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TEST UR MATH!!!

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Post by rica Sat Jun 06, 2009 2:28 am

First topic message reminder :

mga ka cg pips....patunayan ntin na ang mga artist ay magaling din sa math!!!!
common mind buggling problems lng po...para maiba nman.. post ntin lahat dito... thumbsup umpisahan ko na...hehhe
kumbaga, mind exercise lng para magkaron din tau ng kakaibang exercises....weeeeeee


ok....game nah...hehehe

Problem no. 1


There is a bus with 7 girls .

Each girl has 7 bags .

In each bag, there are 7 big cats

Each big cat has 7 little cats.

Each cat has 4 legs .

Question: How many legs are present in the bus?
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Post by Chx Tue Jun 09, 2009 2:51 am

oops.. sorry iba na pala yung tanong.. :p

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Post by mammoo_03 Tue Jun 09, 2009 5:03 am

ang daming mathi thinik dito ah, galing. salamat sa thread mam rica.
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Post by Guest Tue Jun 09, 2009 5:08 am

kietsmark wrote:
KettleRenderer wrote:ang tanong .. naintindihan mo ba un isinagot mo bago ni-copy-paste??? ??? peace man

hindi... saka wala akong pakialam hehehe Twisted Evil tapos na ko sa school e Crazy_Lion

haaaaaaayyy.... akala ko talaga isa kang alamat kietsmark.. nainggit tuloy ako.. binalak ko na talaga magpaka dalubhasa sa mga numero ngaun araw..... pero binasag mo ang binalak ko maghapon ... pero ok lang un .. peace man

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Post by Guest Tue Jun 09, 2009 7:40 am

KettleRenderer wrote:haaaaaaayyy.... akala ko talaga isa kang alamat kietsmark.. nainggit tuloy ako.. binalak ko na talaga magpaka dalubhasa sa mga numero ngaun araw..... pero binasag mo ang binalak ko maghapon ... pero ok lang un .. peace man

sampay ako'y magaling lang sa paglalaba wala ng iba Twisted Evil
to rica... thanks sa thread thumbsup

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Post by nheil29 Fri Jun 12, 2009 3:16 am

about the legs... i thnk its just 9,618
simple as [7(ladies)x7(bags)x7cats x7kittens] x 4legs ..tapos add ka lng 14 sa legs ng 7girls..tama ba????????
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Post by Guest Fri Jun 12, 2009 3:43 am

nheil29 wrote:about the legs... i thnk its just 9,618
simple as [7(ladies)x7(bags)x7cats x7kittens] x 4legs ..tapos add ka lng 14 sa legs ng 7girls..tama ba????????

7 girls = 14 legs
343 big cats = 1372 legs (7 bloody bags each bloody girl x 7 bloody girl)
2401 little cats = 9604 legs
= 10990 bloody legs

total of 10990 bloody legs .. sounds wrong noh...
i wonder how big are the girls that can carry 392 bloody cats and kittens..

medyo magulo kce ang computation... bloody stressful..

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Post by rica Fri Jun 12, 2009 6:48 am

nheil29 wrote:about the legs... i thnk its just 9,618
simple as [7(ladies)x7(bags)x7cats x7kittens] x 4legs ..tapos add ka lng 14 sa legs ng 7girls..tama ba????????

Soution number 1:

7 Small cats = 7 * 4 = 28
1 Big cat = 28 + 4 = 32
7 Big cats = 32 * 7 = 224
1 Bag = 224
7 Bags = 224 * 7 = 1568
1 Girl = 1568 + 2 = 1570
7 Girls = 1570 * 7 = 10990


Solution number two: (eto mas clear po) 2thumbsup

7 ladies * 7 bags(each lady) = 49 (total no. of bags)
49 (bags) * 7 big cats (in each bag) = 343 (total no. of big cats)
343 (big cats) * 7 small cats (with each big cat) = 2401 (total no. of small cats)
2401 (total no. of small cats) + 343 (total number of big cats) = 2744 (total number of cats)
2744 (total no. of CATS) * 4 legs (each cat has 4 legs right?) = 10976 (total number of legs of all cats given)
then lastly add the 14 legs (total number of legs for the 7 ladies) with 10976 (total number of legs of all cats given)

the total is 10, 990 legs....hope its all clear sir and mam.....

sir nheil29, nkalimutan nyo po i-add ung number of kittens sa number of cats...kya po gnun ung lmbas na answer...
thanks for posting po sir nheil...i know napaicip kau ei, kaya kau ng-react ng husto...hehehe..mwahnezz!!

