Epic Systems Interview Question Software Engineer / Developers
How can it be 2 + 4? The 4 secs of the 2nd day includes the 2 secs of 1st day; so you are accounting for 1st days loss twice. Think of it this way - the clock will complete its 2 days when it is actually 2 day & 4 secs. This kind of question is merely to confuse.
Explanation:-
To understand this we have to compare 2 clocks.
At the end of day1, suppose Normal clock shows 12AM. Screwed up clock will be 11 PM 59Min 58Sec. Lost 2 days.
One Interesting fact is that Screwed up clock will reach 12AM only after 2 secs. :)
At the end of Day2, as usual screwed up clock will lose 2 more seconds.
Means, 2+2(make up for day1's lost seconds) +2 = 6secs!
Continuing from Nagabhushana's logic :-
End of Day 1 :- Proper Clock :- 12:00:00
Screwed Clock :- 11:59:58
Proper Clock :- 12:00:02
Screwed Clock :- 12:00:00
End of Day 2 Proper Clock :- 12:00:02 (ur logic Screwed here as u took 12:00:04 )
Screwed Clock :- 11:59:58
Proper Clock :- 12:00:04
Screwed Clock :- 12:00:00
Mostly right, so after 48 hours...
Proper clock - 12:00:00
Screwed Clock - 11:58:56
The screwed up clock falls behind 2 seconds for each 24 hour cycle.
end of 1st day loses 2 sec from day 0
end of 2nd day loses 2 sec from day 1 == 4 sec from day 0
total loss 6 sec
None of you are right!!
The clock loses 2secs a day. At end of 24hrs, 2secs on a real non-faulty clock(RNFC), the faulty slow clock (FSC) shows 24hrs. This means that the loss is per second and can be measured as a whole number only at the end of a 24hr, 2sec period on a RNFC.
When this RNFC reaches 24hrs, 2sec it must be advanced only 23hrs, 59min and 58sec after that to get to 2days. This means that in 24hrs 2sec FSC lost 2 sec. In 23hrs, 59min, 58sec FSC loses slightly less than 2secs/ Therefore in two days measured on RNFC, it loses slightly less than 4secs, in particular ~3.99 seconds.
You idiot AN, don't you know what a day means, go learn how to read the time. Mandar is right..terms like RNFC, FSC only obfuscate matters.
4 seconds It cant be six. I ran this experiment day before!!
I feel 2 sec lost...!
End of first day real time 12:00:00 AM
End of first day Screwed Time : 11:59:58 PM
ok...... now read the Q (how much time will it loose in 2 days ie.real time).
Start of second day real time : 12:00:01 AM (real world 2nd day started)
Start of second day Screwed Time : 23:59:59 PM ( its already 2nd day and our screwed clok is behind 2 second, per day loose 2 sec. This continues till you make your clock 2 sec front not 4 or 6 sec )
End of second day real time 12:00:00 AM (real time we reach end of second day but screwed clock lost 2 sec.......)
End of second day Screwed Time : 11:59:58 PM
Third day...... fourth day...... lose of 2 sec in screwed clock per day in real time.
< 4
the assumption is that on the second day it takes 24hours + 2 seconds for he clock to lose another 2 seconds. so for the full 48 hours, it loses < 4 seconds
question should be, if a clock loses 2 seconds in one day how many seconds will it have lost after 2 days. the answer is 4 because you finding the overall time that has been lost, after a natural 24 hours the clock will have lost 2 seconds and will show 23.59.58 then after another natural 24 hours(48 hours overall) since the clock only goes to 23.59.58 in 24 hours, it will show 23.59.56 as it has lost another 2 seconds in the real 24 hours period. so when in reality where 172800 seconds have past in 48 hours the clock would show 172796 seconds meaning a 4 second loss
Depends on 1) how exactly is the wording of the question and 2) on the available answers (assuming that it is a multiple choice).
If the question is the same as presented here, so we can try to realize what are the assumptions. If the answers (~3.99), (4), and (6) appear, (~3.99) will be the answer because this is how a real-word clock delays (fractions of seconds) and not in "jumps".
If the answers (4) and (6) appear, (6) must be the answer because for an IQ exam, the easy answer (4) shows that something is wrong. In this case, a "game of words" must be considered (delay "per day") and we need to assume that it is a fictitious scenario where the delay d_i of each day i (i = any integer from 1 to infinite+) is given as d_i = (i)*2s. In this case, d_1 = 2s, d_2 = 4s, d_3 = 6s, d_4 = 8s. Therefore, at the end of the second day, the total delay is d_1 + d_2 = 2 + 4 = 6s.
Finally, if the answer (4) appears without the other 2 answers, this is the expected common-sense answer (99.99999..9%) would take this answer without taking more than 5s to think. With nerds... it is another story :)
The answer is 4. Not 6.
Think in this way. If you have a clock, which slower or faster than the normal clock for 2 secs. So what are you gonna do every day? You gotta adjust it every day. You need to add add 2 secs to it if it lost 2 secs everyday. So, say, after 30 days, how many secs you add to it? It is 2 * 30 secs = 60 secs.
It's like word game, the first day is 2 sec late, the 2nd day is 4 sec late, for two days, it's 6..
- ilikedeal February 05, 2010