A 60W bulb means a bulb that when connected to it's rated supply voltage draws 60W. Unless otherwise stated one assumes that the rated supply voltage is the normal supply voltage in your country.

Your teacher is assuming that the bulbs act like resistors. If we make that assumption then we can solve your question in two steps.

1. Use the power ratings of the bulb and it's normal operating voltage to calculate the resistances.
2. Use the resistances calculated in the first step to calculate the behaviour of your actual circuit.

The first part of the question can be answered without actually doing the calculation. The lower rated bulb will have a higher resistance which means it will take a larger proportion of the voltage in the series circuit which means it will receive more power. For the latter part you have to actually work out the sums.

In practice the teachers assumption is highly inaccurate, but like with many textbook problems we have to run with it because we haven't been given any better information. In reality the resistance of an incandescent bulb increases singificantly with temperature. When you first turn a bulb on it draws far more current than it does after warming up.

What does this mean for our problem? it means two things.

1. The total power delivered to the two bulbs will be higher than the nieve calculations would suggest because the bulbs are cooler than they would be in normal operation.
2. The ratio of power delivered will be more extreme than the nieve calculations would suggest because the higher resistance (lower power rating) bulb will heat up less than the lower resistance (higher power rating) bulb.

A 60W bulb means a bulb that when connected to it's rated supply voltage draws 60W. Unless otherwise stated one assumes that the rated supply voltage is the normal supply voltage in your country.

Your teacher is assuming that the bulbs act like resistors. If we make that assumption then we can solve your question in two steps.

1. Use the power ratings of the bulb and it's normal operating voltage to calculate the resistances.
2. Use the resistances calculated in the first step to calculate the behaviour of your actual circuit.

The first part of the question can be answered without actually doing the calculation. The lower rated bulb will have a higher resistance which means it will take a larger proportion of the voltage in the series circuit which means it will receive more power. For the latter part you have to actually work out the sums.

In practice the teachers assumption is highly inaccurate, but like with many textbook problems we have to run with it because we haven't been given any better information. In reality the resistance of an incandescent bulb increases singificantly with temperature. When you first turn a bulb on it draws far more current than it does after warming up.

What does this mean for our problem? it means two things.

1. The total power delivered to the two bulbs will be higher than the nieve calculations would suggest because the bulbs are cooler than they would be in normal operation.
2. The ratio of power delivered will be more extreme than the nieve calculations would suggest because the higher resistance (lower power rating) bulb will heat up more than the lower resistance (higher power rating) bulb.