The answer is B.

I’ve spent a couple of hours trying to figure this out but still couldn’t. Can someone explain in detail how the paths and the values of current/voltage does change by closing and opening the switches? Some solution says voltage decreases but current doesn’t change therefore, bulbY gets dimmer after the change while the other says opposite to it… I’m so stuck with this type of questions related to switch and bulb.

Hey, I think I have a solution to why the answer is B. It’s helpful to sort of create an imaginary vertical line separating Bulb X and Bulb Y. When you do so, you can now deal with either side separately. On the Bulb X side, when you “cut” the third parallel circuit, you increase the voltage in Bulb X (since Vt (12V) is equal to the sum of all the voltages of the parallel circuits). Now, as for Bulb Y, according to the formula of voltage in series, Vt=V1+V2. So, when you increase the voltage value on the Bulb X side, you have to decrease the voltage value on the Bulb Y side to make the sum of the two sides 12 V. This decrease in voltage on the Bulb Y side makes it dimmer and the increase in voltage on the Bulb X side makes it brighter. Hope this helped.