How many hours does it take to fill a 16m x 7m x 7m tank with water?

[mister_tom]

Homework Statement

water fills a tank at a rate of 150 litres during the first hour, 350 litres during the second hour, 550 litres during the 3rd hour and so on. find the number of hours neccesary to fill a rectangular tank 16m x 7m x7m

Homework Equations

l=a+(n-1)d

S= n/2 (a+l)

where:

l = term
a = first no in series
n = nth term
d = common difference
S = sum

The Attempt at a Solution

volume of tank = 784

although it doesn't say i assumed 1m(3) is equal to 1000kg = 1000 litres

therefore water required to fill tank = 784,000 litres

a = 150

d = 200

what i don't understand is in either equation given for arithmetic progression there are 2 unknows?the answer is 88.3 hours

 

[thrill3rnit3]

You have 2 equations in 2 unknowns. Do you know how to solve a system of equations?

 

[mister_tom]

You have 2 equations in 2 unknowns. Do you know how to solve a system of equations?

not entirely, i attempted to replaced l in the 2nd equation to the answer for l in the first to give:

S= n/2 (a+(a+(n-1)d))

but i couldn't seem to get a decent answer from that

 

[Mark44]

Substitute for a and d, which you know, and try a few value of n.

 

[thrill3rnit3]

not entirely, i attempted to replaced l in the 2nd equation to the answer for l in the first to give:

S= n/2 (a+(a+(n-1)d))

but i couldn't seem to get a decent answer from that

You know a (first term), d (common difference), and S (sum, which is 784,000).

You're left with 1 variable, n, which is exactly what you're trying to find (number of terms/hours)

 

[mister_tom]

Substitute for a and d, which you know, and try a few value of n.

i don't understand sorry, why would i need to substitute values that i already know?

 

[mister_tom]

You know a (first term), d (common difference), and S (sum, which is 784,000).

You're left with 1 variable, n, which is exactly what you're trying to find (number of terms/hours)

the problem I am having now is although there's only 1 variable (n) there's 2 of them, and i need to get only 1 for the answer.

this is what i tryed:

S= n/2 (a+(a+(n-1)d))
2S = n (150+(150+(n-1)200))
2S = n (150+150+200n-200)
2S = n (150+(200n-50)
2S = n (100+350n)
2S = 100n+350n²

not sure if this is the right method or if it is where to go from here?

 

[thrill3rnit3]

S is the sum of the series, so it's 150+350+550+...

It should total what is necessary to fill the container tank, right? So what do we substitute for S?

 

[Mark44]

the problem I am having now is although there's only 1 variable (n) there's 2 of them, and i need to get only 1 for the answer.

this is what i tryed:

S= n/2 (a+(a+(n-1)d))
2S = n (150+(150+(n-1)200))
2S = n (150+150+200n-200)
2S = n (150+(200n-50)

Mistake in line below.

2S = n (100+350n)
2S = 100n+350n²

not sure if this is the right method or if it is where to go from here?

This is the right way, but you have a mistake, noted above.

You know S, so what you have (when you fix the error) is a quadratic equation in n. Hopefully, you know how to solve quadratic equations.

 

[mister_tom]

ahhh brilliant thanks guys.

changing my mistake i get:

2 * 784000 = 250n + 200n²

giving:

0 = 200n² + 250n - 1568000

then applying to a quadratic equation i get 87.921, not quite 88.3 but then I am assuming that 1 cubic metre of water = 1000 litres.

thanks again that confused me a lot more than it should have

 

[Mark44]

ahhh brilliant thanks guys.

changing my mistake i get:

2 * 784000 = 250n + 200n²

Looks like what you did is change to a different mistake.

This line looks fine (except for missing right parenthesis)
2S = n (150+(200n-50)
You need only one pair of parentheses, like so:
2S = n (150+ 200n-50)

giving:

0 = 200n² + 250n - 1568000

then applying to a quadratic equation i get 87.921, not quite 88.3 but then I am assuming that 1 cubic metre of water = 1000 litres.

thanks again that confused me a lot more than it should have