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this is the problem. just imagine, a faucet puts out a single drop of water every minute into a glass of water. the quantity of water which the faucet drops, doubles every minute.. take note that in the first minute, there is only a single drop of water. The faucet can fill the entire glass in one hour. The question is, how long will it take for the faucet to fill half of the glass? Thanks in advance..
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59 minutes
IPBLE: Increasing Performance By Lowering Expectations.
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How did u come up with that answer? can u let me know the solution? pls..
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The answer must be>59 minutes, because in the beginning of the 60th minute in the glass will be
(v-1)/2 drops, where v is the number of drops, which full the glass
If you use that the number of drops, falled to the t-th minute (including the t-th minute) is:
IPBLE: Increasing Performance By Lowering Expectations.
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Let's look at doubling:
Drops This Time, Total Drops
1, 1
2, 3
4, 7
8, 15
16, 31
32, 63
64, 127
You can see that the *total* drops nearly doubles each time (nearly ... just one drop different).
So if it doubles every minute, then one minute before it was half. And that one drop innaccuracy will be very small because after 59 doublings that is a LOT of drops (about a quintillion drops, or 50 trillion liters - a lake).
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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What do you call that formula? How did you get the V?
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I think you are asking me.
First, MathsIsFun-perfect explanation.
Second- I'm actually not getting v at all. I'm assuming that the glass fills with v drops, so v/2 drops will be needed to full the half of the glass.
IPBLE: Increasing Performance By Lowering Expectations.
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you could also think of it as a geometric series
and then the current volume of water in the glass would be the summation
now, lets say that a is 10^-20, after one hour the glass is full, so given that a is 10^-20
the glass' capactiy is 0.011529215, half of which is 0.005764607
so we have
Last edited by luca-deltodesco (2006-07-22 23:47:19)
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No!
Can't be 55 minutes.
Let
Last edited by krassi_holmz (2006-07-23 01:17:23)
IPBLE: Increasing Performance By Lowering Expectations.
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luca, I think you are getting 55 minutes, because you're losing too much presicion by this 10^16.
Pure symbolic answer is cutter.
Last edited by krassi_holmz (2006-07-23 01:15:42)
IPBLE: Increasing Performance By Lowering Expectations.
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ive figured out a solution with these problem... but i don't know if this is right..
lets change the given to a smaller values.. lets say that every second, the faucet puts out a drop of water. (same situation).. It takes 8 seconds to fill the glass. the time within which the faucet will fill half of the glass is,
A is the total no. of drops to fill the glass
A=2^8-1=255 2 there is the doubling; 8 is the time for the glass to be filled; and 1 is the
first second or the number which no doubling is made...
As we can see, the total no. of drops is, since there is 1 drop in the first second, multiply it by 2 and add 1 (as i have observed in the sequence)... therefore, 1 on the first second, 1x2+1=3 on 2nd, 3x2+1=7 3rd, 7x2+1=15 4th, and so on up to 127x2+1=255 8th second.
255/2 = 127.5 - the quantity of half of the glass
ooops... i forgot the method on how to determine on what second this 127.5
fall. but logically, take note that we have multiplied and added 1 until we
have reached 255 (and that is in the 8th second). therefore, reversing the
process, (255 - 1)/2=127. Realizing that we have reversed and moved back
1 time, we subtract 1 from 8.. and that is 7...
from this long explanation, i came up to this answer ---> 7... I just don't now if Im right... Lolz...
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You are right... - but the answer is a little bigger than 7 seconds, because in the 7th second, there will be 127 drops in the glass, but there must be 127.5, so half a drop must fall, to reach the half of the glass, so the time to full the half of the glass is bigger than 7 seconds with the time of falling of half a drop, which is really small.
IPBLE: Increasing Performance By Lowering Expectations.
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oh yeah ur right.. but, do i still need to calculate for the remaining time to exactly fill half of the glass? If so, how?
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luca, I think you are getting 55 minutes, because you're losing too much presicion by this 10^16.
Pure symbolic answer is cutter.
yeh, i was looking at my maths, and i couldnt find anything wrong with it, and yet the 55 was just like... eh?
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And one last note-strictly, fractional number of minutes won't be allowed, because we define the function (r) as a sum, and thus, it's only for integers (I'm not saying that it can't be defined over R, as we have done, I'm saying that in this case it should be defined only for integers (because we have drops, not stream))
IPBLE: Increasing Performance By Lowering Expectations.
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