Official MSBSHSE Solutions for Class 7 Maths: Chapter 3 Set 13 HCF and LCM
Explore reliable textbook solutions for Chapter 3 Set 13 HCF and LCM tailored for Class 7 learners. Utilizing these Maths answers ensures thorough preparation and strengthens foundational knowledge before final MSBSHSE evaluations.
Chapter-wise Solutions for Maths: Chapter 3 Set 13 HCF and LCM
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Question 1. Find the LCM:
(i) 12, 15
(ii) 6, 8, 10
(iii) 18, 32
(iv) 10, 15, 20
(v) 45, 86
(vi) 15, 30, 90
(vii) 105, 195
(viii) 12,15,45
(ix) 63,81
(x) 18, 36, 27
Answer:
(i) 12, 15
| 3 | 12 | 15 |
| 2 | 4 | 5 |
| 2 | 5 |
\( \therefore \) LCM of 12 and 15 = 3 x 2 x 2 x 5
= 60
(ii) 6, 8, 10
| 2 | 6 | 8 | 10 |
| 2 | 3 | 4 | 5 |
| 3 | 2 | 5 |
\( \therefore \) LCM of 6, 8 and 10 = 2 x 2 x 3 x 2 x 5
= 120
(iii) 18, 32
| 2 | 18 | 32 |
| 2 | 9 | 16 |
| 2 | 9 | 8 |
| 2 | 9 | 4 |
| 3 | 9 | 2 |
| 3 | 2 |
\( \therefore \) LCM of 18 and 32 = 2 x 2 x 2 x 2 x 3 x 3 x 2
= 288
(iv) 10, 15, 20
| 5 | 10 | 15 | 20 |
| 2 | 2 | 3 | 4 |
| 1 | 3 | 2 |
\( \therefore \) LCM of 10, 15 and 20 = 5 x 2 x 3 x 2
= 60
(v) 45, 86
| 2 | 45 | 86 |
| 3 | 45 | 43 |
| 3 | 15 | 43 |
| 5 | 43 |
\( \therefore \) LCM of 45 and 86 = 2 x 3 x 3 x 5 x 43
= 3870
(vi) 15, 30, 90
| 3 | 15 | 30 | 90 |
| 5 | 5 | 10 | 30 |
| 2 | 1 | 2 | 6 |
| 1 | 1 | 3 |
\( \therefore \) LCM of 15,30 and 90 = 3 x 5 x 2 x 3
= 90
(vii) 105, 195
| 3 | 105 | 195 |
| 5 | 35 | 65 |
| 7 | 13 |
\( \therefore \) LCM of 105 and 195 = 5 x 3 x 7 x 13
= 1365
(viii) 12, 15, 45
| 3 | 12 | 15 | 45 |
| 3 | 4 | 5 | 15 |
| 5 | 4 | 5 | 5 |
| 2 | 4 | 1 | 1 |
| 2 | 2 | 1 | 1 |
| 1 | 1 | 1 |
\( \therefore \) LCM of 12, 15 and 45 = 3 x 3 x 5 x 2 x 2
= 180
(ix) 63, 81
| 3 | 63 | 81 |
| 3 | 21 | 27 |
| 3 | 7 | 9 |
| 7 | 3 |
\( \therefore \) LCM of 63 and 81 = 3 x 3 x 3 x 7 x 3
= 567
(x) 18, 36, 27
| 3 | 18 | 36 | 27 |
| 3 | 6 | 12 | 9 |
| 2 | 2 | 4 | 3 |
| 1 | 2 | 3 |
\( \therefore \) LCM of 18, 36 and 27 = 3 x 3 x 2 x 2 x 3
= 108
In simple words: The Least Common Multiple (LCM) is the smallest positive integer that is divisible by each of the given integers. It is found by multiplying all unique prime factors raised to their highest powers from the prime factorization of each number.
🎯 Exam Tip: Students should practice the division method for finding LCM, ensuring all prime factors are accounted for, and present calculations clearly to avoid errors. This method is fundamental for solving problems involving common multiples.
