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125% of s = 60
=> 1.25s = 60
=> s = 60*4/5
=> s = 48

60 – s = 60 – 48 = 12

let the total no of breads be x.
1st man 2nd man
x- (x/2) – 1/2 – 1/2((x-1)/2) – 1/2 ….. so on.
ans is 31.
first ate : 15.5 + .5 = 16 remaining 15
second ate : 7.5+0.5 = 8 remaining 7.
third ate : 3.5 +0.5 = 4 remaining 3.
fourth ate : 1.5 + 0.5 = 2 remaining 1
fifth ate : 0.5 + 0.5 = 1 remaining 0

430
5x5x6 = 150 cube

150 – 20( corner cube) = 130 cube

130 x 3 (side remaining) = 390
20 x 2 (side remaining) = 40
total side remaining = 390 + 40 = 430

Friends,
The Answer given by ‘Gaurav Sharma’ is correct and
the approach (bottom to top) suggested by ‘Shailesh’ is
good. But with minor correction we can arrive the solution
using this approach:
After 5th loot, No. of breads left = 3
after 4th loot, no. of breads left = (3+0.5)x2 = 7
after 3rd loot, no. of breads left = (7+0.5)x2 = 15
after 2nd loot, no. of breads left = (15+0.5)x2 = 31
after 1st loot, no. of breads left = (31+0.5)x2 = 63
So, before 1st loot, no. of breads left = (63+0.5)x2 = 127

x:y=18:11

when know that 13 is factor of 39 so it will also not be able to divide it differently. so the remainder will be same 20 but as we know that 20 is greater then 13 so we can divide it once again and in the casE of 13 the remainder will be 20-13=7