Question

Difficulty: MediumFactors, Multiples, and Prime Factorization

Three different wooden boards have lengths of 34\frac{3}{4} meters, 910\frac{9}{10} meters, and 158\frac{15}{8} meters, respectively. A carpenter wants to cut them into smaller pieces of equal length such that no material is wasted. What is the maximum possible length of each smaller piece?

  1. A
    452\frac{45}{2} meters
  2. B
    32\frac{3}{2} meters
  3. 340\frac{3}{40} metersAnswer
  4. D
    245\frac{2}{45} meters

Answer

The correct answer is 340\frac{3}{40} meters.
To find the maximum equal length that can be cut from all three boards, we must find the Highest Common Factor (HCF) of their lengths. The formula for the HCF of fractions is the HCF of the numerators divided by the LCM of the denominators. The numerators are 3,9,3, 9, and 1515, yielding an HCF of 33. The denominators are 4,10,4, 10, and 88, yielding an LCM of 4040. Thus, the maximum length is 340\frac{3}{40} meters.

Step-by-Step Solution

1
Identify the mathematical operation required.
The problem asks for the 'maximum possible length' that can evenly divide all three board lengths without remainders. This requires finding the Highest Common Factor (HCF) of the three fractions.
Cutting multiple items into the largest possible equal smaller pieces is a classic application of HCF.
2
Recall the formula for finding the HCF of fractions.
HCF of fractions=HCF of NumeratorsLCM of Denominators\text{HCF of fractions} = \frac{\text{HCF of Numerators}}{\text{LCM of Denominators}}
This is the standard rule for finding the greatest common divisor of a set of rational numbers.
3
Calculate the HCF of the numerators (3,9,153, 9, 15).
The factors of 33 are 1,31, 3. The factors of 99 are 1,3,91, 3, 9. The factors of 1515 are 1,3,5,151, 3, 5, 15. The highest common factor is 33.
The numerator of our final fraction must be the HCF of the given numerators.
4
Calculate the LCM of the denominators (4,10,84, 10, 8).
The prime factorization of 44 is 222^2. For 88, it is 232^3. For 1010, it is 2×52 \times 5. The LCM takes the highest power of each prime present: 23×5=8×5=402^3 \times 5 = 8 \times 5 = 40.
The denominator of our final fraction must be the LCM of the given denominators.
5
Combine the results to find the final HCF.
340\frac{3}{40} meters.
Applying the formula from Step 2 with the values found in Step 3 and Step 4.

Key Concept

HCF and LCM of Fractions
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