three pieces of timber 42 m 49 m and 63 m long have to be divided into the planks of the same length. what is the greatest possible length of each plank?
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Answered by
20
Write 42 ,49 and 63 as product of prime factors and find their hcf
42=2×3*7
49=7×7
63=3×3×7
Therefore
Hcf(42,49,63)=7
Greatest possible plank of equal length = 7m
42=2×3*7
49=7×7
63=3×3×7
Therefore
Hcf(42,49,63)=7
Greatest possible plank of equal length = 7m
Answered by
1
The lengths of three pieces of timber are 42m, 49m and 63m respectively.
We have to divide the timber into equal length of planks.
∴ Greatest possible length of each plank = HCF (42, 49, 63) Prime factorization: 42 = 2 × 3 × 7 49 = 7 × 7 63 = 3 × 3 × 7
∴HCF = Product of the smallest power of each common prime factor involved in the numbers = 7
Hence, the greatest possible length of each plank is 7m.
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