Question

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Answer 1

1) 1(1/3)+2(1/4)

= 4/3+9/4

lcm of 3 and 4 is 12

= (16+27)/12

=43/12

2) Maximum possible length = HCF of length of the ropes

HCF(8, 12) = 4

if you wanted to find HCF

2|8,12

2|4,6

|2,3

from left hand side terms 2×2=4

so 4 is HCM use prime numbers to divide (2,3,5.....) to get HCF

Hence, the maximum possible length of each piece must be 4 m.

I hope you understand