Question:
Find the length of the plank that can be used to measure the lengths 4 m 20 cm and 5 m 4 cm exactly, in the least time.
Find the length of the plank that can be used to measure the lengths 4 m 20 cm and 5 m 4 cm exactly, in the least time.
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Always ensure all units are converted to a single common unit before starting HCF or LCM calculations.
To check your final answer, divide both original numbers by your HCF:
$420 \div 84 = 5$ (exact measurements)
$504 \div 84 = 6$ (exact measurements)
Since both yield whole numbers, $84$ cm is correct!
To check your final answer, divide both original numbers by your HCF:
$420 \div 84 = 5$ (exact measurements)
$504 \div 84 = 6$ (exact measurements)
Since both yield whole numbers, $84$ cm is correct!
Updated On: Jul 7, 2026
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Solution and Explanation
Step 1: Understanding the Question:
This question is a real-world application of "Real Numbers", specifically involving the concept of the Highest Common Factor (HCF).
We are given two different physical lengths: $4$ m $20$ cm and $5$ m $4$ cm.
We need to find the length of a single measuring plank that can measure both of these lengths "exactly".
This means the length of both items must be an exact integer multiple of the plank's length.
To achieve this in the "least time", we need the longest possible measuring unit, which minimizes the number of individual measurements.
Therefore, we need to compute the Highest Common Factor (HCF) of the two given lengths.
Step 2: Key Formula or Approach:
1. First, convert all given physical measurements into a single uniform unit (centimeters) to simplify calculations.
2. Express each number as a product of its prime factors using the Prime Factorization Method.
3. Calculate the HCF by finding the product of the lowest powers of all common prime factors:
\[ \text{HCF} = \prod (p_i)^{\min(e_i)} \]
Step 3: Detailed Explanation:
• Convert the lengths into centimeters:
- First length: $4\text{ m } 20\text{ cm} = (4 \times 100) + 20 = 420\text{ cm}$
- Second length: $5\text{ m } 4\text{ cm} = (5 \times 100) + 4 = 504\text{ cm}$
• Find the prime factorization of $420$:
Divide by 2: $420 \div 2 = 210$
Divide by 2: $210 \div 2 = 105$
Divide by 3: $105 \div 3 = 35$
Divide by 5: $35 \div 5 = 7$
Divide by 7: $7 \div 7 = 1$
\[ 420 = 2^2 \times 3^1 \times 5^1 \times 7^1 \]
• Find the prime factorization of $504$:
Divide by 2: $504 \div 2 = 252$
Divide by 2: $252 \div 2 = 126$
Divide by 2: $126 \div 2 = 63$
Divide by 3: $63 \div 3 = 21$
Divide by 3: $21 \div 3 = 7$
Divide by 7: $7 \div 7 = 1$
\[ 504 = 2^3 \times 3^2 \times 7^1 \]
• Identify the common prime factors and their lowest exponents:
- The common prime factor 2 has exponents 2 and 3; the lowest exponent is $2^2$.
- The common prime factor 3 has exponents 1 and 2; the lowest exponent is $3^1$.
- The common prime factor 7 has exponents 1 and 1; the lowest exponent is $7^1$.
- The factor 5 is not common to both.
• Compute the HCF:
\[ \text{HCF}(420, 504) = 2^2 \times 3^1 \times 7^1 = 4 \times 3 \times 7 = 84 \] Therefore, the maximum length of the measuring plank is $84$ cm.
Step 4: Final Answer:
The length of the plank that can be used is $84$ cm (or $0.84$ m).
This question is a real-world application of "Real Numbers", specifically involving the concept of the Highest Common Factor (HCF).
We are given two different physical lengths: $4$ m $20$ cm and $5$ m $4$ cm.
We need to find the length of a single measuring plank that can measure both of these lengths "exactly".
This means the length of both items must be an exact integer multiple of the plank's length.
To achieve this in the "least time", we need the longest possible measuring unit, which minimizes the number of individual measurements.
Therefore, we need to compute the Highest Common Factor (HCF) of the two given lengths.
Step 2: Key Formula or Approach:
1. First, convert all given physical measurements into a single uniform unit (centimeters) to simplify calculations.
2. Express each number as a product of its prime factors using the Prime Factorization Method.
3. Calculate the HCF by finding the product of the lowest powers of all common prime factors:
\[ \text{HCF} = \prod (p_i)^{\min(e_i)} \]
Step 3: Detailed Explanation:
• Convert the lengths into centimeters:
- First length: $4\text{ m } 20\text{ cm} = (4 \times 100) + 20 = 420\text{ cm}$
- Second length: $5\text{ m } 4\text{ cm} = (5 \times 100) + 4 = 504\text{ cm}$
• Find the prime factorization of $420$:
Divide by 2: $420 \div 2 = 210$
Divide by 2: $210 \div 2 = 105$
Divide by 3: $105 \div 3 = 35$
Divide by 5: $35 \div 5 = 7$
Divide by 7: $7 \div 7 = 1$
\[ 420 = 2^2 \times 3^1 \times 5^1 \times 7^1 \]
• Find the prime factorization of $504$:
Divide by 2: $504 \div 2 = 252$
Divide by 2: $252 \div 2 = 126$
Divide by 2: $126 \div 2 = 63$
Divide by 3: $63 \div 3 = 21$
Divide by 3: $21 \div 3 = 7$
Divide by 7: $7 \div 7 = 1$
\[ 504 = 2^3 \times 3^2 \times 7^1 \]
• Identify the common prime factors and their lowest exponents:
- The common prime factor 2 has exponents 2 and 3; the lowest exponent is $2^2$.
- The common prime factor 3 has exponents 1 and 2; the lowest exponent is $3^1$.
- The common prime factor 7 has exponents 1 and 1; the lowest exponent is $7^1$.
- The factor 5 is not common to both.
• Compute the HCF:
\[ \text{HCF}(420, 504) = 2^2 \times 3^1 \times 7^1 = 4 \times 3 \times 7 = 84 \] Therefore, the maximum length of the measuring plank is $84$ cm.
Step 4: Final Answer:
The length of the plank that can be used is $84$ cm (or $0.84$ m).
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