There are 100 lights which are functional and each is controlled by a

Hi Vyshak,

Congrats on being made a MOD..

Now about the Q, you have understood the Q correctly but missed out that it is a ON/OFF switch and both kids press the switch..
Relook into the answer

Thanks chetan2u. I have edited my solution now. Is it correct now?

I hope to avoid these silly mistakes during the exam!!

Hi,
You are correct now..
Very important to note every important aspect of a Q. Under STRESS in actual exam, it is even more difficult..

A turns every 3rd switch and B turns every 5th switch. LCM of 3 and 5 is 15. Now consider first 15 switches only.
A turns switches numbered 3,6,9,12 and 15.Therefore, 5 are on and 10 off.
B turns switches numbered 5,10 and 15. So, 2 on and 1 off (15th switch was earlier turned on by A)
So, for a set of 15 switches 6 are on and 9 are off.

Repeating the patern for 90 switches yields 36 on and 54 off

From 91 to 100, A turns 93, 96 and 99, and B turns on 95 and 100 so, 36 + 5 = 41 on and 54 + 5 = 59 off.

Answer B

Great question. A trick within a trick haha

Given 100 light bulbs in OFF position.

Child A starts switching them ON, in multiples of 3. Hence {(Last multiple of 3 - First Multiple of 3)/3} + 1 = {(99-3)/3} + 1 = 33
Hence after Child A reaches the end, 33 lights are ON.

Child B starts flipping the switches, in multiples of 5. Hence {(Last multiple of 5 - First Multiple of 5)/5} + 1 = {(100-5)/5} + 1 = 20
Hence Child B flips, 20 switches.

Now LCM of 3 & 5 is 15, and we have 6 multiples of 15 below 100. Hence there are 6 switches that were flipped ON by Child A & later flipped OFF by Child B.

# of lights switched ON = (# of Lights switched ON by A - # of Lights switched OFF by B) + (# of Lights flipped by B - # of Lights switched OFF by B) = (33-6) + (20-6) = 27 + 14 = 41

Answer B

Thanks,
GyM

Good question!

The equation I used to solve it was (multiples of 3 - multiples of 3&5) + (multiples of 5 - multiples of 3&5). There are 33 multiples of 3, 20 multiples of 5, and 6 multiples of 3 and 5 between 0 and 100, so the equation is (33 - 6) + (20 - 6) = 41.

(B) is the correct answer.

Person A turns on 33 lights (99/3)
Person B turn on 20 lights BUT 6 of them are already turned on, so hes turning on (100/20 - 6) new lights and turning off (33-6) old lights

33-6 + 20-6 = 27+ 14 = 41

Superb question. One of the best I have ever seen.