Two metal pipes have lengths 1.5 m and 1.2 m respectively. Both pipes are to be cut into smaller pieces of equal length, with no leftover material from either pipe. What is the greatest possible length (in metres) of each piece so that all the pieces are of the same size and there is no wastage?

Introduction / Context:
This is a classic application of the highest common factor (HCF) or greatest common divisor (GCD) concept in a geometric setting. The question asks for the greatest possible length of equal pieces into which two given pipe lengths can be cut without leaving any remainder. This is equivalent to finding the greatest length that divides both lengths exactly, which is precisely the idea of finding the HCF of the two measurements when expressed in a common unit.

Given Data / Assumptions:

• Length of first pipe = 1.5 m.
• Length of second pipe = 1.2 m.
• Both pipes are to be cut into equal pieces.
• No leftover length (no wastage) is allowed from either pipe.
• All pieces must have the same length.

Concept / Approach:
To find the greatest possible common piece length, we need a length that divides both 1.5 m and 1.2 m exactly. This is a greatest common divisor in the context of measurements. It is easier to work in centimetres to avoid decimals. We convert both lengths to centimetres, find their HCF, and then convert the result back to metres. The HCF ensures that the pieces are as long as possible while still dividing both original lengths exactly.

Step-by-Step Solution:
Convert lengths to centimetres: 1.5 m = 150 cm, 1.2 m = 120 cm.Now we need the HCF of 150 and 120.List prime factors: 150 = 2 * 3 * 5 * 5, and 120 = 2 * 2 * 2 * 3 * 5.Common prime factors are 2, 3, and 5.Their product is 2 * 3 * 5 = 30.So the HCF of 150 and 120 is 30.This means the greatest common length in centimetres is 30 cm.Convert back to metres: 30 cm = 30 / 100 m = 0.3 m.Therefore, each piece can be at most 0.3 m long.

Verification / Alternative check:
We can verify by dividing each original length by 0.3 m directly. For the first pipe: 1.5 / 0.3 = 5 pieces, which is an integer. For the second pipe: 1.2 / 0.3 = 4 pieces, also an integer. No leftover length remains in either case. If we tried a larger value, such as 0.4 m, then 1.2 / 0.4 = 3 pieces works, but 1.5 / 0.4 is not an integer, so 0.4 m is too large and not acceptable. This confirms that 0.3 m is indeed the greatest possible common length.

Why Other Options Are Wrong:
0.13 m and 0.2 m both divide the lengths, but they are smaller than 0.3 m and hence do not represent the greatest possible common length. The values 0.4 m and 0.41 m fail to divide one or both of the original lengths exactly, leaving remainders. Therefore, these cannot be used to cut the pipes into equal pieces without wastage. Only 0.3 m satisfies both conditions: exact division and maximal possible length.

Common Pitfalls:
Some candidates try to work directly with decimals and make rounding errors. Others may find a common divisor but forget to check whether it is the greatest. Mistakes can also occur when converting between metres and centimetres. A systematic approach using integer measurements and prime factorization or the Euclidean algorithm helps avoid these issues and quickly leads to the correct HCF.

Final Answer:
The greatest possible length of each piece is 0.3 m.