Class 7 Maths Chapter 3 Finding Common Ground notes and book solutions from Ganita Prakash Part 2 with easy explanations, important concepts, examples and complete textbook question answers for revision and exam preparation.
Finding Common Ground
Class 7 Mathematics | Ganita Prakash Part II | Chapter 3
Easy Notes, Concepts, Textbook Questions and Kid-Friendly Solutions
Chapter at a Glance
This chapter is mainly about finding what numbers have in common. We learn about factors, multiples, prime factorisation, HCF and LCM. We also discover useful patterns and relationships between HCF, LCM and the product of two numbers.
Highest Common Factor
Lowest Common Multiple
Writing a number as a product of primes
A statement that works for all cases
Easy Concept: Factors
A factor of a number divides the number exactly, leaving no remainder.
For example:
Therefore, the factors of 12 are: 1, 2, 3, 4, 6, 12.
Highest Common Factor
The Highest Common Factor (HCF) of two or more numbers is the highest or greatest factor that is common to all the numbers.
HCF is also called the Greatest Common Divisor (GCD).
Factors of 12 = 1, 2, 3, 4, 6, 12
Factors of 16 = 1, 2, 4, 8, 16
Common factors = 1, 2, 4
Therefore, HCF(12,16) = 4.
Textbook Question
Sameeksha is building her new house. The main room of the house is 12 ft by 16 ft. She feels that the room would look nice if the floor is covered with square tiles of the same size. She also wants to use as few tiles as possible, and for the length of the tile to be a whole number of feet. What size tiles should she buy?
The side of a square tile must divide both 12 ft and 16 ft exactly.
Common factors of 12 and 16 are 1, 2 and 4.
The largest common factor is 4.
Therefore, Sameeksha should buy square tiles of size 4 ft × 4 ft.
A larger square tile cannot fit exactly into both 12 ft and 16 ft. Using the largest tile also means that fewer tiles are required.
Textbook Question
How many tiles of this size should she purchase?
Number of tiles along the 12 ft breadth:
12 ÷ 4 = 3
Number of tiles along the 16 ft length:
16 ÷ 4 = 4
Total tiles = 3 × 4 = 12 tiles.
Textbook Question
What if Sameeksha did not insist on the length of the tile to be a whole number of feet and the length could be a fractional number of feet? Would the answer change?
The largest square tile that can exactly fit both dimensions is still 4 ft × 4 ft.
Allowing fractional lengths does not give a larger square tile that fits exactly into both 12 ft and 16 ft.
So the answer does not change: 4 ft × 4 ft.
Rice Bag Example
Lekhana purchases rice from two farms and sells it in the market. She bought 84 kg of rice from one farm and 108 kg from the other farm. She wants the rice to be packed in bags, so each bag has rice from only one farm and all bags have the same weight that is a whole number of kg. If she wants to use as few bags as possible, what should the weight of each bag be?
Factors of 108 include 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108.
Common factors = 1, 2, 3, 4, 6, 12.
The greatest common factor is 12.
Therefore, each bag should contain 12 kg.
If each bag contains 12 kg:
84 ÷ 12 = 7 bags
108 ÷ 12 = 9 bags
Total = 16 bags.
Choosing the largest common weight gives the fewest bags.
Textbook Question
Do you remember the ‘Jump Jackpot’ game from Grade 6 (see the chapter ‘Prime Time’)? Grumpy places a treasure on a number and Jumpy chooses a jump size and tries to collect the treasure. In each case below, the two numbers upon which treasures are kept are given. Find the longest jump size (starting from 0) using which Jumpy can land on both the numbers having the treasure.
(a) 14 and 30
(b) 7 and 11
(c) 30 and 50
(d) 28 and 42
Is the longest jump size for the numbers the same as their HCF? Explain why it is so.
The longest jump size is the HCF because the jump size must divide both numbers exactly.
| Numbers | HCF / Longest Jump |
|---|---|
| 14 and 30 | 2 |
| 7 and 11 | 1 |
| 30 and 50 | 10 |
| 28 and 42 | 14 |
Prime Numbers
A prime number is a number greater than 1 having exactly two factors: 1 and itself.
Examples: 2, 3, 5, 7, 11, 13, 17, 19.
The number 1 is neither prime nor composite.
Prime Factorisation
Prime factorisation means writing a number as a product of prime numbers.
The prime factors remain the same even if we use a different order.
Textbook Question
Can you write the prime factorisation of 105 and 30 using these two figures?
