DAY 269
You have three boxes labelled PEACHES, ORANGES, and PEACHES AND ORANGES. All three boxes are labelled incorrectly. How can you label the boxes correctly by opening just one box and removing just one piece of fruit?
Answer:
Open the box labelled PEACHES AND ORANGES. Since it is labelled incorrectly, it should contain either all oranges or all peaches. If you pull out an orange, you know that box should be labelled ORANGES. Now switch the labels on the other two boxes and they will all be correct.
Tuesday, 29 June 2010
DAY 268
If I put one canary in each cage, I have one canary too many. If I put two canaries in each cage, I have one cage too many. How many canaries and how many cages do I have?
Answer:
4 canaries and 3 cages...............Let X = canaries and Y = cages.
X - 1 = Y and X/2 + 1 = Y...............So X - 1 = X/2 + 1.
If I put one canary in each cage, I have one canary too many. If I put two canaries in each cage, I have one cage too many. How many canaries and how many cages do I have?
Answer:
4 canaries and 3 cages...............Let X = canaries and Y = cages.
X - 1 = Y and X/2 + 1 = Y...............So X - 1 = X/2 + 1.
Monday, 21 June 2010
Saturday, 19 June 2010
Wednesday, 16 June 2010
Friday, 11 June 2010
DAY 259
Mixed up in a bag are ten five-dollar bills, ten ten-dollar bills, and ten twenty-dollar bills. You can remove the bills, one at a time, without looking, but you must stop when you get three bills of the same denomination. What is the most money you could get out of the bag before you had to stop?
Answer:
$90..........$20 + $20 + $10 + $10 + $5 + $5 + $20
Mixed up in a bag are ten five-dollar bills, ten ten-dollar bills, and ten twenty-dollar bills. You can remove the bills, one at a time, without looking, but you must stop when you get three bills of the same denomination. What is the most money you could get out of the bag before you had to stop?
Answer:
$90..........$20 + $20 + $10 + $10 + $5 + $5 + $20
DAY 257
FRACTIONS AND MARRIAGE
Credit: GAMES Magazine--Bill Graham
NOTE: If you can solve this problem, according to a speaker who presented it at a math conference, then you really understand fractions.
In a town, three-fifths of the women are married to two-thirds of the men. What fraction of the town's adults are married? (Marriage is defined here as one man married to one woman.)
Answer:
Twelve-nineteenths.................If m = the number of men in the town and w = the number of women, then 3/5 (w) = 2/3 (m). Multiplying both sides of the equation by 15 yields the equation 9w = 10m. The smallest whole numbers that could satisfy this equation are w = 10 and m = 9, in which case we can confirm that three-fifths of the women (6) would be married to two-thirds of the men (6). This in turn would mean that 12 of the nineteen townspeople are married. Using solutions that are multiples of 10 and 9 will not alter this fraction.
FRACTIONS AND MARRIAGE
Credit: GAMES Magazine--Bill Graham
NOTE: If you can solve this problem, according to a speaker who presented it at a math conference, then you really understand fractions.
In a town, three-fifths of the women are married to two-thirds of the men. What fraction of the town's adults are married? (Marriage is defined here as one man married to one woman.)
Answer:
Twelve-nineteenths.................If m = the number of men in the town and w = the number of women, then 3/5 (w) = 2/3 (m). Multiplying both sides of the equation by 15 yields the equation 9w = 10m. The smallest whole numbers that could satisfy this equation are w = 10 and m = 9, in which case we can confirm that three-fifths of the women (6) would be married to two-thirds of the men (6). This in turn would mean that 12 of the nineteen townspeople are married. Using solutions that are multiples of 10 and 9 will not alter this fraction.
Thursday, 10 June 2010
DAY 256
FIGURE IT OUT
Credit: GAMES Magazine---Raymond Love
The two columns below add up to the same number and are related in a special way. Can you determine the significance of the juxtaposition of these columns?
FIVE.....................4
SIX........................9
SEVEN..................5
FIVE + SIX + SEVEN = 18
4 + 9 + 5 = 18
Answer:
The spelled-out numbers FIVE, SIX, and SEVEN contain the Roman numerals IV, IX, and V, respectively, corresponding to the 4, 9, and 5 in the right column.
FIGURE IT OUT
Credit: GAMES Magazine---Raymond Love
The two columns below add up to the same number and are related in a special way. Can you determine the significance of the juxtaposition of these columns?
FIVE.....................4
SIX........................9
SEVEN..................5
FIVE + SIX + SEVEN = 18
4 + 9 + 5 = 18
Answer:
The spelled-out numbers FIVE, SIX, and SEVEN contain the Roman numerals IV, IX, and V, respectively, corresponding to the 4, 9, and 5 in the right column.
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