Latest Maharashtra State Board (SSC & HSC) 2026-27 Syllabus Digest & Solutions Updated!
Class 7Mathematics & Statistics2026-27 Syllabus

Chapter 3 HCF and LCM Practice Set 12 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 3 HCF and LCM Practice Set 12. Step-by-step solved exercises, numerical problems, and digest answers.

3 Solved Questions588 words

HCF and LCM Class 7 Practice Set 12 Answers Solutions Chapter 3

Question 1 Maharashtra Board Solution
i. 25, 40 ii. 56, 32 iii. 40, 60, 75 iv. 16, 27 v. 18, 32,48 vi. 105, 154 vii. 42, 45, 48 viii. 57, 75, 102 ix. 56, 57 x. 777, 315, 588
Solution & Step-by-Step Answer:
i. 25, 40 ∴ 25 = 5 × 5 40 = 2 × 2 × 2 × 5 ∴ HCF of 25 and 40 = 5

ii. 56, 32

∴ 56 = 2 × 2 × 2 × 7
32 = 2 × 2 × 2 × 2 × 2
∴ HCF of 56 and 32 = 2 × 2 × 2
∴ HCF of 56 and 32 = 8

iii. 40, 60, 75

∴ 40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 3 × 5
75 = 3 × 5 × 5
∴ HCF of 40, 60 and 75 = 5

iv. 16, 27

∴ 16 = 2 × 2 × 2 × 2 × 1
27 = 3 × 3 × 3 × 1
∴ HCF of 16 and 27 = 1

v. 18, 32,48

∴ 18 = 2 × 3 × 3
32 = 2 × 2 × 2 × 2 × 2
48 = 2 × 2 × 2 × 2 × 3
∴ HCF of 18, 32 and 48 = 2

vi. 105, 154

∴ 105 = 3 × 5 × 7
154 = 2 × 2 × 11
∴ HCF of 105 and 154 = 7

vii. 42, 45, 48

∴ 42 = 2 × 3 × 7
45 = 3 × 3 × 5
48 = 2 × 2 × 2 × 2 × 3
∴ HCF of 42,45 and 48 =3

viii. 57, 75, 102

∴ 57 = 3 × 19
75 = 3 × 5 × 5
102 = 2 × 3 × 17
∴ HCF of 57, 75 and 102 = 3

ix. 56, 57

∴ 56 = 2 × 2 × 2 × 7 × 1
57 = 3 × 19 × 1
∴ HCF of 56 and 57 = 1

x. 777, 315, 588

∴ 777 = 3 × 7 × 37
315 = 3 × 3 × 5 × 7
588 = 2 × 2 × 3 × 7 × 7
∴ HCF of 777, 315 and 588 = 3 × 7
HCF of 777, 315 and 588 = 21

Question 2 Maharashtra Board Solution
Find the HCF by the division method and reduce to the simplest form: i. ii. iii.
Solution & Step-by-Step Answer:
i.

ii.

iii.

Maharashtra Board Class 7 Maths Chapter 3 HCF and LCM Practice Set 12 Intext Questions and Activities

Question 1 Maharashtra Board Solution
In each of the following examples, write all the factors of the numbers and find the greatest common divisor. (Textbook pg. no. 17) i. 28, 42 ii. 51, 27 iii. 25, 15, 35
Solution & Step-by-Step Answer:
i. Factors of 28 = 1,2,4, 7, 14, 28 Factors of 42 = 1,2, 3, 6, 7, 14, 21, 42 ∴ HCF of 28 and 42 = 14

ii. Factors of 51 = 1, 3, 17, 51
Factors of 27 = 1, 3, 9, 27
∴ HCF of 51 and 27 = 3

iii. Factors of 25 = 1, 5, 25
Factors of 15 = 1, 3, 5, 15
Factors of 35 = 1, 5, 7, 35
∴ HCF of 25, 15 and 35 = 5