# Water draining from two tanks.

• Jul 3rd 2011, 01:30 AM
fleurdel
Water draining from two tanks.
Hi all,
There is no indication of base so am wondering how to solve this sum. Thank you for taking the time.(Nod)
http://i55.tinypic.com/nnqykn.jpg
• Jul 3rd 2011, 04:45 AM
Soroban
Re: Is this sum solvable?
Hello, fleurdel!

Quote:

18. There are two tanks filled with water.
They are being emptied by two different taps, each with a constant flow of water.
Tank A is 5 cm taller than Tank B.

Code:

      * - - - *       |: : : :|       | : : : |    * - - - - *       |: : : :|    |: : : : :|     x | : : : |    | : : : : |       |: : : :|    |: : : : :| x-5       | : : : |    | : : : : |       |: : : :|    |: : : : :|       *-------*    *---------*           A              B

The tap of Tank A was opened at 7:00 a.m.
Tank A was empty at 1:00 p.m.

The tap of Tank B was opened at 8:30 a.m.
Tank B was empty at 12:30 p.m.

The water level of both tanks were equal at 11 a.m.

What is the height of tank A?

$\text{Let }a\text{ = rate of flow of the tap on Tank A, relative to the height of the water,}$
. . $\text{measured in cm/hour.}$

$\text{Let }b\text{ = rate of flow of the tap on Tank B, relative to the height of the watr,}$
. . $\text{measured in cm/hour.}$

$\text{In 6 hours, the height of Tank A went from }x\text{ cm to 0 cm.}$
. . $\text{We have: }\:6a \,=\,x$ .[1]

$\text{In 4 hours, the height of Tank B went from }x-5\text{ cm to 0 cm.}$
. . $\text{We have: }\:4b \,=\,x-5$ .[2]

Substitute [1] into [2]: . $4b \:=\:6a - 5 \quad\Rightarrow\quad 6a - 4b \:=\:5$ .[3]

At ll:00 a.m. the two tanks had the same height.

Tank A: in 4 hours, the level dropped $4a\text{ cm.}$
. . The height was: . $x - 4a\text{ cm.}$

Tank B: in 2.5 hours, the level dropped $\tfrac{5}{2}b\text{ cm.}$
. . The height was: . $(x-5)-\tfrac{5}{2}b\text{ cm.}$

We have: . $x - 4a \:=\:(x-5) - \tfrac{5}{2}b \quad\Rightarrow\quad 8a - 5b \:=\:10$ .[4]

Solve [3] and [4]: . $\begin{Bmatrix}6a - 4b &=& 5 \\ 8a - 5b &=& 10 \end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix}a &=& \frac{15}{2} \\ b &=& 10 \end{Bmatrix}$

Substitute into [1]: . $x \:=\:6a \:=\:6\left(\tfrac{15}{2}\right) \:=\:45$

Therefore, Tank A is 45 cm high.

• Jul 18th 2011, 07:22 AM
fleurdel
Re: Water draining from two tanks.
Thank you so much, Soroban.