A jeweler is cutting equal-length pieces of gold wire from three different spools to make … — State PSC Exam

A jeweler is cutting equal-length pieces of gold wire from three different spools to make uniform necklace links without any wastage. The three spools contain gold wire of lengths $\frac{27}{2}$ cm, $\frac{45}{4}$ cm, and $\frac{63}{5}$ cm. What is the maximum possible length of each uniform gold wire piece that can be cut? (Provide your answer as a precise decimal)

Cevap: 0.45 cm

Cevap

0.45

The maximum uniform length is found by computing the Highest Common Factor (HCF) of the fractional wire lengths. Applying the formula yields an HCF of numerators (9) over the LCM of denominators (20), giving

\frac{9}{20}

, which correctly converts to exactly

0.45

cm.

Adım Adım Çözüm

Determine the mathematical operation required for the scenario.

Calculate the Highest Common Factor (HCF) of the three fractional lengths.

Cutting multiple lengths into the maximum possible equal segments without wastage is defined by the HCF.

Apply the rule for finding the HCF of fractions.

HCF = HCF(Numerators) / LCM(Denominators)

This is the standard formula for finding the greatest common divisor of multiple fractional values.

Find the HCF of the numerators 27, 45, and 63.

HCF(27, 45, 63) = 9

9 is the largest integer that divides perfectly into 27, 45, and 63.

Find the LCM of the denominators 2, 4, and 5.

LCM(2, 4, 5) = 20

20 is the smallest integer that is a multiple of 2, 4, and 5.

Combine the results into the final fraction and convert to a decimal.

\frac{9}{20} = 0.45

The question requires the precise decimal representation of the fraction.

Anahtar Kavram

Calculating the Highest Common Factor (HCF) of fractions to solve optimization word problems.

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