# Thread: Word Problem - Basic Math

1. ## Word Problem - Basic Math

Could someone please explain how to solve this? Thanks.

2. ## Re: Word Problem - Basic Math

You need to find out how much is 80% of 10.

3. ## Re: 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.

4. ## Re: 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.

5. ## Re: Word Problem - Basic Math

Originally Posted by Deveno
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.

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.

6. ## Re: Word Problem - Basic Math

Originally Posted by Sprinkledozer
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.
80% of $\displaystyle \frac{10}{32}$ equals $\displaystyle \frac{8}{32}$

This can be calculated:

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

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

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

How much tread wear remains if the tread depth is $\displaystyle \frac{4}{32}$?
We now consider that the legal minimum for tread is $\displaystyle \frac{2}{32}$
$\displaystyle \frac{4}{32}-\frac{2}{32}=\frac{2}{32}$ inches of tread remaining.