Question 2. Find the HCF and LCM of the numbers given below. Verify that their product is equal to the product of the given numbers:
(i) 32, 37
(ii) 46, 51
(iii) 15, 60
(iv) 18, 63
(v) 78, 104
Answer:
(i) 32, 37
32 = 2 x 16
= 2 x 2 x 8
= 2 x 2 x 2 x 4
= 2 x 2 x 2 x 2 x 2 x 1
37 = 37 x 1
\( \therefore \) HCF of 32 and 37 = 1
LCM of 32 and 37 = 2 x 2 x 2 x 2 x 2 x 37
= 1184
HCF x LCM = 1 x 1184
= 1184
Product of the given numbers = 32 x 37
= 1184
\( \therefore \) HCF x LCM = Product of the given numbers.
(ii) 46, 51
46 = 2 x 23 x 1
51 = 3 x 17 x 1
\( \therefore \) HCF of 46 and 51 = 1
LCM of 46 and 51 = 2 x 23 x 3 x 17
= 2346
HCF x LCM = 1 x 2346
= 2346
Product of the given numbers = 46 x 51
= 2346
\( \therefore \) HCF x LCM = Product of the given numbers.
(iii) 15, 60
15 = 3 x 5
60 = 2 x 30
= 2 x 2 x 15
= 2 x 2 x 3 x 5
\( \therefore \) HCF of 15 and 60 = 3 x 5
= 15
LCM of 15 and 60 = 3 x 5 x 2 x 2
= 60
HCF x LCM = 15 x 60
= 900
Product of the given numbers = 15 x 60
= 900
\( \therefore \) HCF x LCM = Product of the given numbers.
(iv) 18, 63
18 = 2 x 9
= 2 x 3 x 3
63 = 3 x 21
= 3 x 3 x 7
\( \therefore \) HCF of 18 and 63 = 3 x 3
= 9
LCM of 18 and 63 = 3 x 3 x 2 x 7
= 126
HCF x LCM = 9 x 126
= 1134
Product of the given numbers = 18 x 63
= 1134
\( \therefore \) HCF x LCM = Product of the given numbers.
(v) 78, 104
78 = 2 x 39
= 2 x 3 x 13
104 = 2 x 52
= 2 x 2 x 26
= 2 x 2 x 2 x 13
\( \therefore \) HCF of 78 and 104 = 2 x 13
= 26
LCM of 78 and 104 = 2 x 13 x 3 x 2 x 2
= 312
HCF x LCM = 26 x 312
= 8112
Product of the given numbers = 78 x 104
= 8112
\( \therefore \) HCF x LCM = Product of the given numbers.
In simple words: The HCF (Highest Common Factor) is the largest number that divides two or more numbers without leaving a remainder, while the LCM is the smallest common multiple. For any two positive integers, the product of their HCF and LCM is always equal to the product of the numbers themselves.
🎯 Exam Tip: Understanding the relationship HCF x LCM = Product of Numbers is crucial. Students should verify this property after calculating HCF and LCM, as it serves as an excellent check for accuracy in calculations involving two numbers.
Maharashtra Board Class 7 Maths Chapter 3 HCF And LCM Practice Set 13 Intext Questions And Activities
Question 1. Write the tables of the given numbers and find their LCM. (Textbook pg. no. 19)
(i) 6, 7
(ii) 8, 12
(iii) 5, 6, 15
Answer:
(i) Multiples of 6 : 6, 12, 18, 24, 30, 36, 42
Multiples of 7 : 7, 14, 21, 28, 35, 42, 49
\( \therefore \) LCM of 6 and 7 = 42
(ii) Multiples of 8 : 8, 16, 24, 32, 40
Multiples of 12 : 12, 24, 36, 48
\( \therefore \) LCM of 8 and 12 = 24
(iii) Multiples of 5 : 5, 10, 15, 20, 25, 30, 35
Multiples of 6 : 6, 12, 18, 24, 30, 36
Multiples of 15 : 15, 30, 45, 60
\( \therefore \) LCM of 5,6 and 15 = 30
In simple words: The Least Common Multiple (LCM) is the smallest number that appears in the list of multiples for all the given numbers. Listing out multiples helps visualize and identify this common multiple.
🎯 Exam Tip: For smaller numbers, listing multiples is a straightforward way to find the LCM. Ensure you list enough multiples to find the first common one, and double-check your arithmetic for each table to avoid errors.
Maths Class 7 Curriculum Solutions: Chapter 3 Set 13 HCF and LCM
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