Question #4451

A merchant has a beaker containing 24 ounces of a precious fluid. He also has empty 5-ounce, 11-

ounce, and 13-ounce beakers. How can he divide the fluid into three equal portions?

ounce, and 13-ounce beakers. How can he divide the fluid into three equal portions?

Expert's answer

Here is an algoritm

1) full 13-ounce beaker

[table]

5-ounce

11-ounce

13-ounce

0

0

13

[/table]

2)from 13-ounce fill the 11-ounce beaker(at 13-oun. Wewill have 2 ounce of water)

[table]

5-ounce

11-ounce

13-ounce

11

2

[/table]

3)empty 11-oun. And fill it from 13 oun. Wich has 2 ounce of w.

[table]

5-ounce

11-ounce

13-ounce

2

[/table]

4)fill 13-oun. And pour water from it to 5-ouncebeaker

[table]

5-ounce

11-ounce

13-ounce

5

2

13-5=8(!!!)

[/table]

5)

Empty 5-ounce and fill it from 11-ounce

[table]

5-ounce

11-ounce

13-ounce

2

0

8

[/table]6)

Fill 11-ounce and add water to 5-ouncebeaker

[table]

5-ounce

11-ounce

13-ounce

5

11-3=8

8

[/table]7)

Return 5ounce to 24-ounce beaker and itwill have also 5+3=8 ounce of water

1) full 13-ounce beaker

[table]

5-ounce

11-ounce

13-ounce

0

0

13

[/table]

2)from 13-ounce fill the 11-ounce beaker(at 13-oun. Wewill have 2 ounce of water)

[table]

5-ounce

11-ounce

13-ounce

11

2

[/table]

3)empty 11-oun. And fill it from 13 oun. Wich has 2 ounce of w.

[table]

5-ounce

11-ounce

13-ounce

2

[/table]

4)fill 13-oun. And pour water from it to 5-ouncebeaker

[table]

5-ounce

11-ounce

13-ounce

5

2

13-5=8(!!!)

[/table]

5)

Empty 5-ounce and fill it from 11-ounce

[table]

5-ounce

11-ounce

13-ounce

2

0

8

[/table]6)

Fill 11-ounce and add water to 5-ouncebeaker

[table]

5-ounce

11-ounce

13-ounce

5

11-3=8

8

[/table]7)

Return 5ounce to 24-ounce beaker and itwill have also 5+3=8 ounce of water

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