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June 2016 p12 q9
907
A water tank holds 2000 litres when full. A small hole in the base is gradually getting bigger so that each day a greater amount of water is lost.
Assume instead that 10 litres of water are lost on the first day and that the amount of water lost increases by 10% on each succeeding day. Find what percentage of the original 2000 litres is left in the tank at the end of the 30th day after filling.
Solution
The amount of water lost each day forms a geometric progression (GP) with the first term, \(a = 10\), and common ratio, \(r = 1.1\).
The sum of the first 30 terms of a GP is given by:
\(S_{30} = \frac{a(r^{30} - 1)}{r - 1}\)
Substituting the values, we have:
\(S_{30} = \frac{10(1.1^{30} - 1)}{1.1 - 1}\)
Calculating this gives \(S_{30} = 1645\).
The total water lost after 30 days is 1645 litres.
The percentage of the original 2000 litres left in the tank is: