If #a/b < c/d# then #(color(blue)ma+color(red)nc)/(color(blue)mb+color(red)nd)# lies between #(a/b,c/d)#, if #color(blue)m>0# and #color(red)n>0#
Observe that #1/3=3/9=(color(blue)1xx1+color(red)1xx2)/(color(blue)1xx4+color(red)1xx5)#
Hence #1/3# lies in the range #(1/4,2/5)#
Similarly as #13/15<1# i.e. #13/15<9/9#, #(color(blue)1xx13+color(red)1xx9)/(color(blue)1xx15+color(red)1xx9)=(13+9)/(15+9)=22/24=11/12#
Hence #11/12# lies in the range #(13/15,1)#.
Although above is an interesting way for such problems, the easiest is to convert every number to decimal form.
Hence numbers are #4/7=0.57#, #1/3=0.33# and #11/12=0.916#
and ranges are #(0,0.25)#, #(0.866,1)# and #(0.25,0.40)#
and as is seen #1/3# lies in the range #(1/4,2/5)#, #11/12# lies in the range #(13/15,1)#, but #4/7# does not lie in any of the given range.
Additional Information #-# If in #(color(blue)ma+color(red)nc)/(color(blue)mb+color(red)nd)#, one of the #m# or #n# is negative, #(color(blue)ma+color(red)nc)/(color(blue)mb+color(red)nd)# gives out a number which is outside the range.
Observe that #(color(blue)((-9))xx0+color(red)4xx1)/(color(blue)((-9))xx1+color(red)4xx4)=(color(blue)3xx13+color(red)((-3))xx1)/(color(blue)3xx15+color(red)((-3))xx1)=(color(blue)((-2))xx1+color(red)3xx2)/(color(blue)((-2))xx4+color(red)3xx5)=4/7#