105 = 3 × 5 × 7
30 = 2 × 3 × 5
Textbook Question
Try finding the prime factorisation of 1200 using the method above.
1200 = 12 × 100
12 = 2 × 2 × 3
100 = 2 × 2 × 5 × 5
Therefore:
Factors from Prime Factorisation
If a number is written as a product of prime factors, we can make its factors by choosing different combinations of these prime factors.
2 × 2 × 7 = 28 is a factor of 840 because:
840 = 28 × 30.
Textbook Question
Similarly, is 2 × 7 = 14 a factor of 840? Why or why not?
Yes.
840 ÷ 14 = 60, which is a whole number.
Therefore, 14 is a factor of 840.
Textbook Question
Is 2 × 2 × 2 a factor of 840? Why or why not?
Yes.
2 × 2 × 2 = 8 and 840 ÷ 8 = 105.
Therefore, 8 is a factor of 840.
Textbook Question
Is 3 × 3 × 3 a factor of 840? Why or why not?
No.
The prime factorisation of 840 contains only one factor 3. It does not contain three 3s.
Therefore, 27 is not a factor of 840.
Textbook Question
Can we use this idea to list down all the possible factors of a number using just its prime factors?
Yes. We can form different factors by selecting different combinations of the prime factors, including 1 and the complete product.
Textbook Question
Find the factors of 225 using prime factorisation.
225 = 3 × 3 × 5 × 5
Therefore the factors are:
Figure it Out
List all the factors of the following numbers:
(a) 90 (b) 105
(c) 132 (d) 360 (this number has 24 factors)
(e) 840 (this number has 32 factors)
(a) 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
(b) 105: 1, 3, 5, 7, 15, 21, 35, 105
(c) 132: 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132
(d) 360: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360
(e) 840: 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840
Textbook Question
After observing a few prime factorisations, Anshu claims “The larger a number is, the longer its prime factorisation will be”.
What do you think of Anshu’s claim?
The claim is not always true.
For example:
96 = 2 × 2 × 2 × 2 × 2 × 3
121 = 11 × 11
Although 121 is larger than 96, its prime factorisation has fewer prime factors.
Therefore Anshu’s claim is false. This is a conjecture that has been disproved using a counterexample.
Easy Rule
To find the HCF using prime factorisation:
- Prime-factorise every number.
- Find the prime factors common to all numbers.
- For each common prime, take the smaller number of occurrences.
- Multiply those common prime factors.
Example 1 Question
Find the common factors, and the HCF of 45 and 75.
45 = 3 × 3 × 5
75 = 3 × 5 × 5
Common prime factors are 3 and 5.
Common factors are: 1, 3, 5, 15.
Therefore:
Example 2 Question
Find the common factors, and the HCF of 112 and 84.
112 = 2 × 2 × 2 × 2 × 7
84 = 2 × 2 × 3 × 7
Common prime part = 2 × 2 × 7.
Example 3 Question
Find the common factors and the HCF of 96 and 275.
96 = 2 × 2 × 2 × 2 × 2 × 3
275 = 5 × 5 × 11
There is no common prime factor.
Therefore the only common factor is 1.
Example 4 Question
Find the HCF of 30 and 72.
30 = 2 × 3 × 5
72 = 2 × 2 × 2 × 3 × 3
Common primes are 2 and 3.
Take the minimum occurrence of each:
2 occurs once in 30 and three times in 72, so take one 2.
3 occurs once in 30 and twice in 72, so take one 3.
Example 5 Question
Find the HCF of 225 and 750.
225 = 3 × 3 × 5 × 5
750 = 2 × 3 × 5 × 5 × 5
Common primes are 3 and 5.
Take one 3 and two 5s.
Figure it Out
Find the common factors and the HCF of the following numbers:
(a) 50, 60 (b) 140, 275
(c) 77, 725 (d) 370, 592
(e) 81, 243
(a) 50, 60:
Common factors = 1, 2, 5, 10
HCF = 10
(b) 140, 275:
Common factors = 1, 5
HCF = 5
(c) 77, 725:
Common factors = 1
HCF = 1
(d) 370, 592:
Common factors = 1, 2, 37, 74
HCF = 74
(e) 81, 243:
Common factors = 1, 3, 9, 27, 81
HCF = 81
Textbook Question
How do we directly find the HCF without listing all the factors?
Prime factorisation gives a quicker method. We compare the prime factors and take the common primes with the minimum number of occurrences.
