by itself.

Case Study Based Questions

Question 11. Tiling the floor of a reading room. The village library is getting a new reading room. The floor of the room is a rectangle, 20 ft long and 28 ft wide. The librarian wants the whole floor covered with square tiles of exactly the same size, with no tile cut and no gap left anywhere near the walls. The side of each tile must be a whole number of feet. The carpenter explains that the side of the tile must divide 20 exactly so that the tiles fit along the length, and must also divide 28 exactly so that they fit along the width. Since tiles cost money to lay, the librarian wants to use as few tiles as possible.
(i) List all the factors of 20.
(ii) Write all the common factors of 20 and 28.
(iii) What should be the side of the largest square tile the carpenter can use?
(iv) How many such tiles are needed to cover the whole floor?
Answer:
(i) Factors of 20: 1, 2, 4, 5, 10, 20.
(ii) Factors of 28: 1, 2, 4, 7, 14, 28. Common factors of 20 and 28: 1, 2 and 4.
(iii) The largest common factor is 4, the HCF of 20 and 28. So the tile side is 4 ft.
(iv) Along the length: 20 ÷ 4 = 5 tiles; along the width: 28 ÷ 4 = 7 tiles. Total = 5 × 7 = 35 tiles.
In simple words: The biggest tile that fits both sides is 4 ft, and 5 rows of 7 tiles cover the floor.

Teacher's Note:
1. Fewest tiles means the largest tile, the HCF.
2. Multiply the tiles along each side.

Question 12. Packing sweets for a festival. For a school festival, Kabamai’s family prepares 36 laddus and 60 barfis. They want to pack them in small boxes to give to the guests. Each box must contain only one kind of sweet, and every box must hold the same number of pieces, whether it is a laddu box or a barfi box. No sweet should be left unpacked and no box should be partly filled. Kabamai’s grandmother says that to make the packing look generous, each box should hold as many pieces as possible. The family starts by writing down the numbers that divide both 36 and 60 exactly.
(i) Should they use the HCF or the LCM of 36 and 60? Give a reason.
(ii) Write all the common factors of 36 and 60.
(iii) How many sweets will each box hold?
(iv) How many boxes will be needed in all?
Answer:
(i) The HCF. The number in a box must divide both 36 and 60 exactly and be as large as possible, so it is the highest common factor.
(ii) 36 = 2 × 2 × 3 × 3 and 60 = 2 × 2 × 3 × 5, so the common factors are 1, 2, 3, 4, 6 and 12.
(iii) HCF = 2 × 2 × 3 = 12, so each box holds 12 sweets.
(iv) Laddu boxes = 36 ÷ 12 = 3; barfi boxes = 60 ÷ 12 = 5. Total = 8 boxes.
In simple words: Each box holds 12 sweets, giving 3 laddu boxes and 5 barfi boxes.

Teacher's Note:
1. 'Divide exactly' and 'as many as possible' signal the HCF.
2. Add the boxes of both kinds.

Question 13. Two temple bells. In a town, a small bell in the temple is rung every 15 minutes and a big bell in the same temple is rung every 20 minutes. One morning both bells are rung together exactly at 9 a.m. The priest tells the children that after this, the small bell will ring at the times that are multiples of 15 minutes after 9 a.m., and the big bell at the times that are multiples of 20 minutes after 9 a.m. The children want to know when they will next hear both bells ringing at the same moment, and how often that will happen during the day.
(i) Write the first four multiples of 15 and the first three multiples of 20.
(ii) Is 60 a common multiple of 15 and 20?
(iii) At what time will the two bells next ring together?
(iv) How many times will the two bells ring together between 9 a.m. and 1 p.m., counting the 1 p.m. ringing but not the 9 a.m. one?
Answer:
(i) Multiples of 15: 15, 30, 45, 60. Multiples of 20: 20, 40, 60.
(ii) Yes. 60 = 15 × 4 and 60 = 20 × 3. In fact it is their LCM.
(iii) They ring together after LCM(15, 20) = 60 minutes, so next at 10 a.m.
(iv) They ring together every hour: at 10 a.m., 11 a.m., 12 noon and 1 p.m., that is 4 times.
In simple words: The bells meet every 60 minutes, so 4 times from 10 a.m. to 1 p.m.

Teacher's Note:
1. Ringing together again is an LCM question.
2. Count carefully which ends are included.

