Brain teaser

[jermo22]

Walt weatherman went to check three rainwater barrels. After collecting a total of 48 ounces, he wanted to make sure each barrel contained 16 ounces.

First he dumped 1/4 of barrel 1 into barrel 2.
Next, he dumped 1/2 of barrel 2 into barrel 3.
Finally, he dumped 1/3 of barrel 3 into barrel 1.

How many ounces dodid each barrel have before he w started pouring.

[Gadion]

At the end of the pouring, each barrel has to have 16 ounces, if I am understanding this correctly...
How did Walt Weatherman go to work?

Working backwards, barrel 1 and barrel 3 both had 16 ounces after he dumped 1/3 of barrel 3 into barrel 1... Thus, before that point barrel 3 had 1/3 too many ounces relative to what it needed to be, and if 2/3 of this is 16 [remember barrel 3 has to end on 16 ounces], it had to contain 24 (if 0.667x = 16 then x = 24) ounces before pouring over, while barrel 1 had to receive 8 ounces to get to 16 and thus it had 8 ounces in it before this point.
->Before pouring 3, barrel 1 had 8 ounces and barrel 3 had 24 ounces.

From above, barrel 3 had 24 ounces after 1/2 of barrel 2 was dumped into it... Thus, before that point barrel 2 had 1/2 too many ounces relative to what it needed to be, and if 1/2 of this is 16 [remember barrel 2 has to end on 16 ounces], it had to contain 32 (if 0.5x = 16, then x = 32) ounces before pouring over, while barrel 3 had to receive 16 ounces to get to 24 ounces <as per above> and thus it had 8 ounces in it before this point.
-> Before pouring 2, barrel 2 had 32 ounces and barrel 3 had 8 ounces.

Now, barrel 2 had 32 ounces after he dumped 1/4 of barrel 1 into it... Thus, before that point, barrel 1 had 1/4 too many ounces relative to what it was after this point, and if 3/4 of this is 8 [remember that it was determined that prior to pouring 3 barrel 1 had 8 ounces in it and it has not been involved in pouring 2], it had to contain 10.667 (if 0.75x = 8 then x = 10.667 rounding to the third decimal) ounces before pouring over, while barrel 2 had to receive 2.667 ounces to get to 32 ounces <as per above> and thus it had 29.333 ounces rounded to the third decimal in it before this point.
-> Before pouring 1, barrel 1 had 10.667 ounces and barrel 2 had 29.333 ounces.

Since barrel 3 wasn't involved in pouring 1, the amount in it prior to pouring 2 was its starting amount, which was 8 ounces.
->Prior to the influence of Walt Weatherman, barrels 1, 2 and 3 contained 10.667 ounces, 29.333 ounces and 8 ounces respectively.
First, 0.250(10.667) + 29.333 = 32 (Barrel 2 has 32, and 0.750(10.667) = 8 so Barrel 1 has 8)
Next, 0.500(32) + 8 = 24 (Barrel 3 has 24, and 0.500(32) = 16 so Barrel 2 has 16)
Finally, 0.333(24) + 8 = 16 (Barrel 1 has 16, and 0.667(24) = 16 so Barrel 3 has 16)


Perhaps we could rather represent this entire scenario by the following set of equations:
{1} 0.250B10 + B20 = B21 {6} 0.750B10 = B31
{2} 0.500B21 + B31 = B32 ({5} 0.500B21 = 16 = B22 since we know that B22 = 16 as per the stipulations of the scenario)
{3} 0.333B32 + B11 = B13 = 16 ({4} 0.667B32 = 16 = B33 since we know that B13 = 16 and B33 = 16 as per the stipulations of the scenario)

where B10 is the amount of ounces in barrel 1 before Mr Weatherman starts interfering with its contents
B20 is the amount of ounces in barrel 2 before Mr Weatherman starts interfering with its contents
B21 is the amount of ounces in barrel 2 after the first pouring over (from barrel 1 into barrel 2)
B31 is the amount of ounces in barrel 3 before Mr Weatherman starts interfering with its contents
B32 is the amount of ounces in barrel 3 after the second pouring over (from barrel 2 into barrel 3)
B11 is the amount of ounces in barrel 1 after the first pouring over (from barrel 1 into barrel 2)
B13 is the amount of ounces in barrel 1 after the third pouring over (from barrel 3 into barrel 1) - the final contents of barrel 1
B33 is the amount of ounces in barrel 3 after the third pouring over (from barrel 3 into barrel 1) - the final contents of barrel 3
B22 is the amount of ounces in barrel 2 after the second pouring over (from barrel 2 into barrel 3) - the final contents of barrel 2

Starting with {4}, we calculate B32 = 24
from {5}, we calculate B21 = 32
substitute B32 = 24 into {3} and we determine B11 = 16 - 0.333(24) = 8
substitute B32 = 24 into {2} and we determine B31 = 24 - 0.500(32) = 8
substitute B31 = 8 into {6} and we determine B10 = 8/0.750 = 10.667
substitute B10 = 10.667 and B21 = 32 into {1} and we determine B20 = 32 - 0.250(10.667) = 29.333

From this we have calculated that B10 = 10.667, B20 = 29.333 and B31 = 8
and thus barrels 1, 2 and 3 have 10.667, 29.333 and 8 ounces of water in them respectively initially.

[Ripster42]

You doing his homework.

He's all like, thanks for the help with my homework dude!

[May90]

{x, y, z} - initial amounts in 1st, 2nd and 3rd barrel.

1st step: {(3/4)x, y+(1/4)x, z}
2nd step: {(3/4)x, (1/2)y+(1/8)x, z+(1/2)y+(1/8)x}
3rd step: {(3/4)x+(1/3)z+(1/6)y+(1/24)x, (1/2)y+(1/8)x, (2/3)z+(1/3)y+(2/24)x}, or

{(19/24)x+(1/6)y+(1/3)z, (1/2)y+(1/8)x, (2/3)z+(1/3)y+(1/12)x}={16,16,16}, or

{(19/8)x+(1/2)y+z, (1/8)x+(1/2)y, (1/8)x+(1/2)y+z}={48,16,24}

Deduct the 2nd equation from the 3rd, you get z=8. So:

{(19/8)x+(1/2)y, (1/8)x+(1/2)y, (1/8)x+(1/2)y={40,16,16}

Deduct the 2nd equation from the 3rd, you get (18/8)x=24, or x=(32/3). Finally, we have:

y=48-x-z = (88/3).

{x,y,z}={(32/3),(88/3),8}.

Unless I've messed up somewhere. Gadion got the same result, but in different notation, so I guess it's right.

[Twoddle]

Got the same answer working backwards:

After step 3: 16,16,16
A third of barrel 3 was poured from barrel 3 to 1 to reach the above so barrel 3 had 16/(2/3) = 24 after step 2, adjust barrel 1 with the difference

After step 2: 8,16, 24
A half of barrel 2 was poured from barrel 2 to 3 to reach the above so barrel 2 had 16/(1/2) = 32 after step 1, adjust barrel 3 with the difference

After step 1: 8,32,8
A quarter of barrel 1 was poured from barrel 1 to 2 reach the above so barrel 1 had 8/(3/4) = 10⅔ at the start, adjust barrel 2 with the difference

Initially they had:
10⅔, 29⅓, 8