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

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- Jul 3rd 2011, 12:30 AMfleurdelWater 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, 03:45 AMSorobanRe: 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?

$\displaystyle \text{Let }a\text{ = rate of flow of the tap on Tank A, relative to the height of the water,}$

. . $\displaystyle \text{measured in cm/hour.}$

$\displaystyle \text{Let }b\text{ = rate of flow of the tap on Tank B, relative to the height of the watr,}$

. . $\displaystyle \text{measured in cm/hour.}$

$\displaystyle \text{In 6 hours, the height of Tank A went from }x\text{ cm to 0 cm.}$

. . $\displaystyle \text{We have: }\:6a \,=\,x$ .[1]

$\displaystyle \text{In 4 hours, the height of Tank B went from }x-5\text{ cm to 0 cm.}$

. . $\displaystyle \text{We have: }\:4b \,=\,x-5$ .[2]

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

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

Tank A: in 4 hours, the level dropped $\displaystyle 4a\text{ cm.}$

. . The height was: .$\displaystyle x - 4a\text{ cm.}$

Tank B: in 2.5 hours, the level dropped $\displaystyle \tfrac{5}{2}b\text{ cm.}$

. . The height was: .$\displaystyle (x-5)-\tfrac{5}{2}b\text{ cm.}$

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

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

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

Therefore, Tank A is 45 cm high.

- Jul 18th 2011, 06:22 AMfleurdelRe: Water draining from two tanks.
Thank you so much, Soroban.