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i_online BizHat Newbie

Joined: 11 May 2006 Posts: 3
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Posted: Thu May 11, 2006 12:14 pm Post subject: Puzzles for you |
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solve this puzzle:
In a country in which people only want boys, every family continues to have children until they have a boy. If they have a girl, they have another child. If they have a boy, they stop. What is the proportion of boys to girls in the country? Assume these are infinite families in the country?
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http://neerajsingh.bizhat BizHat Addict

Joined: 19 Aug 2005 Posts: 403 Location: Bangalore-India
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Posted: Thu May 11, 2006 2:29 pm Post subject: |
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If we consider, n as the total number of families,
Then the ration of number of boys to girls will always be 2:n, where n is the number of family and n>=1
Thanks |
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dsmm BizHat Newbie

Joined: 21 May 2006 Posts: 46 Location: IL
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Posted: Sun May 21, 2006 11:44 pm Post subject: |
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i'm sorry but i'm not very good with math(also with English)
so is his answer were right?
it's seems to be logical .
any way this is a great puzzle, do you have one that not involved with math? |
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indianbaba BizHat MOD

Joined: 21 Aug 2005 Posts: 5505 Location: India
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Posted: Wed May 24, 2006 8:14 am Post subject: |
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nice puzzle.
Keep sharing. |
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http://neerajsingh.bizhat BizHat Addict

Joined: 19 Aug 2005 Posts: 403 Location: Bangalore-India
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Posted: Wed May 24, 2006 9:42 am Post subject: |
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Multiply of 37 by multiplies of 3:
3 x 37 = 111
6 x 37 = 222
9 x 37 = 333
12 x 37 = 444
15 x 37 = 555
18 x 37 = 666
21 x 37 = 777
24 x 37 = 888
27 x 37 = 999
Nice one:
111.111.111 x 111.111.111 =
12.345.678.987.654.321
Trapeze:
1 x 9 + 2 = 11
12 x 9 + 3 = 111
123 x 9 + 4 = 1111
1234 x 9 + 5 = 11111
12345 x 9 + 6 = 111111
123456 x 9 + 7 = 1111111
1234567 x 9 + 8 = 11111111
12345678 x 9 + 9 = 111111111
another Trapeze:
1x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321
and another one:
0x 9 + 8 = 8
9 x 9 + 7 = 88
98 x 9 + 6 = 888
987 x 9 + 5 = 8888
9876 x 9 + 4 = 88888
98765 x 9 + 3 = 888888
987654 x 9 + 2 = 8888888
9876543 x 9 + 1 = 88888888
98765432 x 9 + 0 = 888888888
987654321 x 9 - 1 = 8888888888
9876543210 x 9 - 2 = 88888888888 |
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indianbaba BizHat MOD

Joined: 21 Aug 2005 Posts: 5505 Location: India
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Posted: Wed May 24, 2006 10:16 am Post subject: |
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Nice puzzle.
Keep going neerajsingh. |
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indianbaba BizHat MOD

Joined: 21 Aug 2005 Posts: 5505 Location: India
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Posted: Wed May 24, 2006 11:14 am Post subject: |
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| i_online wrote: | Two MIT math grads bump into each other at Fairway on the upper west side. they haven't seen each other in over 20 years.
The first grad says to the second: "how have you been?"
Second: "great! I got married and i have three daughters now"
First: "really? How old are they?"
Second: "well, the product of their ages is 72, and the sum of their ages is the same as the number on that building over there."
First: "right, ok.. Oh wait! Hmm, I still don't know"
Second: "oh sorry, the oldest one just started to play the piano"
First: "wonderful! My oldest is the same age!"
How old are the daughters? |
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indianbaba BizHat MOD

Joined: 21 Aug 2005 Posts: 5505 Location: India
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Posted: Wed May 24, 2006 11:15 am Post subject: |
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| http://neerajsingh.bizhat wrote: | Well the ages of the daughters are 8 years, 3 years and 3 years
My approach:
1- First factor 72 in three parts.
2- Now we need to consider that one where the sum is something that is on the building.
3- As the second mathmatician got confused, that means there are two factors whose sum equal the same. They are 8*3*3 and 2*6*6. Both equal to 14.
4- Now he needed one more hint to chose one of them. So when the first mathmatician about her Eldest daughter, that means it can be only 8*3*3, because in other case, there is a draw.
Hnce the answer. Thanks for posting the puzzle dear, what a nice way to start your posting in a forum.
Wish you a very wonderful time here among us Bizhatians. |
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indianbaba BizHat MOD

