2. How will you measure height of building when you are at the top of the building?
3.You have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door. what state are the doors in after the last pass? which are open which are closed?
4.A mad bomber is out on the job, making bombs. he has two fuses (pieces of string) of varying thickness which each burn for 30 seconds. unfortunately he wants this bomb to go off in 45 seconds. he can’t cut the one fuse in half because the fuses are different thicknesses and he can’t be sure how long it will burn. how can he arrange the fuses to make his bomb go off at the right time?
5.A one armed surgeon with a hand wound needs to operate on three patients. the surgeon only has two gloves. how can he operate on the three patients in turn without risking exchange of fluids? (remember he only has one arm so he only needs to wear one glove at a time.)
The solution will be posted tomorrow..:)
take 1 candle, light its both ends, which burns for 30min.
ReplyDeletelight 2nd candle fully..it takes 60min...so total 90min.
thanks sreesha..Its quite easy one i guess...
ReplyDeleteSolutions :
ReplyDelete1. First light up the two ends of the 1st candle. When it will burn out light up one end of the second candle. (30+60=90)
2. Drop the stone and find the time taken for the stone to reach the ground. find height using the formula
s = a + gt ( s = height, a= initial velocity=0, g=9.8m/s, t = time taken).
3. For example, after the first pass every door is open. on the second pass you only visit the even doors (2,4,6,8…) so now the even doors are closed and the odd ones are opened. the third time through you will close door 3 (opened from the first pass), open door 6 (closed from the second pass), etc..
question: what state are the doors in after the last pass? which are open which are closed?
solution: you can figure out that for any given door, say door #42, you will visit it for every divisor it has. so 42 has 1 & 42, 2 & 21, 3 & 14, 6 & 7. so on pass 1 i will open the door, pass 2 i will close it, pass 3 open, pass 6 close, pass 7 open, pass 14 close, pass 21 open, pass 42 close. for every pair of divisors the door will just end up back in its initial state. so you might think that every door will end up closed? well what about door #9. 9 has the divisors 1 & 9, 3 & 3. but 3 is repeated because 9 is a perfect square, so you will only visit door #9, on pass 1, 3, and 9… leaving it open at the end. only perfect square doors will be open at the end.
4. light both ends of one of the fuses. when that fuse goes out, 15 seconds has elapsed. then light the other fuse.
5.The surgeon places both gloves on his hand (1 and 2). he operates on patient A. he then takes the top glove off (#2), leaving on the bottom glove (#1) and operates on patient B. then he carefully reverses glove #2, so the clean side is on the outside, and he places it on top of glove #1 which is on his hand, and operates on patient C. this problem is kind of dumb because how’s the surgeon going to change the gloves on his hand when he only has one hand. plus no offense, but how often do you come across a one-armed surgeon (i’m sure there are plenty of one-armed doctors, but a surgeon!?!). anyway, i had to make this problem child friendly and changing the story to the above was the only way to do it. consider for a minute what the initial problem was. the surgeon was just a guy, the patients were women, and the glove was… well, i won’t insult your intelligence