Two pipes of length 1.5 m and 1.2 m are to be cut into equal pieces without leaving any extra length of pipes. The greatest length of the pipe pieces of same size which can be cut from these two lengths will be

The correct option is 0.3 metre.

Step-by-step Explanation:

To find the greatest length of the pipe pieces of the same size that can be cut from two pipes of lengths 1.5 m and 1.2 m without leaving any extra length, we need to find the Highest Common Factor (HCF) or Greatest Common Divisor (GCD) of the two lengths.

Let the lengths of the two pipes be:
Pipe 1: 1.5 m
Pipe 2: 1.2 m

To make the calculations simpler, we can convert these decimal lengths into integers by converting metres to decimetres (since 1 m = 10 dm) or by expressing them as fractions.

Method 1: Converting to a smaller unit (decimetres)
Let us convert the lengths from metres (m) to decimetres (dm):
Length of Pipe 1 = 1.5 m = 1.5 × 10 dm = 15 dm
Length of Pipe 2 = 1.2 m = 1.2 × 10 dm = 12 dm

Now, we find the HCF of 15 and 12 by prime factorization:
Prime factorization of 15 = 3 × 5
Prime factorization of 12 = 2 × 2 × 3 = 2² × 3

The common prime factor is 3. Therefore, the HCF is:
HCF(15, 12) = 3 dm

Converting this back to metres:
3 dm = 3 / 10 m = 0.3 m

Method 2: Using fractional division
We can express the decimal lengths as fractions:
1.5 = 15/10
1.2 = 12/10

The HCF of fractions is calculated as:
HCF of Fractions = (HCF of Numerators) / (LCM of Denominators)

Applying this to our values:
HCF of Numerators (15 and 12) = 3
LCM of Denominators (10 and 10) = 10

Therefore:
HCF(1.5, 1.2) = 3 / 10 = 0.3 m

Thus, the greatest length of the pipe pieces that can be cut is 0.3 metre.