Joined: 21 Aug 2005 Posts: 5505 Location: India
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Posted: Wed May 24, 2006 11:16 am Post subject: |
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| i_online wrote: | | There are three closed and opaque cardboard boxes. One is labeled "APPLES", another is labeled "ORANGES", and the last is labeled "APPLES AND ORANGES". You know that the labels are currently misarranged, such that no box is correctly labeled. You would like to correctly rearrange these labels. To accomplish this, you may draw only one fruit from one of the boxes. Which box do you choose, and how do you then proceed to rearrange the labels? |
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indianbaba BizHat MOD

Joined: 21 Aug 2005 Posts: 5505 Location: India
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Posted: Wed May 24, 2006 11:17 am Post subject: |
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| http://neerajsingh.bizhat wrote: | 1- We will choose the box named "APPLES AND ORANGES" and pick one fruit.
2.1- If the drawn fruit is Apple, then we would rearrange the labels in the following manner.
2.1.1 - Move the label "APPLES" to the "APPLES AND ORABGES".
2.1.2 - Move the label "APPLES AND ORANGES" to "ORANGES"
2.1.3 - Move the label "ORANGES" to "APPLES"
2.2- If the drawn fruit is Orange, then we would rearrange the labels in the following manner.
2.2.1 - Move the label "ORANGES" to the "APPLES AND ORABGES".
2.2.2 - Move the label "APPLES AND ORANGES" to "APPLES"
2.2.3 - Move the label "APPLES" to "ORANGES" |
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indianbaba BizHat MOD

Joined: 21 Aug 2005 Posts: 5505 Location: India
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Posted: Wed May 24, 2006 11:19 am Post subject: |
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This thread made into Sticky. If it is not updated daily it will be removed.
@i_online, in view of making this thread sticky two of your puzzles have been added here and your original posts are moved.
Cheers
indianbaba. |
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http://neerajsingh.bizhat BizHat Addict

Joined: 19 Aug 2005 Posts: 403 Location: Bangalore-India
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Posted: Wed May 24, 2006 12:26 pm Post subject: |
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Thanks indianbaba for making this thread sticky. It would surely help others.
Finger Multiplication
by Sidney Kolpas
Mathematics Teacher April 2002
It requires that the students hold both hands up, with palms facing the student. The student counts his or her thumbs and fingers consecutively
from left to right, with the thumb on the left hand representing the number 1 and the thumb on the right hand representing the number 10. To multiply n times 9, where 1 <= n <= 10 and where n is a whole number, the student bends down the nth finger. The number of fingers to the left of the bent finger represents the tens-place digit of the product, whereas the number of fingers to the right of the bent finger represents the ones-place digit of the product.
The figure below illustrates 7 x 9.
Mr. Kolpas explains why this works in his article in the April 2002 issue of the Mathematics Teacher.
Courtesy : http://www.pen.k12.va.us
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indianbaba BizHat MOD

Joined: 21 Aug 2005 Posts: 5505 Location: India
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Posted: Wed May 24, 2006 12:38 pm Post subject: |
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i never knew that.
That was something marvellous.
Keep sharing. |
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http://neerajsingh.bizhat BizHat Addict

Joined: 19 Aug 2005 Posts: 403 Location: Bangalore-India
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Posted: Wed May 24, 2006 12:39 pm Post subject: |
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Finger Multiplication by Nine
by Paul Horrell
Mathematics Teacher February 2003
The "finger folding" multiplication by 9 can be expanded to include multiplicands up to 99.
| Code: | The algorithm used is similar to the following:
To multiply a two-digit number by 9,
ENTER the multiplicand by forming a V between the tens-digit finger and the next finger to its right (SPREAD).
ENTER the units digit by FOLDing the units-digit finger. |
| Code: | To read the product
Hundreds digit: Count the standing fingers before the SPREAD.
Tens digit: Moving from left to right, count the fingers from the SPREAD to the FOLD; wrap around if necessary.
Units digit: Count the standing fingers to the right of the FOLD. |
| Code: | Example:
48x9
Enter (1 2 3 4 _ 5 6 7 F 9 0)
Read:
Hundreds (1 2 3 4) 4
Tens (5 6 7) 3
Units (9 0) 2
Example with wraparound:
83 x 9
Enter (1 2 F 4 5 6 7 8 _ 9 0)
Read:
Hundreds (1245 6 78) 7
Tens (9 0 1 2) 4
Units (45 6 7890) 7 |
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http://neerajsingh.bizhat BizHat Addict

Joined: 19 Aug 2005 Posts: 403 Location: Bangalore-India
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Posted: Fri May 26, 2006 9:18 am Post subject: |
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Climbing Snail
A snail is at the bottom of a 20 meters deep pit. Every day the snail climbs 5 meters upwards, but at night it slides 4 meters back downwards.
The Question: How many days does it take before the snail reaches the top of the pit? |
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