[MOVED] maintaining 4-gal measure w/ 3-gal, 5-gal jugs

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conair

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In the movie 'Die Hard". The actors are given two jugs. A 3 gal. and a 5 gal. jug. A scale is attattched to an explosive and you must maintain the weight equal to 4 gallons of liquid. There is a water fountain close by. Explain how they could use both jugs and maintain the the 4 gallon weight.Keep in mind the the jugs are unmarked.
 
conair said:
In the movie 'Die Hard". The actors are given two jugs. A 3 gal. and a 5 gal. jug. A scale is attattched to an explosive and you must maintain the weight equal to 4 gallons of liquid. There is a water fountain close by. Explain how they could use both jugs and maintain the the 4 gallon weight.Keep in mind the the jugs are unmarked.
Click to expand...

This is rather simple problem.

Please show us your work/thoughts so that we know where to start.

remember

5 - 3 = 2

3 - 2 = 1

5 - 1 =4

That's all you need....
 
Hello, conair!

This is a classic problem . . . "older than dirt."

The word "maintain" is not correct.
. . They are to create an amount of exactly 4 gallons, using the two given jugs.

In the movie 'Die Hard", the actors are given two jugs.: a 3-gal and a 5-gal jug.
A scale is attattched to an explosive and you must place exactly 4 gallons of liquid on the scale.
Explain how they could use both jugs and create the 4-gallon weight.
Click to expand...

We have a 3-gallon and a 5-gallon jug.
Code: 
                  |     |
                  |     |
                  |     |
      |     |     |     |
      |     |     |     |
      |     |     |     |
      *-----*     *-----*
         3           5

Fill the 3-gallon jug.
Code: 
                  |     |
                  |     |
                  |     |
      |:::::|     |     |
      |::3::|     |     |
      |:::::|     |     |
      *-----*     *-----*
         3           5

Empty it into the 5-gallon jug.
Code: 
                  |     |
                  |     |
                  |     |
      |     |     |:::::|
      |     |     |::3::|
      |     |     |:::::|
      *-----*     *-----*
         3           5

Fill the 3-gallon jug.
Code: 
                  |     |
                  |     |
                  |     |
      |:::::|     |:::::|
      |::3::|     |::3::|
      |:::::|     |:::::|
      *-----*     *-----*
         3           5

Pour as much as possible into the 5-gallon jug.
Code: 
                  |:::::|
                  |::2::|
                  | - - |
      |     |     |:::::|
      |     |     |::3::|
      |::1::|     |:::::|
      *-----*     *-----*
         3           5

Empty the 5-gallon jug.
Code: 
                  |     |
                  |     |
                  |     |
      |     |     |     |
      |     |     |     |
      |::1::|     |     |
      *-----*     *-----*
         3           5

Empty the 3-gallon jug into the 5-gallon jug.
Code: 
                  |     |
                  |     |
                  |     |
      |     |     |     |
      |     |     |     |
      |     |     |::1::|
      *-----*     *-----*
         3           5

Fill the 3-gallon jug.
Code: 
                  |     |
                  |     |
                  |     |
      |:::::|     |     |
      |::3::|     |     |
      |:::::|     |::1::|
      *-----*     *-----*
         3           5

Pour it into the 5-gallon jug.
Code: 
                  |     |
                  |:::::|
                  |::3::|
      |     |     |:::::|
      |     |     | - - |
      |     |     |::1::|
      *-----*     *-----*
         3           5

There . . . We have exactly 4 gallons!

 
I am curious to know how you solved the problem. Can you give me more details?
 
My way:

Fill up 5 gallon - pour into 3 gallon (2 gallon left) - pour out 3 gallon - pour in 2 gallons from the 5 gallon jug (1 more gallon needed to fill up 3 gallon jug) - fill up 5 gallon - pour into 3 gallon to fill that up (1 gallon poured out) - 4 gallons left in the 5 gallon jug.

This is actually "stupid" problem. It is "mixing" definitions of "weight" (balance) and "volume" (gallon)- specially for high-school kids this becomes wrong type of "psuedo-physics" problem.
 
Fill both.

Pour out of 5gallon can by tipping can slowly, until horizontal water line goes
from bottom edge to opposite top edge; that'll leave 2.5 gallons in can.

Do same with 3gallon can, but pour inside the 5gallon can: 2.5 + 1.5 = 4

Anybody beat that :shock:
 
Denis said:
Pour out of 5gallon can by tipping can slowly, until horizontal water line goes
from bottom edge to opposite top edge; that'll leave 2.5 gallons in can.
Click to expand...
Doesn't this assume that the "jug" is neither a typical jug nor a typical bucket, but a cylindrical can? (Jugs come in many shapes, often with handles. Buckets tend to have radii that increase with the distance from the bottom of the bucket.)

I could be wrong, of course....

Eliz.
 
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