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

Hi..

INFO from Q--

1) there are 100 On/Off switches for 100 bulbs
2)A presses every third switch..
3) B presses every fifth switch..
4) All switches were in Off position..

Solution--

1)A switches every third switch so these are turned ON..
total ON= 100/3=33..
2) B presses every 5th switch
so he presses 100/5=20 switch..

But there are switches which are pressed TWICE ..
the CATCH is here..
it is easy to find the switches pressed twice and subtract that from TOTAL.
But that is WRONG. Why?
Because when pressed twice the bulb is again switched Off.
so we will have to subtract this number twice, ONE from A's total and one from B's total..

so lets find switches pressed twice--
LCM of 3 and 5= 15..
so number of switches pressed twice= 100/15= 6

ous answer= 33+20-2*6=41
ans 41

B
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Editing my solution:

Number of switches = 100

Number of switches turned on by A: 3, 6, ... 99 = 33
Number of switches turned on by B: 5, 10, .... 100 = 20

Few switches are turned on by A and later turned off by B: LCM(3,5) = 15x = 15, 30,....90 = 6.

Subtract the above 6 switches from both A and B as they are turned off.

Number of switches that are turned on = (33 - 6) + (20 - 6) = 41

Answer: B

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
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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..
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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
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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.