in my experience throwing a 3v and 4v 30Q cell together in parallel maybe increased the heat by 5 degrees F. MAYBE you could just throw the two in parallel when the one on the board is dying. 
I did the math based on ohm’s law and assuming this 30Q resistance and 10s3p
here’s how MY math goes and maybe someone can correct it:
first finding the resistance of the 3 cells in parallel would be 1/.034 + 1/.034 + 1/.034…and keeping it simple lets call it 1/.1, so answer being .01 for each p group and then add 10 in series getting to .1ohms for the battery. and you have to take into account the other battery’s resistance as well I think so would be .2ohms total for both batteries the voltage would be applied to.
assuming a 5v difference between the old battery and new, so 35v for the dead and 40 for the new ohm’s law says it would be a 25 amp charge current. for (3) 30Q cells the stated max charge current for 3p would be 12amps, but that’s a conservative number assuming any state of charge and at lowest states of charge you can do more. Id do it and see at what state of charge the battery starts to heat up knowing the charge rate will drop quickly as the cells get closer in voltage. a bms would likely be an obstacle