Maharashtra Board Class 6 Maths Chapter 9 HCF LCM Set 24 Solutions

HCF-LCM Class 6 Maths Chapter 9 Practice Set 24 Solutions Maharashtra Board

Std 6 Maths Practice Set 24 Solutions Answers

Question 1. 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
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
In simple words: The HCF (Highest Common Factor) of two or more numbers is the largest number that divides each of them without leaving a remainder. To find it, list all factors of each number and identify the largest factor they share.

🎯 Exam Tip: Clearly show the steps of listing factors for both numbers to earn full marks in HCF problems. Ensure the common factors are correctly identified before stating the highest one.

Question 2. 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?
Answer:
Solution:
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.
In simple words: To find the maximum length of equal-sized square beds, you need to calculate the Highest Common Factor (HCF) of the land's length and width. This HCF will be the largest side length that perfectly divides both dimensions.

🎯 Exam Tip: For problems requiring division into equal parts or finding the largest common measure, HCF is the correct approach. Clearly state the factors and the derived HCF to justify your answer.

Question 3. Two ropes, one 8 metres long and the other 12 metres long are to be cut into pieces of the same length. What will be the maximum possible length of each piece be?
Answer:
Solution:
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.
In simple words: When cutting items into equal, largest possible pieces, you must find the Highest Common Factor (HCF) of their original lengths. This gives the greatest common length for the pieces.

🎯 Exam Tip: Recognize "maximum possible length" or "greatest common measure" as keywords for HCF problems. Show the factorization and identification of the HCF clearly.

Question 4. 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?
Answer:
Solution:
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.
In simple words: To form groups of equal and maximum size from different classes, calculate the HCF of the total students from each class. This ensures efficient grouping and minimizes the number of required guides.

🎯 Exam Tip: Problems involving distributing items into equal groups or minimizing resources (like guides) often require finding the HCF. Clearly explain *why* HCF is used for the "why" part of the question.

Question 5. 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?
Answer:
Solution:
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.
In simple words: To find the maximum equal weight for bags of different seed varieties, calculate the HCF of their total weights. Then, divide each total weight by this HCF to find the number of bags for each variety.

🎯 Exam Tip: When a question asks for the "maximum possible" equal measure for two or more quantities, think HCF. Remember to answer all parts of the question, including the number of resulting units/bags.