A technician is preparing a heavy-duty cable that will be cut into three separate sections… — Armed Services Vocational Aptitude Battery (ASVAB)

A technician is preparing a heavy-duty cable that will be cut into three separate sections. The total length of the cable before any cuts is $15 \frac{4}{5}$ yards. The first section measures $4 \frac{1}{3}$ yards long, and the second section measures exactly $6$ feet long. Assuming no material is lost during cutting, how many yards long must the third section be?

$9 \frac{7}{15}$ Answer
B
$5 \frac{7}{15}$
C
$9 \frac{3}{8}$
D
$6 \frac{1}{3}$

Answer

The third section must be

9 \frac{7}{15}

yards long.

The correct answer is found by first converting the 6 feet into 2 yards. Adding this to the first section gives a total used length of

6 \frac{1}{3}

yards. By finding a common denominator of 15, the total length (

15 \frac{12}{15}

) minus the used length (

6 \frac{5}{15}

) correctly equals

9 \frac{7}{15}

yards.

Step-by-Step Solution

Convert the length of the second section from feet to yards so all units match.

6 \text{ feet} = 2 \text{ yards}

Measurements must be in the same unit (yards) before they can be added or subtracted.

Add the lengths of the first and second sections to find the total length used so far.

4 \frac{1}{3} + 2 = 6 \frac{1}{3} \text{ yards}

This establishes the combined length that needs to be subtracted from the whole.

Find a common denominator for the total cable length and the used cable length.

The common denominator for

5

and

3

15

. Thus,

15 \frac{4}{5} = 15 \frac{12}{15}

and

6 \frac{1}{3} = 6 \frac{5}{15}

Fractions must have common denominators before they can be subtracted.

Subtract the used length from the total length to find the length of the third section.

15 \frac{12}{15} - 6 \frac{5}{15} = 9 \frac{7}{15} \text{ yards}

Subtracting the sum of the known parts from the total length yields the remaining unknown part.

Key Concept

Adding and subtracting mixed numbers with unlike denominators after converting units.

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