Deceptively Difficult?

Question

Ol' clamcrunchies is going to ask this one again for y'all. Glass A is x% full of water, glass B is y% full of wine. If a tablespoon of water is taken from glass A and mixed with the wine in glass B, and then a tablespoon of the water-wine mixture in glass B is mixed with water in glass A, and if the glasses are the same size, then in terms of x and y what percent of the water-wine mixture in glass A is water?

Answer 1

let the volume of each be z then let us assume each table spoon is transfering k units.
Now it will be (yk1000000)/(yz+100k)

Answer 2

Let V be the volume of glasses A and B, and let a tablespoon be k% of V. (Obviously there will have to be at least a tablespoon in glass A.)

The percent water in glass A after the two transfers will be:

100*(x(y+k)-ky) / x(y+k)

Note: -ky in the numerator ensures a reduced %
example: if x=y= 50% and k=5% then we get:
100*10/11 = 90.91% approx.
(Also, 1 tablespoon = 0.5 US fl. oz or 14.8ml, so once you know the volume of the glass in ml or fl. oz, you can quickly compute k, and solve the problem.)

Derivation:

Note: since x, y and k are percentages, before the transfers the volume of water in glass A = xV/100 and the volume of wine in glass B is yV/100. The volume of a tablespoon is kV/100.

kV/100 is the amount of water put into Glass B from glass A , and (kV/100) /((yV/100)+(kV/100)) = k/(y+k) is now the fractional volume of water in the wine-water mixture in glass B.

Returning a tablespoon to glass A:

When we return the tablespoon to A, the fraction that will be water is:

((x-k)V/100+(kV/100)*k/(y+k)) / (xV/100)

=(x-k+(k^2/(y+k)))/x =

(x(y+k)-ky) / (x(y+k))

Multiply by 100 for the percent value.

Answer 3

You really do need to know the volumes that the tablespoon and glasses hold, so for this exercise, let's assume the tablespoon volume is (T) mL, and that it is less than the volume of water in Glass A, naturally. For the glasses, I'll also work with millilitres and then it will be easy to convert to percentages.

Let (V) mL = the volume of Glass A and Glass B.

Glass A is x% full of water, so the volume of water in Glass A = (xV / 100) mL.
Glass B is y% full of wine, so the volume of wine in Glass B = (yV / 100) mL.

Taking (T) mL of water out of Glass A leaves glass A with (xV / 100 - T) mL of water.

Putting (T) ml of water in Glass B means Glass B will contain (T) mL of water and (yV /100) mL of wine, for a total volume of (yV / 100 + T) mL = [(yV + 100T) /100] mL.

After mixing, (T) ml of solution from Glass B will contain [100T^2 / (yV + 100T)] mL of water.

Adding this (T) mL to Glass A will make the total volume of water in Glass A
= (xV / 100 - T) + [100T^2 / (yV + 100T)] mL
= [xV / 100 - yVT / (yV + 100T)] mL.

The total volume in Glass A is now back to the original (xV / 100) mL.

The percentage water in Glass A is then: 100{1 - 100yT / [x(yV + 100T)]} %.
This is the final formula in terms of x and y, but also in terms of V and T.

As an example, if V = 200, x = 75, y = 25 and T = 10,
then the percentage water in Glass A
= 100{1 - 100*25*10 / [75(25*200 + 100*10)]}
= 94.4%

Answer 4

this has been asled before, its not possible to work out unless you know the volume of the teaspoon in relation to the glasses and also assume that the glasses have equal volumes.

Answer 5

Let
X = volume of water in container A
Y = volume of wine in container B
T = volume of teaspoon.

Then I come up with the % of water in A as
100 * (1 - T*Y / (X*(T+Y)))