Question

A carpenter has two water pipes which are 60 dm and 72 dm long, respectively. He cuts them into equal lenghts. What is the greatest possible length of each pipe?

Answer 1

SOLUTION :
The greatest possible length of each pipe is equal to the HCF of 60 dm & 62 dm .
To find the HCF of these numbers use prime factorization method of calculating HCF.

Given :
Length of First water pipe = 60 dm
Length of First water pipe = 72 dm

HCF of 60 & 72
60 = 2² × 3¹ × 5¹
72 = 2³ × 3²

HCF (60,72) = 2² × 3¹ = 4 × 3 = 12
[Product of the smallest power of each common prime factor]

Hence, the greatest possible length of each pipe is 12 dm.

HOPE THIS WILL HELP YOU….

Answer 2

Solution:-

given by:-
length of first water pipe = 60dm.
length of secound water pipe = 79dm.
accourding to question,
He cuts them into equal lenghts.
then ,
take HCF of both pipe's length.

60 = 2×2×3×5
72= 2×2×2×3×3

HCF = 2×2×3 = 12.

HCF = The largest common factor of two or more numbers is called the highest common factor.

we want the greatest possible length of each pipe is 12 dm.


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