The title of the problem tells you how to approach these four questions. (A). A bus leaves Moscow for Tula at noon. An hour later a cyclist leaves Tula for mosco, moving slower than the bus. When the bus and cyclist meet, which one of the two will be farther from Moscow? (B). Which is worth more: a pound of $10 gold peices or half a pound of $20 gold pieces? (C). At six o’clock the wall clock struck 6 times. Checking with my watch, I noticed the time between the first and last strokes was 30 seconds. How long will the clock take to strike 12 at midnight? (D). Three swallows fly outward from a point. When will they all be on the same plane in space? Now check the answers. Did you fall into any of the traps which lurk in these simple problems?
Answer: (A). Neither (B). A pound of metal is always more than half a pound of the same metal. (C). Six strokes took 30 seconds, therefore 12 strokes will take 60 seconds. But when the clock struck six, there were only 5 intervals between strokes, and each interval was 30/5=6 seconds. Between the first and twelfth strokes there will be 11 intervals of 6 seconds each, therefore 12 strokes will take 66 seconds. (D). There is always a plane that contains any 3 points
Solution:
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(A). Neither
(B). A pound of metal is always more than half a pound of the same metal.
(C). Six strokes took 30 seconds, therefore 12 strokes will take 60 seconds. But when the clock struck six, there were only 5 intervals between strokes, and each interval was 30/5=6 seconds. Between the first and twelfth strokes there will be 11 intervals of 6 seconds each, therefore 12 strokes will take 66 seconds.
(D). There is always a plane that contains any 3 points
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(A). Neither
(B). A pound of metal is always more than half a pound of the same metal.
(C). Six strokes took 30 seconds, therefore 12 strokes will take 60 seconds. But when the clock struck six, there were only 5 intervals between strokes, and each interval was 30/5=6 seconds. Between the first and twelfth strokes there will be 11 intervals of 6 seconds each, therefore 12 strokes will take 66 seconds.
(D). There is always a plane that contains any 3 points
(B). A pound of metal is always more than half a pound of the same metal.
(C). Six strokes took 30 seconds, therefore 12 strokes will take 60 seconds. But when the clock struck six, there were only 5 intervals between strokes, and each interval was 30/5=6 seconds. Between the first and twelfth strokes there will be 11 intervals of 6 seconds each, therefore 12 strokes will take 66 seconds.
(D). There is always a plane that contains any 3 points
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I have 10 red socks and 10 blue socks in a drawer. How many socks must I take out to ensure that I have at least one matching pair? One how many to ensure I have at least a pair of red and a pair of blue?
Answer: Three for one pair, and twelve to ensure one pair of each color.
Solution:
Jack has 8 bricks 7 of them weights the same amount and one is slightly heavier. Using a balance scale, how can Jack find the heavier brick in two weighings?
Answer: First he split them in to piles of 3, 3, and 2 bricks. Then he weighs both groups of 3 with each other. If they balance he knows the brick is one of the 2 unweighed bricks and he can weigh them to find the heaver one. If the the stacks of 3 bricks do not balance, he will weigh 2 of the 3 bricks. If they balance he will know the brick left unweighed is heavier, or if they do not balance, he will find the heavier one.
Solution:
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First he split them in to piles of 3, 3, and 2 bricks. Then he weighs both groups of 3 with each other. If they balance he knows the brick is one of the 2 unweighed bricks and he can weigh them to find the heaver one. If the the stacks of 3 bricks do not balance, he will weigh 2 of the 3 bricks. If they balance he will know the brick left unweighed is heavier, or if they do not balance, he will find the heavier one.
Show Answer
First he split them in to piles of 3, 3, and 2 bricks. Then he weighs both groups of 3 with each other. If they balance he knows the brick is one of the 2 unweighed bricks and he can weigh them to find the heaver one. If the the stacks of 3 bricks do not balance, he will weigh 2 of the 3 bricks. If they balance he will know the brick left unweighed is heavier, or if they do not balance, he will find the heavier one.
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