Given:
a.) A new tire had a tread depth of 5 over 8 in. (5/8)
b.) After driving one year the tire had 17 over 32 in. of tread remaining. (17/32)
First, let's make the two fractions 5/8 and 17/32 into similar terms. (With the same denominator)
The LCM of 8 and 32 is 32. Therefore, let's transform 5/8 into its equivalent fraction with a denominator of 32.
We get,
[tex]\text{ }\frac{5}{8}\text{ = }\frac{5\text{ x 4}}{32}\text{ = }\frac{20}{32}[/tex]Therefore, the original tire depth is 5/8 or 20/32 in.
Next, let's determine the difference between the original and new depth after one year.
[tex]\text{ }\frac{5}{8}\text{ - }\frac{17}{32}\text{ = }\frac{20}{32}\text{ - }\frac{17}{32}\text{ = }\frac{20\text{ - 17}}{32}\text{ = }\frac{3}{32}[/tex]3/32 in. of the thread was worn out. Let's now determine its equivalent percentage.
[tex]\text{ Percentage worn = }\frac{\text{ Depth worn out}}{\text{ Original depth}}\text{ = }\frac{\frac{3}{32}}{\frac{20}{32}}\text{ x 100}[/tex][tex]\frac{3}{32}\text{ }\div\text{ }\frac{20}{32}\text{ = }\frac{3}{32}\text{ x }\frac{32}{20}\text{ = }\frac{3}{20}[/tex][tex]\text{ }\frac{3}{20}\text{ x 100 = 0.15 x 100}[/tex][tex]\text{ = 15\%}[/tex]Therefore,
The answer is 15%.15% of the thread was worn after a year.
The answer is 15%.