Figure it Out
1. Find the HCF of the following numbers:
(a) 24, 180 (b) 42, 75, 24
(c) 240, 378 (d) 400, 2500
(e) 300, 800
(a) HCF(24,180) = 12
(b) HCF(42,75,24) = 3
(c) HCF(240,378) = 6
(d) HCF(400,2500) = 100
(e) HCF(300,800) = 100
Textbook Question
2. Consider the numbers 72 and 144. Suppose they are factorised into composite numbers as: 72 = 6 × 12 and 144 = 8 × 18. Seeing this, can one say that these two numbers have no common factor other than 1? Why not?
No.
We cannot decide common factors just by looking at these particular composite factorisations.
For example:
72 = 2 × 2 × 2 × 3 × 3
144 = 2 × 2 × 2 × 2 × 3 × 3
They have many common factors. In fact:
So the statement would be incorrect.
Easy Concept: Multiples
A multiple of a number is obtained by multiplying that number by a whole number.
Multiples of 6:
Multiples of 8:
Lowest Common Multiple
The Lowest Common Multiple (LCM) is the smallest positive number that is a multiple of all the given numbers.
Common multiples include 24, 48, 72, …
Therefore LCM = 24.
Textbook Question
Anshu and Guna make torans out of strips of cloth. Multiple strips are placed one next to another to make a toran. Anshu uses strips of length 6 cm and Guna uses strips of 8 cm length. If both have to make torans of the same length, what is the smallest possible length, the torans could be?
We need the smallest number that is a multiple of both 6 and 8.
LCM(6,8) = 24.
Therefore, the shortest toran can be 24 cm long.
Textbook Question
What about the largest common multiple? Does such a number exist?
No.
There is no largest common multiple because common multiples continue forever.
For example, common multiples of 6 and 8 include: 24, 48, 72, 96, 120, …
Textbook Question
A sweet shop gives out free gajak to school children on Mondays. Today is a Monday and Kabamai enjoyed eating the gajak. But she visits the sweet shop once every 10 days. When is the next time she would be able to get free gajak from Sweet shop? (Answer in number of days.)
The shop gives free gajak every 7 days. Kabamai visits every 10 days.
We need LCM(7,10).
Therefore, she will get free gajak again after 70 days.
Textbook Question
Do you remember the ‘Idli-Vada’ game from Grade 6 (see chapter ‘Prime Time’)? Two numbers are chosen and whenever players come to their multiples, ‘idli’ or ‘vada’ should be called out depending on whose multiple the number is. If the number happens to be a common multiple, then ‘idli-vada’ should be called out. In each problem below, the two numbers corresponding to ‘idli’ and ‘vada’ are given. Find the first number for which ‘idli-vada’ will be called out:
(a) 4 and 6 (b) 7 and 11
(c) 14 and 30 (d) 15 and 55
Is the answer always the LCM of the two numbers? Explain.
| Numbers | First common multiple | LCM |
|---|---|---|
| 4 and 6 | 12 | 12 |
| 7 and 11 | 77 | 77 |
| 14 and 30 | 210 | 210 |
| 15 and 55 | 165 | 165 |
Yes. The first common multiple is exactly the LCM.
LCM Using Prime Factorisation
To find the LCM using prime factorisation:
- Prime-factorise all the numbers.
- Take every prime that occurs in any of the numbers.
- For each prime, take the maximum number of occurrences.
- Multiply them.
LCM → maximum occurrences
Example 6: Textbook Question
Find the LCM of 14 and 35.
14 = 2 × 7
35 = 5 × 7
For the LCM we need 2, 5 and 7.
Example 7: Textbook Question
Find the LCM of 96 and 360.
96 = 2 × 2 × 2 × 2 × 2 × 3
360 = 2 × 2 × 2 × 3 × 3 × 5
Take the maximum number of each prime:
Five 2s, two 3s and one 5.
Figure it Out
Find the LCM of the following numbers:
(a) 30, 72 (b) 36, 54
(c) 105, 195, 65 (d) 222, 370
(a) LCM(30,72) = 360
(b) LCM(36,54) = 108
(c) LCM(105,195,65) = 1365
(d) LCM(222,370) = 1110
HCF When One Number Is a Multiple of the Other
If one number is a factor of another number, then the smaller number is the HCF.
For example:
HCF(6,18) = 6
Textbook Question
If m is a number, what could be the other number?
The other number can be any positive multiple of m, such as:
2m, 3m, 4m, 5m, …
For all such pairs, the HCF is m.
Textbook Question
If 7k is a number, what could be the other number?
