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32. C - wrap counterclockwise, with the green fringe on the outside. 37. 2 putts. The first putt goes 11 meters at an angle and ends up 11 meters from the hole. The second putt goes 11 meters and in. 53. This is a poorly-worded question and I refuse to muddle through it. Ditch the "counting this portrait as one" business and it becomes a much more elegant puzzle that still has a correct answer! 47. Jeez, who edited these word problems? Assuming the three candidates are part of the 40-voter pool (and why wouldn't they be, they're locals), 14 votes for the winning candidate including the candidate himself/herself, with each of the other two getting 13 votes.
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Mar 14, 2016 01:21
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114: Assuming that (1) A and B worked at an equal pace and (2) C would have worked at the same pace had he not been 'injured', then C's share of the work is three person-days. A did two-thirds of C's job, B did one-third. A gets six coins, B gets three. 109: Dad catches up with daughter after ten seconds of walking, dog runs a total of fifty meters during that time.
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Jul 19, 2016 01:34
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094: Yay, more assumptions. On the two cubes on the rightmost side, the open end of the 'C' always points towards a 'D' face. The cube just below the mystery cube therefore has a 'D' on its topmost face, and the mystery cube has a 'D' on its bottommost face. Topmost face is a 'B'.
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Jul 26, 2016 03:29
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077: The stars must be 'heavy', or else there is no solution. If the moons are 'light', weight on the left is 5x1 + 3x2 and weight on the right is 2x1 + 1x2 + 2x3, so 1 weight (a moon) is needed. If the moons are 'medium', weight on the left is 3x1 + 5x2 and weight on the right is 1x1 + 2x2 +2x3 so 2 weight (still a moon) is needed. Moon. 121: Counting the upper left corner as (1,1) and the upper right corner as (1,4), then a lamp must go in (3,4) to light the lowest diagonal pathproperly and not overlapping any other lamp use. Second lamp then goes at (4,3) to light the rightmost diagonal; third at (2,2) to light the leftmost diagonal; fourth back at (1,1) to clear the rest. (1,1); (2,2); (4,3); (3,4). 122: Five lamps will be needed. The first goes at (5,5) to light the bottom diagonal. Second and third go at (3,4) and (4,2) to get the other two lit. Next goes at (2,1) to illuminate a column, and last at (1,3). (1,3); (2,1); (3,4); (4,2); (5,5). 097: link. Reasoning: Columns 1 and 2 needed 3 stars each and had 1 star per column fixed, so the 4-star square filled the free space. Row 1 needed 3 stars, had 2 and row 2 needed 3 stars, had 1, so the 3-star free piece that filled its gap also moved. Next, placed the other 3-star square into the bottom right corner (necessity; placing it anywhere else would create a row or column with 4+ stars). This required that the 2-star piece go into bottom center to fill rows 5 and 6 properly, and the rest became fairly trivial.
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Aug 4, 2016 03:55
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100: Not entirely sure about the mechanics of this one either, but on the assumption of "You can advance only one plank or three planks at a time, and can never step into a hole on either side", then here is a possible solution.
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Aug 8, 2016 22:36
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134: link to a path that clears the puzzle. 088: Left 1, Down 2, Down 3, Right 1, Right 2, Right 3, Left 1, Right 2, Down 3, Up 1, Down 2, Left 3 gets the job done in 12 jumps.
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Aug 15, 2016 01:40
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W28: H S C D C D H S D C S H S H D C
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Aug 22, 2016 03:25
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W30: From the third row, the numbers are 0, 1, 2 and 3 in some order. From the fourth row either the third digit is 0 or the fourth digit is 2. Assuming the third digit is 0 means that from the second row the first digit must be 3... but this leads to a contraction when examining the first row. So the solution is of the form xxx2. First row then says we have xx32, second row says x032, and the final answer is 1032. W32: Swap the 1 and the 2, then rotate the 6 to get a '9', and you have 129 which is 43 * 3.
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Aug 23, 2016 02:36
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140 is somewhat elegant once considered properly: block D must be on the base, so block B must be on top. The other blocks can be fitted into a correct config, but actually doing so isn't needed to get a solution. edit: update on the bottom of page 5! Bacon In A Wok fucked around with this message at 22:50 on Aug 24, 2016 |
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Aug 24, 2016 22:43
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151: Dan's answers to questions 6 and 9 are right - he had the same answers as Lisa on all other 8 questions, and scored 20 points more than her. So Mary's score of 70 is right answers on all but 6, 9, and one of 1, 3, 7 or 8 (doesn't matter which - John's answers to these four questions match Mary's). John's answers match Mary's for seven questions (where John got five correct, two incorrect) and differ for three questions (where John got one correct, two incorrect). John's score is 60. ... also, what kind of school has neckbeard Dan and babyface Lisa in the same class, anyway? 152: Total number of tournament rounds for an n-player round-robin is of the form 1+2+...+(n-1). 59 rounds played with one player skipping out on some rounds suggests a starting pool of 12 players, 66 rounds in a complete tournament (11 games per player) so Monty played 4 games before leaving.
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Aug 26, 2016 01:02
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