This is one amazing word problem that gave me a headake so did the other word problem as well... :(
You have been given a death sentence in a foreign country. The night before your execution, you are brought two boxes, one containing fifty red marbles and one containing fifty blue marbles. All the marbles are the same size, weight, and shape. You are told that on the following day, the executioner will be blindfolded and will put his hand into one of the boxes and remove a marble. If the marble is blue, your life will be spared. If the marble is red, it's off with your head.
You can arrange the marbles in the two boxes however you like, but you can't leave out any marbles, and you can't leave either box empty. Also, the executioner is known to take his time choosing a marble, so you can't improve your chances by stacking all the blue marbles at the top.
How can you arrange the marbles in the boxes so as to maximize your chances of keeping your head?
There are ninety-seven birds sitting in a tree. Shotgun Sharon swaggers up to the tree and fires three whole rounds of ammunition at the birds. She kills twice as many in the first round as in the second round, and twice as many in the second round as in the third round. In the third round she kills three birds. How many birds are left in the tree after Sharon fires off the third round?
2006-10-25 09:13:00 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Marbles: don't mix and match anything. Simply present the box with the blue marbles for him to choose from
Birds:
3+6+12=21 birds killed
97-21=76 live birds
2006-10-25 09:22:12 · answer #1 · answered by Anonymous · 0⤊ 0⤋
#1 - Put 50% of each color in each box. Since the executioner takes his time digging, put the majority of blue at the bottom or middle of the box.
#2 - 3 x 2 = 6 x 2 = 12, so that totals 21. 97 - 21 = 76 birds left in the tree
2006-10-25 16:27:54 · answer #2 · answered by Gonko B 2 · 0⤊ 0⤋
76 birds and mix all marbles except one in one box and put a blue one in th other. the excecutioner could be curious to finding out if that one marble is a red one or a blue one, so he might choose that one. Even if he doesnt choose the box with the one blue marble in it, the chances of u getting saved stay the same, but curiosity could help.
2006-10-25 16:25:15 · answer #3 · answered by ysomebody 2 · 0⤊ 0⤋
I've got question #2:
Round 3 - 3 killed
Round 2 - 3x2 = 6 killed
Round 1 - 6x2 =12 killed
Total killed = 3+6+12 = 21
Total left: 97-21 = 76
2006-10-25 16:26:43 · answer #4 · answered by whabtbob 6 · 0⤊ 0⤋
I can't think of an answer for the first one, but the second one is pretty simple. she killed 3 in the third round, 6 in the second round, and 12 in the first round. 12+6+3=21. 97-21=76
2006-10-25 16:22:22 · answer #5 · answered by tamana 3 · 0⤊ 0⤋
as far as the birds are concerned she killed 3 with the 3rd shot, 2x3 with the 2nd shot (6) and 2x6 with the first shot (12), so she kills a total of 21 birds. Now being a bird hunter I would venture the assumption that there are no birds left in the tree after she fired the 3rd round. They all flew away.
2006-10-25 16:21:06 · answer #6 · answered by tigger_32_kitty_27 2 · 1⤊ 0⤋
The box picked is a 50-50 proposition over which you have no control. I think you should make each box a 50-50 proposition.
The rounds are R1, R2, R3
R3 = 3
R2 = 2R3 = 2(3) = 6
R1 = 2R2 = 2(6) = 12
21 kills
76 remaining
2006-10-25 16:27:04 · answer #7 · answered by kindricko 7 · 0⤊ 0⤋
put exactly half of each color marble in each box, that way you have a 50% chance to live in either box he chooses.
2006-10-25 16:16:37 · answer #8 · answered by beans 1 · 0⤊ 0⤋