Question 1132291
We are given that {{{1/m + 4/n = 1/12}}} such that {{{m}}} is odd and {{{m< 60}}}.
Let us try and find out the value of {{{n}}} in terms of {{{m}}}

=> {{{1/m = 1/12 - 4/n}}}
=> {{{4/n = 1/12 - 1/m}}}
=>{{{4/n = (m - 12)/12m}}}
=>{{{4/((m - 12)/12m)=n}}}
=>{{{n=48m/(m - 12)}}}


So, we can say that {{{m > 12}}}
Possible odd values of{{{ m}}} such that {{{12 < m < 60 }}} are 
  

{{{13 }}}| {{{15}}} | {{{17}}} | {{{19}}} | {{{21}}} | {{{23}}} | {{{25 }}}|{{{27}}} | {{{29}}} | {{{31}}}|{{{33}}}|{{{35}}}|{{{37}}}|{{{39}}}|{{{41}}}|{{{43}}}|{{{45}}}|{{{47}}}|{{{49}}}|{{{51}}}|{{{53}}}|{{{55}}}|{{{57}}}|{{{ 59}}}| ({{{24}}} integers)

check possible pairs:

{{{m= 13}}}, {{{n =  624}}}
{{{m = 15}}}, {{{n = 240}}}
{{{m = 17}}},{{{ n =816/5}}} (irreducible)=> not an integer
same is if {{{m=19}}}=>irreducible
{{{m = 21}}}, {{{n = 112}}}
{{{m = 23}}},{{{ n =1104/11}}}  =>irreducible
{{{m = 25}}}, {{{n =1200/13}}} (irreducible)
{{{m = 27}}}, {{{n =432/5}}} (irreducible)
{{{m = 29}}},{{{ n =1392/17}}} (irreducible)
{{{m = 31}}}, {{{n =1488/19}}} (irreducible)
{{{m = 33}}}, {{{n =528/7}}} (irreducible)
{{{m = 35}}}, {{{n =1680/23}}} (irreducible)

and  rest are irreducible


so, you have 3 pairs that satisfy given condition

{{{m= 13}}},{{{ n =  624}}}=>{{{13}}},{{{624}}}
{{{m = 15}}},{{{ n = 240}}}=>{{{15}}},{{{240}}}
{{{m = 21}}}, {{{n = 112}}}=>{{{21}}},{{{112}}}