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Donald Cartmill

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  1. The % of the pay out x money in the pot is correct. In the 2nd method you are taking the same % as before of a larger number composed of real $ and imaginary money. The larger % will always produce the larger quotient and therefore when the missing $ in each case is subtracted the larger quotient will appear to have earned more. This appearance can actually be reversed depending on the amount each fails to include. Example:A) puts in $25; B) puts in $15 ; C) puts in $0 but agrees to add $19 later ; D) is such a pitiful golfer He only owes $1 (to be paid later ) All agree the payout will be 0.6 % and 0.4% : They play and C) wins the 0.6 ; and D) wins the 0.4. OK there is $40 actual so C) gets 0.6 x 40 = $24; D) gets 0.4 x 40 =$16. Now if they were to calculate using the imaginary amount it would be C) 0.6 x 60 = $36 ; D) 0.4 x 60 =$24. Now you take $19 away from C) $36 wins $17 ; D) $24 - $1 wins $ 23 A bit wordy and over done Bottom line is there are 2 different %'s to establish payout. The larger the disparity of "imaginary $ "between the two winners determines what the false winning amount is in each instance
  2. The hat should contain $20 ( but only $10 ) By proportionality then $12 becomes $6 and $8 becomes $4. The 2nd method does NOT work as the following ex: will show Assuming The winner is to receive $16 and the trash to get $4. $16 - $5 =$11 ( only $10 in the hat ) The trash $ 4- $5 Means he would owe $1 Where in contrast by proportionality $16 represents 8/10 of $20 and would receive $8 of the $10 ,and 4/20 = 2/`10 = $2
  3. Problem does NOT state how the first $ 10 was split , but if split according to initial premise dad should get $6 out of every $10 The 1st $10 is split Dad = 6 ; Larry =4 Now they split the 2nd $10 $6 $4 Therefore at this point Dad has $12 Larry has $8 Subtracting out the $5 in each case Dad wins $7 Larry wins $3 Nowhere does it say Larry wins $4. The mention of $4 out of the 2nd $ 10 is misdirection
  4. Started with 760 buttons; Problem is solved by working backward. X/2 -10 = 5 X - 20 = 10 , X = 30. X/5 -10 =30 ; X - 50 =150 X = 200 X/4 -10 =200 X -40 = 800 X = 760
  5. I want to share with you the solution to any odd numbered grid,where by the columns , rows ,and diagonals each has the same total. we have all seen the very simple 3 x 3 grid filled with # 1 thru 9 ; Much more difficult would be a 5 x 5 grid with numbers 1 thru 25 ; A 7 x 7 grid would be virtually impossible ,but with the few simple rules below any odd numbered grid can be filled as fast as you can write and all cols ,rows and diagonals total the same 1a) Always start in the square to right of center square !) you are always traveling diagonally down to the right; when you exit the right side,you move down one row then jump left to farthermost open square 2) moving diagonally down to the right.you will exit the bottom..move one column right and jump up to the highest open square 3)When you are trying to move down and right and you encounter a numbered square ...you go in the upper left corner and out the upper right corner ending on the same row you started 3 a) If the above move puts you out side the main grid then you jump back to the left as far as poss on that row 3 b) If you are still inside the big grid,then you move down to the right as before 3 c) when you exit on the main diagonal Move to the left as far as possible in the bottom row 4) The above are all repetitive moves and allow you fill in any odd numbered grid such that all columns ,all rows and the diagonals add up to the same sum
  6. A monk started out at 7:00 am at the bottom of the mountain and climbed to the top arriving at 7:00 pm He prayed all thru the night and at 7:00 am started down the mountain arriving at the bottom at 7:00 pm. Is there a point on the mountain trail where the monk was at the same time going up and coming down Proof please
  7. Fold rope in 1/2 then again in 1/4 allow 1/4 to burn =15 minutes
  8. 60-40= 20 car lengths; therefor man walked only 10 car lengths. ; mans rate is 3 mph; t = d / speed ; We have to look at the 1st car in each direction .as a unit of length Now assume 10 cl to be 1 mile 1m / 3 mph =.333 hrs; traveling in same direction train traveled = 40 + 10 = 50 lts =5 miles ; s = d / t 5 m/ .33 hrs = 15 mph train traveling opposite direction =60 - 10 = 50 lts = 5m; 5m / .33 hrs = 15 mph So let us assume a different value for 10cl = 0.5 miles ; 0.5 / 3mph = 016666667 hrs ; 50 cl = 2.5 m; 2.5 m / 0.16666666667 = 15 mph Regardless of what value in miles is assumed for 10 car lengths the answer is 15 mph If you cannot assume both are traveling at the same speed,then it is unsolvable. or actually a solutions for every increment of time he walks i.e. he walks 15 seconds and in that time 40 passes one way and 60 passes the other way...Obviously they would really be flying and one going approximately 50% faster
  9. 60-40= 20 car lengths; therefor man walked only 10 car lengths. ; mans rate is 3 mph; t = d / speed ; Now assume 10 cl to be 1 mile . / 3 mph =.333 hrs; traveling in same direction train traveled = 40 + 10 = 50 lts =5 miles ; s = d / t 5 m/ .33 hrs = 15 mph train traveling opposite direction =60 - 10 = 50 lts = 5m; 5m / .33 hrs = 15 mph So let us assume 10cl = 0.5 miles ; 0.5 / 3mph = 016666667 hrs ; 50 cl = 2.5 m; 2.5 m / 0.16666666667 = 15 mph Regardless of what value in miles is assumed for 10 car lengths the answer is 15 mph 60-40= 20 car lengths; therefor man walked only 10 car lengths. ; mans rate is 3 mph; t = d / speed ; Now assume 10 cl to be 1 mile . / 3 mph =.333 hrs; traveling in same direction train traveled = 40 + 10 = 50 lts =5 miles ; s = d / t 5 m/ .33 hrs = 15 mph train traveling opposite direction =60 - 10 = 50 lts = 5m; 5m / .33 hrs = 15 mph So let us assume 10cl = 0.5 miles ; 0.5 / 3mph = 016666667 hrs ; 50 cl = 2.5 m; 2.5 m / 0.16666666667 = 15 mph Regardless of what value in miles is assumed for 10 car lengths the answer is 15 m time = distance / speed; "t " will be in hours. speed = distance / time Regardless of what value in miles is assumed for 10 car lengths the answer is 15 mph
  10.     90 -70= 20 deg using side of square as "a" , we have tan20 = o/a and then O= atan 20 or 0.3639a; The right leg of the  triangle and  the top portion of the right side               20 +45 =65 ; 90-65  =     25 degrees  the other angle at top left and therefore the 3rd angle is 65;                                 tan 25 = o/a  or  o = a tan25;        o = 0.4663a.   since a  is the length of a side then the bottom leg of the lower right triangle = 1 - 0.4663 = 0.5336a ;          the vertical leg of the lower triangle is 1 -   0.3639a  =     0.6360a.  we will designate the left angle of the bottom triangle as Y ,  So we can say Tan Y = o/a or 0.6360a / 0.5336a;      tan Y = 1.1917;  Y = 50    Adding the bottom angles we have 65 + 50 = 115 :   180 - 115 =65 degrees              X = 65 degrees

  11. Machine A worked 40 hours. 45x = 40(x + 5 ); 45x = 40x +200.; 5x = 200 ;x = 40
  12. The area would be that of the original triangle,however that area would vary as determined by the length of the 3rd side
  13. operates on the 1st ; Turns the gloves inside out for # 2; operates bare handed on the 3rd
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