Question 14. Cutting ribbons for name badges. A craft teacher has one red ribbon of length 30 m and one blue ribbon of length 45 m. For a school function she wants to cut both ribbons into pieces that are all of the same length. Each piece must be a whole number of metres long, no piece may be made of two colours joined together, and no bit of ribbon should be wasted. To make the badges look good, she wants each piece to be as long as possible. Her helper suggests first writing down the lengths that divide both 30 and 45 exactly.
(i) Why is this an HCF problem and not an LCM problem?
(ii) Write the prime factorisation of 30 and of 45.
(iii) Find the greatest possible length of each piece.
(iv) How many pieces will she get altogether?
Answer:
(i) The piece length must divide 30 and 45 exactly (no waste), and we want the greatest such length. That is the highest common factor.
(ii) 30 = 2 × 3 × 5 and 45 = 3 × 3 × 5.
(iii) Common primes 3 and 5, each once: HCF = 3 × 5 = 15. Each piece is 15 m.
(iv) Red: 30 ÷ 15 = 2 pieces; blue: 45 ÷ 15 = 3 pieces. Total = 5 pieces.
In simple words: The longest equal piece is 15 m, giving 2 red and 3 blue pieces.

Teacher's Note:
1. Cutting into equal parts uses factors, not multiples.
2. Add the pieces from both ribbons.

Question 15. Three buses from a depot. Three buses leave the same depot and run on three different routes. The bus on Route A leaves every 12 minutes, the bus on Route B every 15 minutes and the bus on Route C every 20 minutes. One morning all three buses leave the depot together at exactly 6:00 a.m. A student waiting at the depot gate wants to photograph all three buses leaving side by side, so he needs to know when that will happen again. The depot manager tells him that this moment comes at a time that is a common multiple of 12, 15 and 20 minutes after 6:00 a.m.
(i) Write the prime factorisation of 12, 15 and 20.
(ii) Find the LCM of 12, 15 and 20.
(iii) At what time will the three buses next leave the depot together?
(iv) How many times will all three leave together between 6:00 a.m. and 12 noon, not counting the 6:00 a.m. start?
Answer:
(i) 12 = 2 × 2 × 3, 15 = 3 × 5, 20 = 2 × 2 × 5.
(ii) Two 2s, one 3 and one 5: LCM = 2 × 2 × 3 × 5 = 60.
(iii) 60 minutes after 6:00 a.m., that is at 7:00 a.m.
(iv) Once every hour: at 7, 8, 9, 10, 11 a.m. and 12 noon, that is 6 times.
In simple words: The three buses meet every 60 minutes, so 6 times up to 12 noon.

Teacher's Note:
1. Leaving together again is an LCM question.
2. List the times to count correctly.

Question 16. Planting saplings in rows. On Van Mahotsav day, a school receives 72 mango saplings and 108 guava saplings to plant in the school garden. The gardener plans to plant them in straight rows. Each row must have only one kind of sapling, and every row, whether of mango or of guava, must contain the same number of saplings. No sapling should be left out and no row should be left short. The gardener wants the rows to be as long as possible so that fewer rows are needed and watering becomes easier. He begins by factorising 72 and 108.
(i) Write the prime factorisation of 72 and of 108.
(ii) Find the HCF of 72 and 108.
(iii) How many saplings will be planted in each row?
(iv) How many rows of mango and how many rows of guava will there be?
Answer:
(i) 72 = 2 × 2 × 2 × 3 × 3 and 108 = 2 × 2 × 3 × 3 × 3.
(ii) Common primes: 2 (twice, the minimum) and 3 (twice, the minimum). HCF = 2 × 2 × 3 × 3 = 36.
(iii) The row size must divide both and be as large as possible, so 36 saplings per row.
(iv) Mango rows = 72 ÷ 36 = 2; guava rows = 108 ÷ 36 = 3; 5 rows in all.
In simple words: Rows of 36 saplings give 2 mango rows and 3 guava rows.

Teacher's Note:
1. Longest rows means the HCF.
2. Divide each total by the row size.

