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Chapter 9 HCF-LCM Practice Set 24 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 9 HCF-LCM Practice Set 24. Step-by-step solved exercises, numerical problems, and digest answers.

5 Solved Questions• 1017 words

6th Standard Maths Practice Set 24 Answers Chapter 9 HCF-LCM

Question 1 Maharashtra Board Solution
Find the HCF of the following numbers. i. 45, 30 ii. 16, 48 iii. 39, 25 iv. 49, 56 v. 120, 144 vi. 81, 99 vii. 24, 36 viii. 25, 75 ix. 48, 54 x. 150, 225
Solution & Step-by-Step Answer:
i. Factors of 45 = 1, 3, 5, 9,15, 45 Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30 ∴ HCF of 45 and 30 = 15

ii. Factors of 16 = 1, 2, 4, 8, 16
Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
∴ HCF of 16 and 48 = 16

iii. Factors of 39 = 1, 3, 13, 39
Factors of 25 = 1, 5, 25
∴ HCF of 39 and 25 = 1

iv. Factors of 49 = 1, 7, 49
Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56
∴ HCF of 49 and 56 = 7

v. Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120
Factors of 144 = 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144
∴ HCF of 120 and 144 = 24

vi. Factors of 81 = 1, 3, 9, 27, 81
Factors of 99 = 1, 3, 9, 11, 33, 99
∴ HCF of 81 and 99 = 9

vii. Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36
∴ HCF of 24 and 36 = 12

viii. Factors of 25 = 1, 5, 25
Factors of 75 = 1, 3, 5, 15, 25, 75
∴ HCF of 25 and 75 = 25

ix. Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Factors of 54 = 1, 2, 3, 6, 9, 18, 27, 54
∴ HCF of 48 and 54 = 6

x. Factors of 150 = 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150
Factors of 225 = 1, 3, 5, 9, 15, 25, 45, 75, 225
∴ HCF of 150 and 225 = 75

Question 2 Maharashtra Board Solution
If large square beds of equal size are to be made for planting vegetables on a plot of land 18 metres long and 15 metres wide, what is the maximum possible length of each bed?
Solution & Step-by-Step Answer:
Length of the land = 18 m Width of the land = 15 m The maximum length of each bed will be the greatest common factor of 18 and 15. Factors of 18 = 1, 2, 3, 6, 9, 18 Factors of 15 = 1, 3, 5, 15 ∴ HCF of 18 and 15 = 3 ∴ The maximum possible length of each bed is 3 metres.
Question 3 Maharashtra Board Solution
Two ropes, one 8 metres long and the other 12 metres long are to be cut into pieces of the same length. What will the maximum possible length of each piece be?
Solution & Step-by-Step Answer:
Length of first rope = 8 m Length of second rope = 12 m The maximum length of each piece will be the greatest common factor of 8 and 12. Factors of 8 = 1, 2, 4, 8 Factors of 12 = 1, 2, 3, 4, 6, 12 ∴ HCF of 8 and 12 = 4 ∴ The maximum possible length of each piece is 4 metres.
Question 4 Maharashtra Board Solution
The number of students of Std 6th and Std 7th who went to visit the Tadoba Tiger Project at Chandrapur was 140 and 196 respectively. The students of each class are to be divided into groups of the same number of students. Each group can have a paid guide. What is the maximum number of students that can be there in each group? Why do you think each group should have the maximum possible number of students?
Solution & Step-by-Step Answer:
Number of students of Std 6th = 140 Number of students of Std 7th = 196 The maximum number of students in each group will be the greatest common factor of 140 and 196. Factors of 140 = 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70, 140 Factors of 196 = 1, 2, 4, 7, 14, 28, 49, 98, 196 ∴ HCF of 140 and 196 = 28 ∴ Maximum students in each group are 28. Each group should have maximum number students so that there will be minimum number of groups and hence minimum number of paid guides.
Question 5 Maharashtra Board Solution
At the Rice Research Centre at Tumsar there are 2610 kg of seeds of the basmati variety and 1980 kg of the indrayani variety. If the maximum possible weight of seeds has to be filled to make bags of equal weight what would be the weight of each bag? How many bags of each variety will there be?
Solution & Step-by-Step Answer:
Weight of basmati rice = 2610 kg Weight of indrayani rice = 1980 kg The weight of each bag will be the greatest common factor of 2610 and 1980. Factors of 2610 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 29, 30, 45, 58, 87, 90, 145, 174, 261, 290, 435, 522, 870, 1305, 2610 Factors of 1980 = 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 15, 18, 20, 22, 30, 33, 36, 44, 45, 55, 60, 66, 90, 99, 110, 132, 165, 180, 198, 220, 330, 396, 495, 660, 990, 1980 ∴ HCF of 2610 and 1980 = 90 Maximum weight of each bag = 90 kg Number of bags of basmati rice = 2610 ÷ 90 = 29 Number of bags of indrayani rice = 1980 ÷ 90 = 22 Maximum weight of each bag is 90 kg. The number of bags of basmati rice is 29, and the number of bags of indrayani rice is 22.
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Chapter 9 HCF-LCM Practice Set 25