Class
Std.-VI
New
Elementary Mathematics Syllabus D-1
Chapter 1
Exercise 1.5 Page- 13
Problem 1:
Find the Highest Common Factor of 60 and 96.
Solution: Consider the numbers 60 and 96
The
list of factors of 60 is 1, 2 , 3 , 4 , 5 , 6 , 10 ,12 , 15 , 20 ,30 and 60
The
list of factors of 96 is 1, 2 , 3 , 4 , 6 , 8 ,12 , 16 , 24 ,32 , 48 and 96
The
list of all the common factors of 60 and 96 is 1, 2 , 3 , 4, 6 and 12
Therefore
the largest of these common factors is 12 and thus we call 12 the Highest
Common Factors (H C F) of 60 and 96
Another
method: The list of all factors of 60 = 2 × 2 × 3 × 5 and
96 = 2 × 2 × 2 × 2 × 2 × 3
Therefore H C F
= 2 × 2 × 3 = 12 Ans.
Problem 2: Find H C F of 36 , 24
, 144 , and 96 .
Solution: The list of factor of 36 = 2 × 2 × 3 × 3
24 = 2 × 2 × 2 × 3
144
= 2 × 2 × 2 × 2 × 3 × 3
96 = 2 × 2 × 2 × 2 × 2 × 3
Therefore
the H C F is = 2 × 2 × 3 = 12 Ans.
Exercise 1.6 Page- 16
Problem 1:
Find the Least Common Multiple(LCM) of 12 and 18.
Solution: Consider the
numbers 12 and 18
The
list of all multiples of 12 is 12, 24 , 36 , 48 , 60 ,72 ……
The
list of all multiples of 18 is 18 , 36 , 54 , 72 , 108 ………
The
list of all the common multiples of 12 and 18 is 36 and 72 ……
Therefore
the smallest of these common multiples is 36 and thus we call 36 the Least
Common multiples (LCM) of 12 and 18
Another
method: The list of all factors of 12 = 2 × 2 × 3 and
18 = 2 × 3 × 3
Therefore
L C M = 2 × 2 × 3 × 3 = 36 Ans
Problem 2:
Find the Least Common Multiple (LCM) of 24, 18 and 36
Solution: The list of all factors of 24
= 2 × 2 × 2 × 3
18
= 2 × 3 × 3
36 = 2 × 2 × 3 × 3
Therefore L C M
= 2 × 2 × 2 × 3 × 3 = 72 Ans
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