 # Three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length. Question:

Three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length. What is the greatest possible length of each plank? How many planks are formed?

Solution:

The lengths of three pieces of timber are 42 m, 49 m and 63 m, respectively.
We have to divide the timber into equal length of planks.
∴ Greatest possible length of each plank = HCF(42, 49, 63)
Prime factorisation:

$42=2 \times 3 \times 7$

$49=7 \times 7$

$63=3 \times 3 \times 7$

∴ HCF = Product of smallest power of each common prime factor in the numbers = 7
Therefore, the greatest possible length of each plank is 7 m.
Now, to find the total number of planks formed by each of the piece, we divide the length of each piece by the HCF, i.e. by 7.
We know that;

$7 \times 6=42$

$7 \times 7=49$

$7 \times 9=63$

Therefore, total number of planks formed $=6+7+9=22$

Hence, total 22 planks will be formed.