The other number can be a multiple of 7k, such as:
14k, 21k, 28k, 35k, …
For example, HCF(7k,14k) = 7k.
Figure it Out
1. Make a general statement about the HCF for the following pairs of numbers. You could consider examples before coming up with general statements. Look for possible explanations of why they hold.
(a) Two consecutive even numbers
(b) Two consecutive odd numbers
(c) Two even numbers
(d) Two consecutive numbers
(e) Two co-prime numbers
Share your observations with the class.
(a) Two consecutive even numbers: Their HCF is at least 2, but it can be greater than 2. For example, HCF(6,8)=2 and HCF(14,16)=2.
(b) Two consecutive odd numbers: Their HCF is always 1.
(c) Two even numbers: Their HCF is at least 2.
(d) Two consecutive numbers: Their HCF is always 1.
(e) Two co-prime numbers: Their HCF is always 1.
Textbook Question
2. The LCM of 3 and 24 is 24 (it is one of the two given numbers).
(a) Find more such number pairs where the LCM is one of the two numbers.
(b) Make a general statement about such numbers. Describe such number pairs using algebra.
Examples:
LCM(4,20)=20
LCM(5,30)=30
LCM(7,35)=35
This happens when one number is a factor of the other.
In algebra, if one number is n and the other is kn, where k is a positive integer, then:
Textbook Question
3. Make a general statement about the LCM for the following pairs of numbers. You could consider examples before coming up with these general statements. Look for possible explanations of why they hold.
(a) Two multiples of 3
(b) Two consecutive even numbers
(c) Two consecutive numbers
(d) Two co-prime numbers
(a) Two multiples of 3: Their LCM is also a multiple of 3.
(b) Two consecutive even numbers: Their HCF is 2, so their LCM is their product divided by 2.
(c) Two consecutive numbers: Their HCF is 1, so their LCM is their product.
(d) Two co-prime numbers: Their HCF is 1, so their LCM is their product.
Math Talk
What happens to the HCF of two numbers if both numbers are doubled? Take some pairs of numbers and explore. Are you able to see why the HCF will also double?
When both numbers are doubled, both prime factorisations get one additional factor 2.
Therefore, their HCF also doubles.
Example:
HCF(270,50) = 10.
Doubling gives 540 and 100.
HCF(540,100) = 20.
Textbook Question
Consider the following two multiples of 14 — 14 × 6, 14 × 9. What is their HCF?
14 × 6 = 84
14 × 9 = 126
HCF(84,126) = 42.
Therefore:
Textbook Question
Here are some more numbers where both numbers are multiples of the same number. Find their HCF:
(a) 18 × 10, 18 × 15
(b) 10 × 38, 10 × 21
(c) 5 × 13, 5 × 20
(d) 12 × 16, 12 × 20
In which of these cases is the HCF the same as the common multiplier, like problem (b) where the HCF is 10? Explore a few more examples of this type to understand when this happens.
(a) HCF(180,270) = 90
(b) HCF(380,210) = 10
(c) HCF(65,100) = 5
(d) HCF(192,240) = 48
The HCF is the same as the common multiplier when the two remaining multipliers are co-prime.
For example, 38 and 21 are co-prime, so HCF(10×38,10×21)=10.
Efficient HCF Procedure
Instead of writing every factor, we can repeatedly divide both numbers by common prime factors.
For example:
84 and 180
Divide both by 2 → 42 and 90
Divide both by 2 → 21 and 45
Divide both by 3 → 7 and 15
Now there is no common prime factor.
Textbook Question
Find the HCF in the following cases.
2 300, 150
5 150, 75
5 30, 15
3 6, 3
2, 1
The common divisors used are 2, 5, 5 and 3.
Textbook Question
Find the HCF in the following cases.
2 630, 770
5 315, 385
7 63, 77
9, 11
The common factors used are 2, 5 and 7.
Textbook Question
This procedure not only gives the HCF but can also be used to find the LCM! Can you see how?
For the LCM, multiply all the common factors used in the divisions and also multiply the final numbers left in the last row.
For 300 and 150:
For 630 and 770:
Textbook Question
Can you see why this works?
The factors collected during the common divisions contain the common prime factors. The numbers left at the end supply the remaining prime factors needed by each original number.
Together, these factors make the smallest number that contains all the prime factors required by both numbers.
Textbook Question
You can try this method for these pairs of numbers.
(a) 90 and 150 (b) 84 and 132
(a) 90 and 150
HCF = 30
LCM = 450
(b) 84 and 132
HCF = 12
LCM = 924
HCF and LCM Together
For two positive integers a and b:
This is a very useful relationship for checking answers.