Question 17. A jumping game on a number track. Two friends play a jumping game on a track marked with the numbers 1 to 100. Mira starts at 0 and jumps 6 steps at a time, so she lands on 6, 12, 18 and so on. Rahul also starts at 0 but jumps 8 steps at a time, landing on 8, 16, 24 and so on. Whenever both of them land on the same number, that number is called a meeting number and they get one point. They want to find out all the meeting numbers on the track before the game begins, so that they can plan their scoring.
(i) Write the first five multiples of 6 and the first four multiples of 8.
(ii) Find the smallest meeting number using the LCM of 6 and 8.
(iii) List all the meeting numbers on the track from 1 to 100.
(iv) If the track were extended to 150 steps, what would be the largest meeting number on it?
Answer:
(i) Multiples of 6: 6, 12, 18, 24, 30. Multiples of 8: 8, 16, 24, 32.
(ii) 6 = 2 × 3 and 8 = 2 × 2 × 2, so LCM = 2 × 2 × 2 × 3 = 24. The smallest meeting number is 24.
(iii) Meeting numbers are multiples of 24: 24, 48, 72 and 96.
(iv) Multiples of 24 up to 150: 24, 48, 72, 96, 120, 144. The largest is 144.
In simple words: They meet at every multiple of 24: 24, 48, 72, 96, and 144 on a 150-step track.

Teacher's Note:
1. Every common multiple is a multiple of the LCM.
2. The next multiple, 168, is past 150.

Teacher's Note:
1. Use the table rows to test the rule first.
2. Working backwards: divide the product by the known number.

Question 19. Three traffic lights on a road. Three traffic lights stand on the same stretch of road. The first light changes every 48 seconds, the second every 72 seconds and the third every 108 seconds. One morning all three lights change together at exactly 8:00:00 a.m. A traffic warden keeping a record wants to know the next moment when all three will change together, and how often this happens in an hour, so that he can note it down in his duty register. He writes the prime factorisation of each of the three times and builds the lowest common multiple from them.
(i) Write the prime factorisation of 48, 72 and 108.
(ii) Find the LCM of 48, 72 and 108 in seconds.
(iii) Convert your answer to minutes and seconds, and write the clock time when the three lights next change together.
(iv) Counting from 8:00:00 a.m., how many times in the next one hour will all three lights change together again?
Answer:
(i) 48 = 2 × 2 × 2 × 2 × 3, 72 = 2 × 2 × 2 × 3 × 3, 108 = 2 × 2 × 3 × 3 × 3.
(ii) Four 2s (from 48) and three 3s (from 108): LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 16 × 27 = 432 seconds.
(iii) 7 × 60 = 420 and 432 − 420 = 12, so 432 s = 7 minutes 12 seconds. They next change together at 8:07:12 a.m.
(iv) One hour = 3600 s, and 3600 ÷ 432 = 8 with remainder 144. So they change together 8 times, the last at 8 × 432 = 3456 s, that is 8:57:36 a.m.
In simple words: The lights meet every 432 s (7 min 12 s), which fits 8 times into one hour.

Teacher's Note:
1. Take the highest count of each prime from any of the three numbers.
2. The remainder tells us the ninth meeting is after 9 a.m.

Question 20. Counting the eggs on a poultry farm. A poultry farmer collects a basketful of eggs. When she packs them in trays of 6, two eggs are left over. When she packs them in trays of 8, again two eggs are left over, and when she packs them in trays of 10, two eggs are left over once more. She remembers only that the basket held more than 100 eggs but fewer than 300. Her daughter, who has just learnt about the LCM, says that if two eggs are first set aside, the number that remains must divide exactly into trays of 6, 8 and 10, so it must be a common multiple of 6, 8 and 10.
(i) Find the LCM of 6, 8 and 10.
(ii) Write the rule connecting the number of eggs with this LCM.
(iii) Find all the possible numbers of eggs in the basket.
(iv) The farmer later finds that the eggs could also be packed exactly into trays of 11 with none left over. How many eggs were there?
Answer:
(i) 6 = 2 × 3, 8 = 2 × 2 × 2, 10 = 2 × 5. LCM = 2 × 2 × 2 × 3 × 5 = 120.
(ii) After setting 2 eggs aside, the rest is a common multiple of 6, 8 and 10, so a multiple of 120. So number of eggs = a multiple of 120, plus 2.
(iii) 120 + 2 = 122 and 240 + 2 = 242 lie between 100 and 300 (the next, 362, is too large). So the basket held 122 or 242 eggs.
(iv) 122 ÷ 11 is not whole, but 242 = 11 × 22. So there were 242 eggs.
In simple words: The egg count is 2 more than a multiple of 120, and only 242 also divides by 11.

Teacher's Note:
1. Remove the remainder first, then use the LCM.
2. Use the extra clue to choose between the options.