Answer
See work below.
Work Step by Step
a. Relate the power to the voltage, for fixed resistance.
$$P=V^2/R$$
$$\frac{P_{105}}{P_{117}}=\frac{(105V)^2/R}{(117V)^2/R}=0.805$$
There is a 19.5 percent decrease in power.
b. At the voltage of 105V, and lower power, the resistor won’t get as hot as it does at 117V.
This means that the resistance is lower. This would result in a higher power output at 105 volts than what we calculated with the assumption that R was the same at both voltages. The actual decrease in power would be smaller than what we calculated in part A.
