A large tank filled with water is to be emptied by removing 1/5 th of the water present in it everyday.after how many days will there be closest to 25% of the water left in the tank??
Answers
This question is simply about powers and logs.
The long way to do this is to say on day 0 the tanks water level is 1oo
Then subtract 20 to get the total percent of water in the tank for day 1, 80.
Then continue this with 80, taking away 16 (a 1/5 of 80) and continue with this till you reach less then 25. Then compare the last two days to see which has a value closest to 25 to answer the question.
However this is long and dull, so instead you should look at multiplying, as subtracting 1/5 is the same as multiplying by 0.8.
(This can clearly be seen as 100 * 0.8 = 80 )
knowing this we no longer have to do this question in various stages, as we just need to use powers of 0.8.
The power that 0.8 is to, is the amount of days which have passed (after one day, (0.8^1 ) the amount left is 80). So we can say 100*0.8^X = 25.
simplifying this we get 0.8^X = 0.25.
Now we can use logs to find a value for X (the number of days that’ve passed to get to 25% of the tanks starting water level).
log (0.8^X) = X log (0.8) = log (0.25)
Therefore log (0.25) / log (0.8) = X = 6.21 ( 3sf)
Rounding this value for X, we can quite confidently say that after 6 days, the tank will be closest to having 25% of its water left.
Easy, but more complicated then the easier (and far duller) route of subtracting 1/5 each time. However by doing the question using powers and logs, we can generalize this to work in any scenario.
As 0.8 = 1 - your removed fraction and 0.25 = your final percentage of water left.
Therefore we can say that (1-A)^X = (B) when A is your removed fraction and B is your final percentage.
Therefore X, the amount of repetitions of removal needed to get to your desired final percentage, would equal
log (B) / log ( 1-A) or log (1-A,B) (log of B in base 1-A)
So if a question stated “you have a bucket with 1 liter of water, you remove 1/1000000 of the water per second, how would it take have only 400 milliliters left in the bucket” you’d be able to answer it easily.
A = 1/1000000 and B = 1000/400 = 0.4
Therefore the amount of seconds it would take would be log (1-0.000001, 0.4), which equals 916290 seconds ( 15272 minutes, 255 hours or 10.6 days). To do this question in the first more basic way wouldn’t just take hours, but it would take days. Even if you managed to subtract a 1/1000000 of the amount in the bucket each second, it would still take 916290 seconds (10.6 days) for you to reach the answer.
Using this method a question like that doesn’t even take 10.6 seconds. In fact using this method would make almost all removal of fraction questions super easy. All you need is alittle log.