Question 871083
I assume that calculating approximate decimal values with a calculator and comparing them is considered cheating.
The traditional way to compare fractions, is to express them all with a common denominator.
Sometimes, depending on the particular numbers, there is an easier way, but I do not see an easier way this time.
{{{27=3*9=3*3*3}}} is a multiple of {{{3}}} and a multiple of {{{9}}},
but {{{27}}} has no factor in common with prime number {{{5}}} ,
so the least common multiple (the smallest common denominator) is
{{{135=27*5=3*9*5=3*(9*5)=3*45=9*(3*5)=9*15}}}
 
{{{0.6=6/10=3/5=3*27/(5*27)=81/135}}}
{{{7/9=7*15/(9*15)=105/135}}}
A. {{{23/27=23*5/(27*5)=115/135>103/135=7/9}}} ,
so {{{23/27}}} is not between {{{81/135=0.6}}} and {{{105/135=7/9}}}
B. {{{20/27=20*5/(27*5)=100/135}}}
If there is only one right answer,
since {{{81/135<100/135<105/135}}} ,
{{{0.6<highlight(20/27)<7/9}}}
 
In fact the other two numbers are not between {{{81/135=0.6}}} and {{{105/135=7/9}}} :
C. {{{14/27=14*5/(27*5)=70/135<81/135=0.6}}} and
D. {{{1/3=1*45/(3*45)=45/135<81/135=0.6}}}