# Word Problem - Basic Math

• Feb 1st 2013, 03:24 PM
Sprinkledozer
Word Problem - Basic Math
Could someone please explain how to solve this? Thanks.
• Feb 1st 2013, 03:44 PM
emakarov
Re: Word Problem - Basic Math
You need to find out how much is 80% of 10.
• Feb 1st 2013, 03:49 PM
Deveno
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.
• Feb 1st 2013, 06:24 PM
Sprinkledozer
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.
• Feb 2nd 2013, 11:55 AM
Sprinkledozer
Re: Word Problem - Basic Math
Quote:

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.
• Feb 2nd 2013, 07:25 PM
Paze
Re: Word Problem - Basic Math
Quote:

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 $\frac{10}{32}$ equals $\frac{8}{32}$

This can be calculated:

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

$\frac{10-8}{32}=\frac{2}{32}$ inches
How much tread wear remains if the tread depth is $\frac{4}{32}$?
We now consider that the legal minimum for tread is $\frac{2}{32}$
$\frac{4}{32}-\frac{2}{32}=\frac{2}{32}$ inches of tread remaining.