The key to problems like this is that rates add.
Let R = the rate for the larger auger and r = rate for the smaller auger. Both rates have the unit: bins/hr
The fact that together they can fill the bin in 6 hr is captured by 6 = 1/(R + r) or R + r = 1/6
The fact that larger can fill the bin in 5hr less than the smaller is captured by 5 = 1/r - 1/R
This gives two equations with two unknowns. The first equation can be solved for r and this substituted in the second to give a quadratic equation for R. The solution is R = 1/10, this, in turn, gives r = 1/15
Thus the time it takes the large auger alone to fill the bin is 1/R = 10 hrs.