A field generator consumes diesel fuel at a rate of 2 \frac{1}{4} gallons every 30 minutes… — Armed Services Vocational Aptitude Battery (ASVAB)

A field generator consumes diesel fuel at a rate of $2 \frac{1}{4}$ gallons every $30$ minutes. If the fuel tank currently holds $33 \frac{3}{4}$ gallons, how many hours can the generator run before the tank is empty?

$7 \frac{1}{2}$ Answer
B
$15$
C
$\frac{2}{15}$
D
$18$

Answer

7 \frac{1}{2}

To find how long the generator can run, first align the time units. Since the generator uses

2 \frac{1}{4}

gallons every 30 minutes, it uses twice that amount in one hour, which is

4 \frac{1}{2}

gallons per hour. Then, divide the total fuel (

33 \frac{3}{4}

gallons) by the hourly rate (

4 \frac{1}{2}

gallons/hour). Converting to improper fractions, this is

\frac{135}{4} \div \frac{9}{2}

. Multiplying by the reciprocal gives

\frac{135}{4} \times \frac{2}{9} = \frac{270}{36}

, which simplifies to

7 \frac{1}{2}

hours.

Step-by-Step Solution

Find the fuel consumption rate per hour.

2 \frac{1}{4} \times 2 = 4 \frac{1}{2}

gallons per hour

The rate is given for 30 minutes, which is half an hour. Multiply by 2 to find the hourly rate.

Convert the mixed numbers to improper fractions.

33 \frac{3}{4} = \frac{135}{4}

and

4 \frac{1}{2} = \frac{9}{2}

Fractions must be in improper form before dividing.

Divide the total fuel by the hourly rate to find the number of hours.

\frac{135}{4} \div \frac{9}{2} = \frac{135}{4} \times \frac{2}{9} = \frac{270}{36} = 7 \frac{18}{36} = 7 \frac{1}{2}

hours

Dividing total capacity by the rate of consumption gives the total time.

Key Concept

Fraction Division with Unit Conversion

Estimated Time:1m 30s

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