Question 304802
0.27, 1/4, 3/8, 2/11, 11%
<pre><font size = 4 color = "indigo"><b>
Change them all to fractions:

{{{0.27}}} means the fraction {{{27/100}}}

{{{1/4}}} is already a fraction

{{{3/8}}} is already a fraction 

{{{2/11}}} is already a fraction 

{{{"11%"}}} means the fraction {{{11/100}}}

Now we line up the five fractions:

{{{27/100}}}, {{{1/4}}}, {{{3/8}}}, {{{2/11}}}, {{{11/100}}}

We get the LCD of all these, 

100=2*2*5*5
4 = 2*2
8 = 2*2*2
11= 11
100=2*2*5*5

So the LCD must contain each factor the most number of times it
occurs as a factor in any one of those, so the LCD must contain 
2 three times, because 2 occurs at most three times as a factor 
in 8. It must contain 5 two times because 5 occurs twice as a 
factor in 4 and 100, and it must contain 11 one time since 11 
occurs at most one time in 11 itself.

So LCD=2*2*2*5*5*11=2200.

Now write them all with the common denominator of 2200:

{{{0.27=27/100=(27*22)/(100*22)=594/2200}}}, 
{{{1/4= (1*550)/(4*550)=550/2200}}},
{{{3/8= (3*275)/(8*275)=825/2200}}},
{{{2/11=(2*200)/(11*200)=400/2200}}},
{{{"11%"=11/100=(11*22)/(100*22)=242/2200}}}.

The smallest of these is the one with the smallest
numerator (when expressend with the same common
denominator):

So {{{"11%"}}} is the smallest of the five.

You might find it easier to convert them all to decimals
instead of fractions:

{{{0.27}}} is already a decimal  

{{{1/4=0.25}}}

{{{3/8=0.375}}} 

{{{2/11="0.1818181818···"}}}  

{{{"11%"=0.11}}} 

The smallest has the smallest tenths digit.
The {{{2/11}}} and {{{"11%"}}} both have smallest tenths 
digits of 1.  So it's between those two.  So we go for the 
one of those two with the smaller hundredths digit, and the 
hundredths digit of {{{"11%"}}} is 1 while the hunmdredths 
digit of {{{2/11}}} is 8. So the smallest of the five is
{{{"11%"}}}.

Edwin</pre>