Finding the Greatest Length for Cutting Pipes

The problem asks us to find the greatest possible length of equal pieces that can be cut from two pipes measuring 1.5 m and 1.2 m, without any length left over. This is a problem that requires finding the greatest common divisor (GCD) of the two lengths.

The greatest common divisor (GCD) is the largest positive integer that divides two or more integers without leaving a remainder. In the context of lengths, we are looking for the largest possible length that is a common factor of both 1.5 m and 1.2 m.

Step 1: Convert Decimals to Integers

To easily find the GCD, it's often helpful to work with integers. We can convert 1.5 m and 1.2 m into centimeters or by multiplying by a factor that makes them whole numbers. Multiplying both lengths by 10 converts them to 15 (representing 15 dm or a relative unit) and 12 (representing 12 dm or a relative unit).

• Length 1: 1.5 m \(\times\) 10 = 15
• Length 2: 1.2 m \(\times\) 10 = 12

Step 2: Find the GCD of the Integers

Now we need to find the greatest common divisor of 15 and 12. We can do this by listing the factors of each number or by using prime factorization.

Let's list the factors:

• Factors of 15: 1, 3, 5, 15
• Factors of 12: 1, 2, 3, 4, 6, 12

The common factors of 15 and 12 are the numbers that appear in both lists. These are 1 and 3.

The greatest among the common factors is 3. So, the GCD of 15 and 12 is 3.

Alternatively, using prime factorization:

• Prime factorization of 15: \(3 \times 5\)
• Prime factorization of 12: \(2 \times 2 \times 3\) = \(2^2 \times 3\)

The common prime factor is 3. The lowest power of the common prime factor is \(3^1\). Therefore, GCD(15, 12) = 3.

Step 3: Convert the GCD back to the Original Unit

Since we multiplied the original lengths by 10 in Step 1, we must now divide the GCD we found by 10 to get the answer in meters.

GCD in meters = \(\frac{3}{10}\) m = 0.3 m.

Explanation of Why GCD is Used

We need to cut the pipes into equal pieces. This means the length of each piece must be a factor of the length of the first pipe (1.5 m) and also a factor of the length of the second pipe (1.2 m). To make the pieces as long as possible, we need to find the greatest length that is a common factor of both original lengths. This is precisely the definition of the Greatest Common Divisor (GCD).

If the piece length is 0.3 m:

• From the 1.5 m pipe, we can cut \(\frac{1.5}{0.3} = 5\) pieces.
• From the 1.2 m pipe, we can cut \(\frac{1.2}{0.3} = 4\) pieces.

In both cases, the division results in a whole number, meaning there is no extra length left over. Any length greater than 0.3 m that is a common factor is not possible because 0.3 m is the greatest common factor.

Step 4: Compare with Options

The calculated greatest length of the pipe pieces is 0.3 m.

Let's look at the options:

Only 0.3 m is a common factor for both 1.5 m and 1.2 m. Also, we determined it is the greatest common factor.

Conclusion

The greatest length of the pipe pieces of the same size which can be cut from these two lengths without leaving any extra length is 0.3 m.

Revision Table: Pipe Cutting Problem

Additional Information: Understanding GCD and HCF

The term Greatest Common Divisor (GCD) is also known as the Highest Common Factor (HCF). Both terms refer to the same mathematical concept.

• Factor: A factor of a number is a number that divides into it exactly, without leaving a remainder. For example, factors of 12 are 1, 2, 3, 4, 6, and 12.
• Common Factor: A common factor of two or more numbers is a number that is a factor of all of them. For example, common factors of 12 and 18 are 1, 2, 3, and 6.
• Greatest Common Divisor (GCD) / Highest Common Factor (HCF): This is the largest of the common factors. For 12 and 18, the GCD/HCF is 6.

Applications of GCD/HCF often involve problems where you need to divide items of different quantities or lengths into the largest possible equal-sized groups or pieces, just like in this pipe cutting problem.

There are several methods to find the GCD:

• Listing Factors: List all factors of each number and find the largest common one. This is practical for small numbers.
• Prime Factorization: Find the prime factorization of each number. The GCD is the product of the common prime factors raised to the lowest power they appear in any of the factorizations.
• Euclidean Algorithm: A more efficient method for larger numbers, based on the principle that the GCD of two numbers does not change if the larger number is replaced by its difference with the smaller number.

In our pipe problem, we used both listing factors and prime factorization for the integers 15 and 12, confirming the GCD is 3, which translates to 0.3 m for the pipe lengths.