A military aviation fuel facility has a bulk storage tank that is full. First, \frac{2}{7}… — Armed Services Vocational Aptitude Battery (ASVAB)

A military aviation fuel facility has a bulk storage tank that is full. First, $\frac{2}{7}$ of the total capacity is pumped out to supply an active squadron. Then, $\frac{2}{3}$ of the remaining fuel is transferred to an underground reserve tank. The fuel that is left in the bulk tank is then divided equally to fill 15 identical fuel transport vehicles. If each transport vehicle holds $1 \frac{3}{4}$ tons of fuel when filled from this process, what was the original total capacity of the bulk storage tank in tons?

A
$36$
B
$43 \frac{3}{4}$
$110 \frac{1}{4}$ Answer
D
$551 \frac{1}{4}$

Answer

The original total capacity of the bulk storage tank was

110 \frac{1}{4}

tons.

By working step-by-step through the fractions, we determine that the final amount of fuel distributed to the trucks represents 5/21 of the original total capacity. Dividing this evenly among 15 trucks means each truck gets 1/63 of the original total. Since each truck holds 7/4 tons, the original capacity is exactly 63 times 7/4, which simplifies to 441/4, or 110 1/4 tons.

Step-by-Step Solution

Calculate the fraction of total fuel remaining after supplying the squadron.

1 - \frac{2}{7} = \frac{5}{7}

The active squadron takes

\frac{2}{7}

of the total fuel, so the rest is left in the tank.

Calculate the fraction of total fuel left after transferring to the underground reserve.

\frac{1}{3} \times \frac{5}{7} = \frac{5}{21}

Since

\frac{2}{3}

of the remaining fuel is transferred out, exactly

\frac{1}{3}

of that remaining fuel is kept in the tank.

Calculate the fraction of the total capacity that each of the 15 transport vehicles receives.

\frac{5}{21} \div 15 = \frac{5}{21} \times \frac{1}{15} = \frac{1}{63}

The remaining

\frac{5}{21}

of the total fuel is distributed equally among 15 vehicles.

Set up an equation to find the total capacity (

C

) based on the known weight in one vehicle.

\frac{1}{63} \times C = 1 \frac{3}{4}

Each vehicle receives

\frac{1}{63}

of the total capacity, which we are given as

1 \frac{3}{4}

tons (or

\frac{7}{4}

tons).

Solve the equation for the original total capacity.

C = \frac{7}{4} \times 63 = \frac{441}{4} = 110 \frac{1}{4}

tons

Multiply the fraction

\frac{7}{4}

by the reciprocal of

\frac{1}{63}

to isolate and solve for

C

Key Concept

Solving multi-step fraction problems involving parts of remainders and distribution into equal batches.

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