Textbook Question
Which is greater — the LCM of two numbers or their product?
You could analyse the above statement using examples. Then try to reason or prove, why the LCM is never greater than the product of the numbers. [Hint: Is the product also a common multiple of the two numbers?]
The product of two numbers is always a common multiple of both numbers.
Since the LCM is the smallest common multiple, it cannot be greater than the product.
Textbook Question
Explore whether the LCM is a factor of the product in the following cases. If yes, identify the number that the LCM should be multiplied by to get the product. Do you see any pattern? Use these numbers:
(a) 45, 105 (b) 275, 352 (c) 222, 370
| Numbers | HCF | LCM | Number multiplying LCM to give product |
|---|---|---|---|
| 45, 105 | 15 | 315 | 15 |
| 275, 352 | 11 | 8800 | 11 |
| 222, 370 | 74 | 1110 | 74 |
The number multiplying the LCM is the HCF.
Textbook Question
Do you see that, in each case, the number by which the LCM is multiplied to get the product is actually the HCF?
Thus, our observations seem to suggest the following:
HCF × LCM = Product of the two numbers.
Why does this happen? Can you give an explanation or proof?
[Hint: Consider the prime factorisation of the given numbers. Among their prime factors, some are common to both factorisations, and the rest occur in only one of them. Between the HCF and the LCM, see how the common and non-common prime factors get distributed. In the product, observe how these two kinds of prime factors occur. Compare them.]
Suppose the common prime factors are written in both numbers and the remaining prime factors occur only in one number.
The HCF contains the common prime factors with their minimum occurrences.
The LCM contains all prime factors with their maximum occurrences.
When HCF and LCM are multiplied, every prime factor appears exactly as many times as it appears in the product of the original two numbers.
Textbook Question
Explore whether this property holds when 3 numbers are considered.
The simple formula HCF × LCM = product is a property for two numbers. It does not generally extend in the same simple form to three numbers.
For three numbers, we need to examine their prime factorisations carefully.
Question 1
In the two rows below, colours repeat as shown. When will the blue stars meet next?
The blue stars repeat:
Top row: every 6 stars → positions 4, 10, 16, …
Bottom row: every 4 stars → positions 4, 8, 12, 16, …
They meet at 4, and the next common position is:LCM(6,4)=12
Since they first meet at position 4: 4+12=16
Answer: The blue stars will meet next at the 16th position.
Question 2
(a) Is 5 × 7 × 11 × 11 a multiple of 5 × 7 × 7 × 11 × 2?
(b) Is 5 × 7 × 11 × 11 a factor of 5 × 7 × 7 × 11 × 2?
(a) No.
The first expression does not contain all the factors needed for the second expression, particularly another 7 and a factor 2.
(b) No.
For the first expression to be a factor of the second, the second number must contain all its prime factors with sufficient occurrences. It does not contain two 11s.
Question 3
Find the HCF and LCM of the following (state your answers in the form of prime factorisations):
(a) 3 × 3 × 5 × 7 × 7 and 12 × 7 × 11
(b) 45 and 36
(a)
First number = 3 × 3 × 5 × 7 × 7
Second number: 12 × 7 × 11 = 2 × 2 × 3 × 7 × 11
HCF = 3 × 7
LCM = 2 × 2 × 3 × 3 × 5 × 7 × 7 × 11
(b)
45 = 3 × 3 × 5
36 = 2 × 2 × 3 × 3
HCF = 3 × 3
LCM = 2 × 2 × 3 × 3 × 5
Question 4
Find two numbers whose HCF is 1 and LCM is 66.
One possible pair is:
HCF(2,33) = 1
LCM(2,33) = 66
Another possible pair is 3 and 22.
Question 5
A cowherd took all his cows to graze in the fields. The cows came to a crossing with 3 gates. An equal number of cows passed through each gate. Later at another crossing with 5 gates again an equal number of cows passed through each gate. The same happened at the third crossing with 7 gates. If the cowherd had less than 200 cows, how many cows did he have? (Based on the folklore mathematics from Karnataka.)
The number of cows must be divisible by 3, 5 and 7.
LCM(3,5,7) = 105.
The next multiple is 210, which is greater than 200.
Question 6
The length, width, and height of a box are 12 cm, 18 cm, and 36 cm respectively. Which of the following sized cubes can be packed in this box without leaving gaps?
(a) 9 cm (b) 6 cm
(c) 4 cm (d) 3 cm
(e) 2 cm
The cube side must divide 12, 18 and 36 exactly.
