A civil supplies department procures three batches of edible oil weighing 0.\overline{6} m… — State PSC Exam

A civil supplies department procures three batches of edible oil weighing $0.\overline{6}$ metric tonnes, $1.\overline{3}$ metric tonnes, and $2.\overline{2}$ metric tonnes. The department intends to repackage the entire procured quantity into smaller, identical standardized containers. To minimize the number of containers used without mixing the batches, what should be the maximum capacity of each standardized container?

A
$6.\overline{6}$ metric tonnes
$0.\overline{2}$ metric tonnesCevap
C
$0.\overline{6}$ metric tonnes
D
$0.1$ metric tonnes

Cevap

The maximum capacity of each standardized container should be

0.\overline{2}

metric tonnes.

The maximum capacity of the standardized containers is determined by finding the Highest Common Factor (HCF) of the oil volumes. First, the given recurring decimals are converted to fractions:

\frac{2}{3}

\frac{4}{3}

, and

\frac{20}{9}

. The HCF of these fractions is found by taking the HCF of the numerators (2, 4, 20), which is 2, and dividing it by the LCM of the denominators (3, 3, 9), which is 9. This calculation yields

\frac{2}{9}

, which is equivalent to the recurring decimal

0.\overline{2}

Adım Adım Çözüm

Convert the recurring decimals representing the batch weights into fractions in their simplest form.

0.\overline{6} = \frac{6}{9} = \frac{2}{3}

1.\overline{3} = \frac{13-1}{9} = \frac{12}{9} = \frac{4}{3}

, and

2.\overline{2} = \frac{22-2}{9} = \frac{20}{9}

Mathematical operations involving divisors and multiples are most accurate when recurring decimals are expressed exactly as fractions.

Identify the mathematical requirement to find the maximum standardized container capacity.

We must determine the Highest Common Factor (HCF) of the three fractional quantities.

To divide all batches completely into identical containers of maximum capacity without leftovers, the container size must be the greatest common divisor of all the given volumes.

Apply the standard formula for calculating the HCF of multiple fractions.

\text{HCF of fractions} = \frac{\text{HCF of numerators}}{\text{LCM of denominators}}

This formula ensures that the resulting fraction will evenly divide each of the original fractional quantities.

Calculate the HCF of the numerators and the LCM of the denominators for

\frac{2}{3}

\frac{4}{3}

, and

\frac{20}{9}

Numerators (2, 4, 20): The

\text{HCF}

is 2. Denominators (3, 3, 9): The

\text{LCM}

is 9.

These values provide the numerator and denominator for the final maximum capacity.

Form the final fraction and convert it back to a recurring decimal to match the given options.

\text{Maximum capacity} = \frac{2}{9} = 0.\overline{2}

metric tonnes.

Converting back to recurring decimal format answers the problem in the original units.

Anahtar Kavram

Fractions and Decimals - Converting Recurring Decimals and Finding HCF of Fractions

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