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dgreening

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Everything posted by dgreening

  1. I am not sure if this satisfies the requirements about not impacting the eco-system.
  2. very elegant solution(s)! My hat is off to both of you!
  3. I think you are both on the right track.
  4. Hi I have been reading this site for sometime, but only recently started actually posting anything. I enjoy puzzles [crossword, KenKen and logic are my current faves], world travel and good novels audio or visual. Looking forward to more great puzzlers!
  5. The OP does not constrain the boy's arrival to be on the minute. Between 7:58 and 7:59, for example, p decreases linearly from 1 to 0.8; it does not drop abruptly to 0.8 at 7:58. Similarly for the next two minute-intervals. The boy does not arrive later than 8:01. Make a graph of p vs boy's arrival time and find its average value. good point! Now that I re-read the problem I see your point. In fact the boy does not necessarily arrive on the minute and the bus does not necessarily arrive on the minute either. I think your approach is better.
  6. I agree, we were working in parallel.
  7. I asked some others and after talking it over, I realized that we cannot mix m [distance] with m3 of volume. The m/ m3 is actually shorthand for m / m3 of fuel. Dividing through would leave the expression 1/ m2 of fuel. Bottom line: I think this was a trick question. I am going to forget about this one.
  8. I think DeGe's strategy is sound. The term "toggle" may be adding confusion. Prisoner 1 always turns the light OFF [or leaves it off] Every other prisoner is allowed to turn the light ON only once [after they turn it on once, the leave it unchanged]. Other prisoners are not allowed to ever turn the light off. after day 1 prisoner 1 counts the times he turns it off and after 99 he can safely assume that all have been in the room
  9. not sure what to make of "Well, I think that you have to add units to the numerator (as well as the denominator) to make any sense of things. in your example, 25 miles/gal = 118,400,000/m2=118,400,000 parsecs/parsecs-m2. Now that makes physical sense!"
  10. when I first started mulling this over, I realized that miles/ gal or km/ liter are skipping several steps. We really mean that gal of gas is sufficient to propel this vehicle so many miles. I took a typical value for a US car [25 mpg] and started converting it. Values are approx 25 miles ~ 32 km 1 Gal ~ 3.7 Liters ~ 3.7/1000 Meters 3 25 Mile/ gal = 32 km/ 0.0037 meters 3 ~ 32,000/ 0.0037 /meters 2 therefore 25 miles/ gal ~ 118,400,000 /meter 2 I am at a loss for how this related to the parallel piped, though I cannot find anything else that makes sense.
  11. I think the smallest square would be a 3 x 3. If we start with the most compact area to contain 4 unit squares, it would be a 2 x 2 square. The addition of another unit square will force it to add at least one row or column - thus forcing you to a 2x3 configuration. To meet the criteria of the problem, we must expand that to a 3 x 3 There are multiple configurations [many are just reflections of others # # # # # - - - - or # # # # - - # - - or # - # - # - # - # or ...
  12. We also do not want the Dr to come into contact with the blood from any of the 3 patients One approach
  13. I hit the wrong key 20 = - T1 ( 0.917 mi/ min) + T2 ( 0.667 mi/ min) [Eqn 1] 100 = T1 +T2 [Eqn 2] Multiplying all terms in Eqn 1 by {- 1/0.667} yeilds -30 = 1.375 T1 - T2 Adding this to Eqn2 results in 70 = 2.375 T1 Therefore T1 = 70 /2.375 = 29.47 min, so D1 = 29.47 min * ( 0.917 mi/ min) = 27.02 D2 = D1 + 20 = 47.02 Therefore Total distance = 74.04 Miles
  14. X = miles at 55 mph X+20 = miles driven at 40 mph and 55 mi/ hr = 0.917 mi/ min 40 mi/ hr = 0.667 mi/ min D = Total Distance = D1 (at 55 mph) + D2 (at 40 mph) T = Total Time = T1 (at 55 mph) + D2 (at 40 mph) D1 = T1 ( 0.917 mi/ min) D2 = T2 ( 0.667 mi/ min) Therefore D = D1 + D2 = T1 ( 0.917 mi/ min) + T2 ( 0.667 mi/ min) We also know that D1 +20 = D2 These can be rearranged into 2 simultaneous equations in the form 20
  15. The problem statement says "His truck can only hold 1000 apples." ‚Äčthe update says that the store is 1000 miles away If the driver starts with a full load [1000 apples] and drives 1000 miles, he will have 0 apples left upon arrival. Thus, unless he can ferry apples to some point(s) in between and then move these forward, there is no way to every deliver even 1 apple. If we assume that there is a point halfway [500 miles] then he could do the following: leave the warehouse with 1000 apples drive 500 miles [now has 500 apples] leave the apples at the halfway point [now has 500 apples] drive back to the ware house leave the warehouse with 1000 apples drive 500 miles [now has 500 apples] Picks up 500 the apples at the halfway point [now has 1000 apples in the truck and 0 apples at the halfway point]] Drives the remaining 500 miles to the store and delivers 500 apples Thus for 2 trips[500 and 1000 miles one way (3000 miles total round trip)] 500 apples can be delivered to deliver 3000 apples would require: an interim storage area at 500 miles 6 "short" round trips between the warehouse and the interim storage area [1000 miles R/T each] 6 "long" round trips from the warehouse [stopping at the interim storage area to reload] to the store [2000 miles R/T each] Notes: you can break this up many ways it is not clear what the optimal interim storage area distance [from the ware house] is [and if more than 1 helps]
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