Common factors include 1, 2, 3 and 6.
From the given options, the possible cube sizes are:
Therefore, the correct options are (b), (d) and (e).
Question 7
Among the numbers below, which is the largest number that perfectly divides both 306 and 36?
(a) 36 (b) 612
(c) 18 (d) 3
(e) 2 (f) 360
HCF(306,36) = 18.
Question 8
Find the smallest number that is divisible by 3, 4, 5 and 7, but leaves a remainder of 10 when divided by 11.
LCM(3,4,5,7) = 420.
We need a multiple of 420 that gives remainder 10 on division by 11.
Try:
420 × 5 = 2100.
2100 ÷ 11 gives remainder 10.
Question 9
Children are playing ‘Fire in the Mountain’. When the number 6 was called out, no one got out. When the number 9 was called out, no one got out. But when the number 10 was called out, some people got out. How many children could have been playing initially?
(a) 72 (b) 90
(c) 45 (d) 3
(e) 36 (f) None of these
The number of children should be a multiple of 10 but should not be a multiple of 6 or 9.
Check the options:
90 is divisible by 6 and 9.
72 is divisible by 6.
45 is divisible by 9.
36 is divisible by 6 and 9.
3 is not a multiple of 10.
Question 10
Tick the correct statement(s). The LCM of two different prime numbers (m, n) can be:
(a) Less than both numbers
(b) In between the two numbers
(c) Greater than both numbers
(d) Less than m × n
(e) Greater than m × n
For two different primes:
The product of two different primes is greater than each individual prime.
Therefore only:
is correct.
Question 11
A dog is chasing a rabbit that has a head start of 150 feet. It jumps 9 feet every time the rabbit jumps 7 feet. In how many leaps does the dog catch up with the rabbit?
In one leap of each:
Dog covers 9 ft.
Rabbit covers 7 ft.
Dog gains:
Head start = 150 ft.
Number of leaps:
Therefore, the dog catches the rabbit after 75 leaps.
Question 12
What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10? Do you remember the answer from Grade 6, Chapter 5?
Prime factorisation gives:
2 = 2
3 = 3
4 = 2 × 2
5 = 5
6 = 2 × 3
8 = 2 × 2 × 2
9 = 3 × 3
10 = 2 × 5
Take the maximum occurrences:
Answer = 360.
Question 13
Here is a problem posed by the ancient Indian Mathematician Mahaviracharya (850 C.E.). Add together 8/15, 1/20, 7/36, 11/63 and 1/21. What do you get? How can we find this sum efficiently?
We find the LCM of 15, 20, 36, 63 and 21.
LCM = 1260.
Convert each fraction:
8/15 = 672/1260
1/20 = 63/1260
7/36 = 245/1260
11/63 = 220/1260
1/21 = 60/1260
Add:
Therefore, the answer is 1.
The efficient method is to find the LCM of the denominators first.
Important Terms
Factor: A number that divides another number exactly.
Multiple: A number obtained by multiplying a number by a whole number.
Prime number: A number greater than 1 with exactly two factors.
Prime factorisation: Writing a number as a product of prime numbers.
HCF: Highest Common Factor.
LCM: Lowest Common Multiple.
Conjecture: A mathematical statement made from observations that still needs proof.
Counterexample: An example that shows a conjecture is false.
Generalisation: A statement or rule that works for all suitable cases.
HCF Rule
LCM Rule
Important Property
Special Cases
If two numbers are consecutive, their HCF is 1.
If two numbers are co-prime, their HCF is 1.
If one number is a factor of another:
LCM = larger number
| Topic | Easy Rule |
|---|---|
| HCF | Greatest common factor |
| LCM | Smallest common multiple |
| HCF by prime factorisation | Minimum occurrences of common primes |
| LCM by prime factorisation | Maximum occurrences of all required primes |
| Consecutive numbers | HCF = 1 |
| Co-prime numbers | HCF = 1 |
| One number divides the other | HCF = smaller number, LCM = larger number |
| Two numbers | HCF × LCM = Product |
Prepare for your exams with this easy and student-friendly worksheet based on Class 7 Maths Chapter 3 from Ganita Prakash Part 2. Practice all the questions carefully to revise the chapter, strengthen your concepts, and improve your mathematical understanding.
Class 7 Maths Part 2 Chapter 3 Worksheet
Practice worksheet for Class 7 Mathematics Part 2 Chapter 3 with important questions, extra practice and answer key.
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