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A Cup Of Coffee II (Posted on 2012-03-06) Difficulty: 3 of 5
This is in continuation of A Cup Of Coffee.

You have a five cup mug, a three cup mug, a water supply, a sink with a drain, and a packet of instant coffee which when dissolved in one cup of water produces coffee of strength 100%.

The packet may be used at any time, but the entire contents of the packet must be dissolved into a single mug when it is used.

What integer values of c (from 1 to 25 inclusively) is possible if the task is to fix 4 cups of coffee at exactly c% strength? Prove that these are indeed the only possible values of c.

No Solution Yet Submitted by K Sengupta    
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re(2): possible solution | Comment 10 of 11 |
(In reply to re: possible solution by Charlie)

Charlie,

I see where I erred in interpretation of the instructions. I mistakenly 'read' that the packet dissolved into one "mug" of water produced coffee of strength 100%.
=======================================
The following is the corrected solution showing how to accomplish the different integer strengths of 25, 20, 16, 5, 4, 2, 1 and 0.

F# = fill #-cup mug from the water supply
T# = transfer as much as possible from the #-mug to the other
D# = dump the water/coffer from the #-cup mug into the sink
P# = add the packet of instant coffee to the #-cup mug
( X ) = a non-integer amount of coffee that is not utilized
    3( c%) 5( c%)
 1. 0(  0) 0(  0)
 2. 0(  0) 5(  0) F5
 3. 3(  0) 2(  0) T5
 4. 0(  0) 2(  0) D3
 5. 2(  0) 0(  0) T5
 6. 2(  0) 5(  0) F5
 7. 3(  0) 4(  0) T5 <===== 0%
 8. 3(  0) 4( 25) P5 <===== 25%
 9. 2(  0) 5( 20) T3
10. 3( X ) 4( 20) T5 <===== 20%
11. 0(  0) 4( 20) D3
12. 3(  0) 4( 20) F3
13. 2(  0) 5( 16) T3
14. 3( X ) 4( 16) T5 <===== 16%
15. 0(  0) 4( 16) D3
16. 3( 16) 1( 16) T5
17. 0(  0) 1( 16) D3
18. 3(  0) 1( 16) F3
19. 0(  0) 4(  4) T3 <====== 4%
20. 3(  4) 1(  4) T5
21. 0(  0) 1(  4) D3
22. 3(  0) 1(  4) F3
23. 0(  0) 4(  1) T3 <====== 1%


    3( c%) 5( c%)
 1. 0(  0) 0(  0)
 2. 0(  0) 5(  0) F5
 3. 3(  0) 2(  0) T5
 4. 0(  0) 2(  0) D3
 5. 2(  0) 0(  0) T5
 6. 2(  0) 5(  0) F5
 7. 2(  0) 5( 20) P5
 8. 3( X ) 4( 20) T5 <===== 20%
 9. 0(  0) 4( 20) D3
10. 3( 20) 1( 20) T5
11. 0(  0) 1( 20) D3
12. 3(  0) 1( 20) F3
13. 0(  0) 4(  5) T3 <====== 5%


    3( c%) 5( c%)
 1. 0(  0) 0(  0)
 2. 0(  0) 5(  0) F5
 3. 0(  0) 5( 20) P5
 4. 3( 20) 2( 20) T5
 5. 3( 20) 0(  0) D5
 6. 0(  0) 3( 20) T3
 7. 3(  0) 3( 20) F3
 8. 1(  0) 5( 12) T3
 9. 3(  8) 3( 12) T5
10. 1(  8) 5( X ) T3
11. 1(  8) 0(  0) D5
12. 0(  0) 1(  8) T3
13. 3(  0) 1(  8) D3
14. 0(  0) 4(  2) T3 <====== 2%

Edited on March 7, 2012, 3:59 pm
  Posted by Dej Mar on 2012-03-07 14:14:06

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