How do you divide #\frac { 2\frac { 1} { 2} } { \frac { 5} { 8} }#?

2 Answers
Mar 22, 2017

#(2 1/2)/(5/8)=4#

Explanation:

Dividing a fraction, say #a/b#, by another fraction say #c/d#, is equivalent to multiplying by its multiplicative inverse or reciprocal i.e. #d/c#. Hence #(a/b)/(c/d)# or #a/b-:c/d# is equal to #a/bxxd/c#.

Hence, as #2 1/2=5/2#,

#(2 1/2)/(5/8)#

= #5/2xx8/5#

= #cancel5^1/cancel2^1xxcancel8^4/cancel5^1#

= #4/1=4#

Mar 22, 2017

Convert to improper fractions; get common denominators; cancel the common denominators away. Divide as normal.

OR

Convert to improper fractions; invert and multiply. (i.e. #(" "a/b" ")/(c/d)=a/b xx d/c.#)

The answer is 4.

Explanation:

The question is asking: how many times can you take #5/8# from #2 1/2#?

You can think of it this way: say you have a piece of string #2 1/2# inches long, and you want to cut it into pieces that are only #5/8# inch long. How many of those smaller pieces could you get?

To do this division, we need to convert both the top fraction and the bottom fraction to the same denominator (or "units").

#2 1/2" "=" "4/2 + 1/2" "=" "color(red)(5/2)#

and

#color(red)(5/2)" "=" "5/2xx1" "=" "5/2 xx 4/4" "=" "20/8#

So our question is now

#" "(20/8 )/(5/8)#

So, how many times can we take #5/8# from #20/8#? This is the same as asking how many times we can take 5 from 20, since the #8^"th""s"# are the same for both numbers.

In other words:

#(" "20/8" ")/(5/8)=20/5#,

which is 4.

(If I was counting by #5/8#ths, I would need 4 of them to get to #2 1/2#:

#5/8, 10/8, 15/8, 20/8#

where #20/8 = 2 1/2#, as shown above.)