Three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length. The greatest possible length of each plank and the number of such planks formed are cm, cm respectively.

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Detailed Solution

Three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length. The greatest possible length of each plank and the number of such planks formed are 7cm, 22 cm respectively.
Given the length of three timbers as, 42 m, 49 m, and 69 m.
The length of the plank will be the HCF of the lengths of the three timbers, i.e., 42, 49 and 63.
So, we will proceed with the prime factorization of 42, 49 and 63 as follows,

42 = 2 × 3 × 7 49 = 7 × 7 63 = 3 × 3 × 7

The only number common is 7.
Therefore, HCF (42, 49, 63) = 7.
So, the greatest possible length of the plank in which the three pieces of timbers will be divided is obtained as 7m.
Now total number of such planks formed will be obtained by dividing the total length of the three timbers by the HCF, i.e., 7.
So, we have the total number of planks as,

= 42 + 49 + 63 7 = 1547 = 22

So, the total number of planks of 7 m each is 22.
Hence, the three pieces of timber of respective lengths will be divided into 22 planks, each of length 7 m.

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