1. ## challenging ratio problem

Dear Sir,
Would appreciate very much if someone can help me to solve the below problem.
thanks
Kingsman

Boxes A and B are each filled with some sand. If sand from Box B is poured into Box A until the sand in Box A reaches the brim, there will be 8 litres of sand left in Box B. If sand from Box A is poured into Box B until the sand in Box B reaches the brim, there will be 26 litres of sand left in Box A. The ratio of the volume of Box A to the volume of Box B is 5:3. How many more litres of sand are needed to fill box Boxes to their brim?

2. I got it. I'll start posting it right now. Might take around 15 minutes.

3. Volume of A : Volume of B = 5:3

So let's say that Volume of A = $5k$, Volume of B = $3k$.

Right now there are $x$ liters of sand in box A, $y$ liters of sand in box B.

So let's pour sand from from box B to box A. Let's say that we poured $g$ liters of sand.

$x+g=5k$, $y-g=8$.

Now, pour from box A to box B. Let's say we poured $l$ liters of sand.

$x-l=26$, $y+l=3k$.

Let's do some algebra now.

$x+g-(x-l)=5k-26$

$g+l=5k-26$.

$y+l-(y-g)=3k-8$

$g+l=3k-8$.

$\therefore 5k-26=3k-8$.

$k=9$.

So volume of box A is $45$, and volume of box B is $27$.

Let's rewrite the equations:

$x+g=45$, $y-g=8$.

$x-l=26$, $y+l=27$.

Here we go again:
$x+g-(x-l)=19$

$\therefore g+l=19$.

$y-g+(y-l)=35$

$2y-(g+l)=35$

$2y-19=35$

$y=27$.

Now, let's find $x$.

$y-g=8$

$27-g=8$

$\therefore g=19$.

$x+g=45$

$x=26$.

So, in box A there are 26 liters of sand, and in box B there are 27 liters of sand.

So the amount needed to fill both boxes are:

$45+27-(26+27)=19$.

$\therefore 19$.

Tell me if this is wrong. I might've done some calculations wrong since I did most of them in my head.

4. Thanks very much for the solution but I wonder whether it can be done shorter by not using algebraic approach; instead by taking advantage of some special relationship or properties in the question.
Thanks
Kingsman

5. We can't solve this by not using algebra at all; you would be a genius if you can just guess and check.

It wouldn't be possible to set up any equations whatsoever if I didn't use any relationship or properties in the problem.

There could be other solutions, but basic idea will probably be the same.

6. Can be shortened; using Lanierms' variables :

boxA= 5k: sand in boxA = x : sand required to fill boxA = 5k - x

boxB= 3k: sand in boxB = y : sand required to fill boxB = 3k - y

"If sand from Box A is poured into Box B until the sand in Box B reaches the brim,
there will be 26 litres of sand left in Box A" :
x - (3k - y) = 26 ; x - 3k + y = 26 ; x + y = 3k + 26 [1]

"If sand from Box B is poured into Box A until the sand in Box A reaches the brim,
there will be 8 litres of sand left in Box B" :
y - (5k - x) = 8 ; y - 5k + x = 8 ; x + y = 5k + 8 [2]

[1][2]: 5k + 8 = 3k + 26
k = 9
So boxA = 45, boxB = 27

......

7. Hello, kingman!

Boxes A and B are each filled with some sand.
If sand from Box B is poured into Box A until the sand in Box A reaches the brim,
there will be 8 litres of sand left in Box B.
If sand from Box A is poured into Box B until the sand in Box B reaches the brim,
there will be 26 litres of sand left in Box A.
The ratio of the volume of Box A to the volume of Box B is 5:3.
How many more litres of sand are needed to fill both boxes to their brim?

Box A has volume $\,5k$ and contains $\,a$ liters of sand.
Box B has volume $\,3k$ and contains $\,b$ liters of sand.

Code:
         5k
*      *
|      |       3k
* - - -*    *      *
|::::::|    |      |
|::::::|    * - - -*
|:: a :|    |::::::|
|::::::|    |:: b :|
|::::::|    |::::::|
*-------    *-------
A           B

Sand from box B is poured into box A, filling box A.
The amount of sand transferred is: . $5k-a$ liters.

Code:
         5k
* - - -*
|:5k-a:| ←     3k
* - - -*    *      *
|::::::|    |      |
|::::::|    |      |
|:: a :|    |      |
|::::::|    * - - -*
|::::::|    |:: 8 :|
*-------    *------*
A           B

This leaves: $b - (5k-a) \:=\:8$ liters in box B.

. . We have: . $a + b - 5k \:=\:8$ .[1]

Sand from box A is poured into box B, filling box B.
The amount of sand transferred is: . $3k-b$ liters.

Code:
         5k
*      *
|      |       3k
|      |    * - - -*
|      |    |:3k-b :
* - - -*    * - - -*
|::::::|    |::::::|
|::26::|    |:: b :|
|::::::|    |::::::|
*-------    *-------
A           B

This leaves $a - (3k-b) \,=\,26$ liters in box A.

. . We have: . $a + b - 3k \:=\:26$ .[2]

Subtract [2] - [1]: . $2k \,=\,18 \quad\Rightarrow\quad k \,=\,9$

Substitute into [1]: . $a + b - 5(9) \:=\:8 \quad\Rightarrow\quad a + b \:=\:53$

. . The boxes originally contains $53$ liters of sand.

The total capacity of the two boxes is: . $5k + 3k \:=\:8k \:=\:8(9) \:=\:72$ liters.

Therefore: . $72 - 53 \:=\:19$ liters of sand will fill both boxes.

8. Hi Soroban,
Thanks very much for the solution; and I really admire your great ability for good insight in seeing great gem in a question and most important providing very simple graphical solution for them thus rendering meaning and appreciation for readers.
Bye
Kingsman