Could someone please explain how to solve this? Thanks.

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- Feb 1st 2013, 02:24 PMSprinkledozerWord Problem - Basic Math
Could someone please explain how to solve this? Thanks.

- Feb 1st 2013, 02:44 PMemakarovRe: Word Problem - Basic Math
You need to find out how much is 80% of 10.

- Feb 1st 2013, 02:49 PMDevenoRe: Word Problem - Basic Math
80% = 80/100 = 8/10 = 4/5.

so what is (4/5)(10/32)?

one you have that, say as a/b, you need to express this in 32nds.

if (by luck) there is a number c such that bc = 32, then a/b = ac/32. the "legal limit" of remaining tread is then (10 - ac)/32. - Feb 1st 2013, 05:24 PMSprinkledozerRe: Word Problem - Basic Math
Which question are you referring to?

8/32 = 1/4.

Shouldn't the question be: how do you express 1/16 as 32nds, since 2/32 = 1/16?

2/32 is the answer to both questions btw. - Feb 2nd 2013, 10:55 AMSprinkledozerRe: Word Problem - Basic Math
- Feb 2nd 2013, 06:25 PMPazeRe: Word Problem - Basic Math
80% of $\displaystyle \frac{10}{32}$ equals $\displaystyle \frac{8}{32}$

This can be calculated:

$\displaystyle \frac{10}{32}\cdot 0,80\rightarrow\frac{10\cdot 0,80}{32}=\frac{8}{32}$

Your question states: If 80% of the tread**has worn off.**

That means that we subtract 80% from the original number:

$\displaystyle \frac{10-8}{32}=\frac{2}{32}$ inches

As for the second part of your problem. It asks:

*How much tread wear remains if the tread depth is $\displaystyle \frac{4}{32}$?*

We now consider that the legal minimum for tread is $\displaystyle \frac{2}{32}$

That means that we have to find the difference between the two:

$\displaystyle \frac{4}{32}-\frac{2}{32}=\frac{2}{32}$ inches of tread remaining.