Water fills a tank at a rate of 150 liters during the first hour, 350 liters during the second hour, 550 liters during the third hour, and so on.
Find the number of hours necessary to fill a rectangular tank with dimensions 16 m x 7 m x 7 m.
Answer (3)
Volume of tank = (16m)(7m)(7m) = 784 m³
Conversion of m³ to L: (784 m³) × (1000L / 1m³) = 784,000 L
Rate in the 1st hour: 150 liters/hr
Rate in the 2nd hour: 350 liters/hr
Rate in the 3rd hour: 550 liters/hr
It is apparent that the Fill Rate is increasing by 200 liters/hr every subsequent hour . . . so that can be represented by the following equation
where: t = number of hours
Fill rate (i.e. volume of water filled into tank within the specified hour) = 150 + 200(t - 1)
For t = 1 . . . Fill rate = 150 L/hr For t = 2 . . . Fill rate = 350 L/hr For t = 3 . . . Fill rate = 550 L/hr
Because after every hour there has been more water added to the tank, this problem can be represented as a geometric sequence in order to account for the compounding of the volume after each time step, but it can also be tabulated (which seems to me to be the more direct/simple approach), so I will build a table that accounts for the increasing Fill Rate and the compounding of water volume after each time step . . .
(see attached)
The answer (after all of this) is . . . t = 88 hrs 17 1/2 mins (approx)
Putting this as an arithmetic sequence gives:
u n = 150 + 200 ( n − 1 )
The sum of the series = 16 x 7 x 7 = 784 m^3 = 784 000 L
The sum of an arithmetic series can be written as:
[S_n=n/2 [2a+(n-1)d] = 784 000 /2[2(150)+(n-1)200] = 784 000 [300+200(n-1)=1 568 000 \300n+200n^2-200n = 1 568 000 \200n^2+100n- 1 568 000 = 0 \2n^2 +n- 15680 = 0
= 88.2...,-88.7]
n has to be positive, so we get
n = 88.2 hours (3 s.f.)
The tank's capacity is 784,000 liters. Using the arithmetic series for water delivery, it takes approximately 88.2 hours to fill the tank completely. This calculation accounts for the increasing amount of water filled each hour.
;
