Difficulty: Easy
Correct Answer: 0.3 m
Explanation:
Introduction:
This problem is a practical application of highest common factor (HCF) or greatest common divisor (GCD) in a geometry and measurement context. The idea is that if two lengths have to be cut into equal smaller pieces without leaving any leftover material, the maximum length of each small piece must exactly divide both original lengths.
Given Data / Assumptions:
Concept / Approach:
The greatest possible equal length that divides both pipe lengths evenly is their HCF. Since the numbers are decimals, it is easier to convert them into centimetres and work with whole numbers. After finding the HCF of the two whole numbers, we convert the result back to metres to answer the question in the required units.
Step-by-Step Solution:
Convert to centimetres:1.5 m = 150 cm, 1.2 m = 120 cm.Find the HCF of 150 and 120.Prime factors: 150 = 2 × 3 × 5 × 5, and 120 = 2 × 2 × 2 × 3 × 5.Common factors: one 2, one 3, one 5 → HCF = 2 × 3 × 5 = 30.So the greatest possible piece length is 30 cm.Convert back to metres: 30 cm = 0.3 m.
Verification / Alternative check:
Check that 0.3 m divides both lengths exactly: 1.5 ÷ 0.3 = 5 pieces; 1.2 ÷ 0.3 = 4 pieces. Both are whole numbers, and using any longer piece (for example 0.4 m) will not divide both lengths exactly. Thus 0.3 m is the largest possible common piece length with no wastage.
Why Other Options Are Wrong:
0.13 m and 0.41 m do not divide either length neatly. 0.4 m divides 1.2 m but not 1.5 m. 0.2 m divides both but is smaller than 0.3 m, so it is not the greatest possible length. Only 0.3 m satisfies all conditions.
Common Pitfalls:
Some learners attempt trial and error with decimals instead of converting to centimetres and using HCF, which can be error prone. Others may mistakenly take the least common multiple instead of the greatest common divisor, which would correspond to a larger length that does not fit exactly into both pipes.
Final Answer:
The greatest possible length of each equal pipe piece is 0.3 m.
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