Question

A new tire had a tread depth of 5 over 8 in, After driving one year the tire had 17 over 32 in. of tread remaining. What percent of the tread was worn?

Answer 1

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,

`\text{ }(5)/(8)\text{ = }\frac{5\text{ x 4}}{32}\text{ = }(20)/(32)`

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.

`\text{ }(5)/(8)\text{ - }(17)/(32)\text{ = }(20)/(32)\text{ - }(17)/(32)\text{ = }\frac{20\text{ - 17}}{32}\text{ = }(3)/(32)`

3/32 in. of the thread was worn out. Let's now determine its equivalent percentage.

`\text{ Percentage worn = }\frac{\text{ Depth worn out}}{\text{ Original depth}}\text{ = }((3)/(32))/((20)/(32))\text{ x 100}`
`(3)/(32)\text{ }/\text{ }(20)/(32)\text{ = }(3)/(32)\text{ x }(32)/(20)\text{ = }(3)/(20)`
`\text{ }(3)/(20)\text{ x 100 = 0.15 x 100}`
`\text{ = 15\%}`

Therefore,

The answer is 15%.15% of the thread was worn after a year.

The answer is 15%.