Melanie has a piece of cloth (12)AND1/3 yards long. How many 3/4 yard-long pieces can be cut from the cloth? 15 pieces 12 pieces 17 pieces

Ninja, I am not sure how well you thought this through.   It is correct but maybe there was a bit of luck involved here.  

"Since a 3/4 yard piece can't be cut out of a 1/3 yard-long piece, let's forget about this 1/3 yard piece"

This quote concerns me a little. There might be some left over from the 12 yard peice and together with the 1/3 yard maybe it would make another one.  Of course if you looked at the left over bit afterwards this would be alright.  That is probably what you were intending to do.

Otherwise it would be just

 

$$12\frac{1}{3}\div \frac{3}{4}$$

 

$${\frac{\left({\mathtt{12}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}} = {\frac{{\mathtt{148}}}{{\mathtt{9}}}} = {\mathtt{16.444\: \!444\: \!444\: \!444\: \!444\: \!4}}$$

Obviously you only want the whole bits so the answer is 16pieces

We have a piece of cloth 12 and 1/3 yards long.

Since a 3/4 yard piece can't be cut out of a 1/3 yard-long piece, let's forget about this 1/3 yard piece and just figure out how many 3/4 yard pieces can be cut out of a 12 yard piece.

12/1 ÷ 4/3

To divide two fractions, multiply by the reciprocal.

12/1 x 4/3

48/3

16

This is a good point Melody, I should be carefull about this.

I was implying for this specific question that the extra 1/3 yard piece wouldn't be nessecary for this equation, beuase we are looking for whole pieces.

But the technical correct answer is actually 13.44444 pieces...

 

Thanks for the correction

No Ninja that is not what I was saying at all.

Your (first) answer of 16 pieces was totally correct.

I DO NOT think that an answer of 16.4 repeater peices is any where near as good an answer as just 16pieces

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Here is a different question so that you can see what I really meant.

How many 0.9 yard peices can be cut from a peice of cloth 6 and 8/10 yards long.

If I take your explanation literally then I can discard the 8/10 because I can't get a whole 0.9 yard out of it.

This will leave me with 6 divided by 0.9 = 6.6 repeater = 6 whole lengths

BUT

if you did not discard the 8/10 then you have 6.8/0.9= 7.5repeater = 7 whole lenths

SO you see in this second example you can see why you cannot discard the fractional or decimal bit until the very end.  Do you get what i am saying?

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