True or False:
5832/17496 + 2943/17658 = 6729/13458
Answer:
True............1/3 + 1/6 = 1/2
Also:
4392/175618 = 1/4
2769/1345 = 1/5
2394/16758 = 1/7
3187/25496 = 1/8
6381/57429 = 1/9
Wednesday 27 January 2010
DAY 207
A man and his wife were expecting a baby, and the man wanted to give "the boy" 2/3 of his estate, and leave 1/3 to his wife. But in case "the boy" turned out to be "the girl," then 2/3 of his estate would go to his wife, and 1/3 to his daughter. As it turned out, they had twins---one boy and one girl. In keeping with his original plans, how should Dad settle his estate between his wife and two children?
HINT:
He wanted to give the mother twice as much as the daughter, and the son twice as much as the mother.
Answer:
1/7 to the daughter, 2/7 to his wife, and 4/7 to his son.
`
HINT:
He wanted to give the mother twice as much as the daughter, and the son twice as much as the mother.
Answer:
1/7 to the daughter, 2/7 to his wife, and 4/7 to his son.
`
Tuesday 26 January 2010
DAY 206
A cyclist rode a mile in three minutes with the wind, and returned in four minutes against the wind. Assuming that he pedals at a constant rate, how long would it take him to cycle one mile on a windless day?
A cyclist rode a mile in three minutes with the wind, and returned in four minutes against the wind. Assuming that he pedals at a constant rate, how long would it take him to cycle one mile on a windless day?
Answer:
3 3/7 minutes..........If he can cycle 1 mile in three minutes with the wind (1/3 mile per minute), then he could cycle 1 1/3 (4/3) miles in four minutes with the wind. He returns against the wind in the same four minutes, so he can cycle 2 1/3 miles in eight minutes. The effect of the wind can now be ignored. So, on a windless day, he can cycle 7/3 miles in 8 minutes, or 8 divided by 7/3, one mile in 3 3/7 minutes.
3 3/7 minutes..........If he can cycle 1 mile in three minutes with the wind (1/3 mile per minute), then he could cycle 1 1/3 (4/3) miles in four minutes with the wind. He returns against the wind in the same four minutes, so he can cycle 2 1/3 miles in eight minutes. The effect of the wind can now be ignored. So, on a windless day, he can cycle 7/3 miles in 8 minutes, or 8 divided by 7/3, one mile in 3 3/7 minutes.
Tuesday 19 January 2010
DAY 204
WHICH WITCH
Credit: GAMES Magazine - January, 1989
WHICH WITCH
Credit: GAMES Magazine - January, 1989
Karen Nimmons
If six witches can wind sixty watches in sixteen weeks, how many watches would sixty witches wind in six weeks?
Answer:
225............One witch can wind 10 watches in sixteen weeks. So one witch can wind 10/16 or 5/8 watches in one week. One witch could wind 6 x 5/8 or 30/8 watches in six weeks. Sixty witches could wind 60 x 30/8 or 225 watches in six weeks.
If six witches can wind sixty watches in sixteen weeks, how many watches would sixty witches wind in six weeks?
Answer:
225............One witch can wind 10 watches in sixteen weeks. So one witch can wind 10/16 or 5/8 watches in one week. One witch could wind 6 x 5/8 or 30/8 watches in six weeks. Sixty witches could wind 60 x 30/8 or 225 watches in six weeks.
Friday 8 January 2010
DAY 203
COUNTING TEASE
Credit: GAMES Magazine - September, 2001
COUNTING TEASE
Credit: GAMES Magazine - September, 2001
D.A. Lavallee
Using the above title as a clue, what number logically comes next in the following sequence?
2, 3, 10, 12, 13, ___
Answer:
20..........The sequence is the set of positive integers that start in English with the letter T, written in ascending order. [Counting Tease (T's)]
Using the above title as a clue, what number logically comes next in the following sequence?
2, 3, 10, 12, 13, ___
Answer:
20..........The sequence is the set of positive integers that start in English with the letter T, written in ascending order. [Counting Tease (T's)]
Thursday 7 January 2010
DAY 202
IT ALL ADDS UP
Credit: GAMES Magazine
July, 1985
IT ALL ADDS UP
Credit: GAMES Magazine
July, 1985
Michael W. Ecker
You and an opponent alternate picking numbers from 1 to 6, and a running total is kept. If the object of the game is to make the score total exactly 50, and you go first, what number should you pick to insure a win for yourself?
Answer:
Fifty is 1 more than a multiple of 7, so pick 1 as your number. Then, no matter what your opponent chooses, you say 7 minus that number. On your eighth turn, you will cause the score to hit 50 exactly.
You and an opponent alternate picking numbers from 1 to 6, and a running total is kept. If the object of the game is to make the score total exactly 50, and you go first, what number should you pick to insure a win for yourself?
Answer:
Fifty is 1 more than a multiple of 7, so pick 1 as your number. Then, no matter what your opponent chooses, you say 7 minus that number. On your eighth turn, you will cause the score to hit 50 exactly.
DAY 201
Credit: GAMES Magazine--October, 1997--Mildred K. Henderson (adapted)
If your birthday were January 9, and you were born in the 20th century, the common abbreviated form of your birthday would always state the current year: 1/9/01 - 1/9/99. This phenomenon would stop in the year 2000. (1/9/00 would not be the current year.) When would it start again, and what would your birthday have to be?
Answer:
Your birthday would have to be February 1 and the abbreviation curiosity would start on Feb. 1, 2100---2/1/00. This would not work in the year 2000 or any year in the 21st century because there is no "0" (zero) date in February.
Credit: GAMES Magazine--October, 1997--Mildred K. Henderson (adapted)
If your birthday were January 9, and you were born in the 20th century, the common abbreviated form of your birthday would always state the current year: 1/9/01 - 1/9/99. This phenomenon would stop in the year 2000. (1/9/00 would not be the current year.) When would it start again, and what would your birthday have to be?
Answer:
Your birthday would have to be February 1 and the abbreviation curiosity would start on Feb. 1, 2100---2/1/00. This would not work in the year 2000 or any year in the 21st century because there is no "0" (zero) date in February.
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