I got it. I'll start posting it right now. Might take around 15 minutes.
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?
Volume of A : Volume of B = 5:3
So let's say that Volume of A = , Volume of B = .
Right now there are liters of sand in box A, liters of sand in box B.
So let's pour sand from from box B to box A. Let's say that we poured liters of sand.
, .
Now, pour from box A to box B. Let's say we poured liters of sand.
, .
Let's do some algebra now.
.
.
.
.
So volume of box A is , and volume of box B is .
Let's rewrite the equations:
, .
, .
Here we go again:
.
.
Now, let's find .
.
.
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:
.
.
Tell me if this is wrong. I might've done some calculations wrong since I did most of them in my head.
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.
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
......
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 and contains liters of sand.
Box B has volume and contains 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: . liters.
Code:5k * - - -* |:5k-a:| ← 3k * - - -* * * |::::::| | | |::::::| | | |:: a :| | | |::::::| * - - -* |::::::| |:: 8 :| *------- *------* A B
This leaves: liters in box B.
. . We have: . .[1]
Sand from box A is poured into box B, filling box B.
The amount of sand transferred is: . liters.
Code:5k * * | | 3k | | * - - -* | | |:3k-b : * - - -* * - - -* |::::::| |::::::| |::26::| |:: b :| |::::::| |::::::| *------- *------- A B
This leaves liters in box A.
. . We have: . .[2]
Subtract [2] - [1]: .
Substitute into [1]: .
. . The boxes originally contains liters of sand.
The total capacity of the two boxes is: . liters.
Therefore: . liters of sand will fill both boxes.
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