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Difficulty: HardFractions and Decimals

An industrial chemical plant stores a specialized solvent in three large cylindrical vats. The first vat contains 53.3‾53.\overline{3} liters, the second contains 71.1‾71.\overline{1} liters, and the third contains 26.6‾26.\overline{6} liters of the solvent. The plant manager wants to completely transfer the solvent from all three vats into identical, smaller drums such that every drum is completely filled, no solvent is left over in any vat, and solvents from different vats are not mixed. What should be the maximum possible capacity of each drum?

  1. 8.8‾8.\overline{8} litersAnswer
  2. B
    26.6‾26.\overline{6} liters
  3. C
    71.1‾71.\overline{1} liters
  4. D
    213.3‾213.\overline{3} liters

Answer

The maximum possible capacity of each drum is 8.8‾8.\overline{8} liters.
To find the maximum identical capacity that leaves no remainder, we must calculate the Highest Common Factor (HCF) of the three volumes. By converting the recurring decimals to fractions (1603\frac{160}{3}, 6409\frac{640}{9}, 803\frac{80}{3}) and applying the fraction HCF formula (HCF of numeratorsLCM of denominators\frac{\text{HCF of numerators}}{\text{LCM of denominators}}), we obtain 809\frac{80}{9}, which perfectly translates to 8.8‾8.\overline{8} liters.

Step-by-Step Solution

1
Convert the given recurring decimals representing the solvent volumes into their simplest fractional forms.
53.3‾=53+39=160353.\overline{3} = 53 + \frac{3}{9} = \frac{160}{3}, 71.1‾=71+19=640971.\overline{1} = 71 + \frac{1}{9} = \frac{640}{9}, and 26.6‾=26+69=80326.\overline{6} = 26 + \frac{6}{9} = \frac{80}{3}.
Fractional forms are required to accurately and properly compute the highest common factor (HCF) of non-integer values.
2
Identify the mathematical operation required based on the physical constraints described in the problem.
We must calculate the Highest Common Factor (HCF) of the three volumes: 1603\frac{160}{3}, 6409\frac{640}{9}, and 803\frac{80}{3}.
The solvent must be divided equally without any remainders, meaning the drum size must be a common factor of all three initial volumes, and the problem asks for the 'maximum possible capacity'.
3
Apply the standard formula for finding the HCF of multiple fractions.
The formula is: HCF=HCF of numeratorsLCM of denominators\text{HCF} = \frac{\text{HCF of numerators}}{\text{LCM of denominators}}.
This formula ensures the resulting fraction will evenly divide each of the original fractions without leaving a remainder.
4
Calculate the HCF of the numerators (160160, 640640, 8080) and the LCM of the denominators (33, 99, 33).
HCF(160,640,80)=80\text{HCF}(160, 640, 80) = 80, and LCM(3,9,3)=9\text{LCM}(3, 9, 3) = 9. Thus, the HCF of the fractions is 809\frac{80}{9}.
8080 is the largest integer dividing 160160, 640640, and 8080. 99 is the smallest integer divisible by 33, 99, and 33.
5
Convert the resulting fraction back into a recurring decimal.
809=8+89=8.8‾\frac{80}{9} = 8 + \frac{8}{9} = 8.\overline{8} liters.
The final calculated capacity should match the formatting style of the given options.

Key Concept

Calculating the Highest Common Factor (HCF) of recurring decimals by converting them to fractions and using the fraction HCF rule.
Estimated Time:2m 30s
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Topics

Fractions and Decimals

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