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A friend gives a neighbor [tex]$13 \frac{2}{3}$[/tex] feet of twine from a brand new spool, leaving [tex]$38 \frac{2}{5}$[/tex] feet of twine on the spool. How many total feet of twine were originally on the spool?

A. [tex][tex]$51 \frac{1}{15}$[/tex][/tex]
B. [tex]$51 \frac{4}{5}$[/tex]
C. [tex]$52 \frac{1}{15}$[/tex]
D. [tex][tex]$52 \frac{1}{5}$[/tex][/tex]


Sagot :

To determine the total feet of twine originally on the spool, we need to sum the amount of twine given to the neighbor and the amount of twine left on the spool.

1. Start with the amount of twine given to the neighbor: [tex]\( 13 \frac{2}{3} \)[/tex]
- Convert this mixed number to an improper fraction:
[tex]\[ 13 \frac{2}{3} = 13 + \frac{2}{3} = \frac{39}{3} + \frac{2}{3} = \frac{41}{3} \][/tex]

2. Next, consider the twine left on the spool: [tex]\( 38 \frac{2}{5} \)[/tex]
- Convert this mixed number to an improper fraction:
[tex]\[ 38 \frac{2}{5} = 38 + \frac{2}{5} = \frac{190}{5} + \frac{2}{5} = \frac{192}{5} \][/tex]

3. Now, find a common denominator to add these two fractions. The least common multiple of 3 and 5 is 15.

4. Convert both fractions to have the common denominator of 15:
[tex]\[ \frac{41}{3} = \frac{41 \times 5}{3 \times 5} = \frac{205}{15} \][/tex]
[tex]\[ \frac{192}{5} = \frac{192 \times 3}{5 \times 3} = \frac{576}{15} \][/tex]

5. Add the two fractions:
[tex]\[ \frac{205}{15} + \frac{576}{15} = \frac{205 + 576}{15} = \frac{781}{15} \][/tex]

6. Convert the improper fraction back to a mixed number:
[tex]\[ \frac{781}{15} = 52 \text{ with a remainder of } 781 - (52 \times 15) = 11 \implies 52 \frac{11}{15} \][/tex]

Therefore, the total feet of twine originally on the spool is [tex]\(\boxed{52 \frac{11}{15}}\)[/tex].

Upon examining the options given:

A [tex]\(51 \frac{1}{15} \)[/tex]
B [tex]\(51 \frac{4}{5} \)[/tex]
C [tex]\(52 \frac{1}{15} \)[/tex]
D [tex]\(52 \frac{1}{5} \)[/tex]

The correct answer is:

C [tex]\(52 \frac{1}{15} \)[/tex]

Considering the fact that we have incorrectly matched the fraction, the actual computed fraction should be [tex]\(\frac{1}{15}\)[/tex]. Therefore the correct choice is [tex]\(\boxed{52 \frac{1}{15}}\)[/tex].
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