2thumbsup 2thumbsup 2thumbsup

wait for my next problem, mapapaicip ulit kau...hehehe


Last edited by rica on Fri Jun 12, 2009 6:55 am; edited 1 time in total
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Post by rica Fri Jun 12, 2009 6:52 am

KettleRenderer wrote:
nheil29 wrote:about the legs... i thnk its just 9,618
simple as [7(ladies)x7(bags)x7cats x7kittens] x 4legs ..tapos add ka lng 14 sa legs ng 7girls..tama ba????????

7 girls = 14 legs
343 big cats = 1372 legs (7 bloody bags each bloody girl x 7 bloody girl)
2401 little cats = 9604 legs
= 10990 bloody legs

total of 10990 bloody legs .. sounds wrong noh...
i wonder how big are the girls that can carry 392 bloody cats and kittens..

medyo magulo kce ang computation... bloody stressful..

hope the second solution po will make it all clear madam kettle....
thanks for posting po...para mas ma clear ko po ung solution sa lahat....thanks ...mwah!!! thumbsup thumbsup 2thumbsup
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Post by Guest Fri Jun 12, 2009 11:25 am

Rica, for me, my solution is clearer. . .i just made it easier for me. .just for me. . .
Tama na. . Tama na please. . Lets stop counting the bloody legs of those bloody cats. .
I just cant imagine how big are those bloody gurls that can cary 49 bloody cats and 343 kittens.. (392 felines)

Iba naman ang bilangin natin.. Lalake naman. . Joke! . Or pera. Para malalaman natin ang hindi pwde utangan!! . . Next na!

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Post by rica Fri Jun 12, 2009 10:29 pm

KettleRenderer wrote:Rica, for me, my solution is clearer. . .i just made it easier for me. .just for me. . .
Tama na. . Tama na please. . Lets stop counting the bloody legs of those bloody cats. .
I just cant imagine how big are those bloody gurls that can cary 49 bloody cats and 343 kittens.. (392 felines)

Iba naman ang bilangin natin.. Lalake naman. . Joke! . Or pera. Para malalaman natin ang hindi pwde utangan!! . . Next na!

hehehe...very good idea mam kettle.....cge po, ill take ur advice....mag-iicip ako ng problem na my involve na boys and money...hehehe..ok?thanks for suggesting...mwahnezz!!!! 2thumbsup 2thumbsup hippie
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Post by destijl_art Sat Jun 13, 2009 2:52 am

rica wrote:
nheil29 wrote:about the legs... i thnk its just 9,618
simple as [7(ladies)x7(bags)x7cats x7kittens] x 4legs ..tapos add ka lng 14 sa legs ng 7girls..tama ba????????

Soution number 1:

7 Small cats = 7 * 4 = 28
1 Big cat = 28 + 4 = 32
7 Big cats = 32 * 7 = 224


Solution number two: (eto mas clear po) 2thumbsup

7 ladies * 7 bags(each lady) = 49 (total no. of bags)
49 (bags) * 7 big cats (in each bag) = 343 (total no. of big cats)
343 (big cats) * 7 small cats (with each big cat) = 2401 (total no. of small cats)
2401 (total no. of small cats) + 343 (total number of big cats) = 2744 (total number of cats)
2744 (total no. of CATS) * 4 legs (each cat has 4 legs right?) = 10976 (total number of legs of all cats given)
then lastly add the 14 legs (total number of legs for the 7 ladies) with 10976 (total number of legs of all cats given)

the total is 10, 990 legs....hope its all clear sir and mam.....

sir nheil29, nkalimutan nyo po i-add ung number of kittens sa number of cats...kya po gnun ung lmbas na answer...
thanks for posting po sir nheil...i know napaicip kau ei, kaya kau ng-react ng husto...hehehe..mwahnezz!!

2thumbsup 2thumbsup 2thumbsup

wait for my next problem, mapapaicip ulit kau...hehehe

Tanong lang po s magaling s math,

Db po eh 1-bag = 224
at meron po tyong 7 babae n may bag (224 legs each bag x 7 bags =1568
so db dapat po idadagdag nlang ntin ung legs nung 7 women (7 women x 2 legs = 14) + 1,568 legs = 1,582
kc s tingin ko po kung multiply p nang 7 ung 1,568 lumalabas n bawat isang babe eh may 7 bags n hawak d po b?
as shown at the figures below:
1 Bag = 224
7 Bags = 224 * 7 = 1568
1 Girl = 1568 + 2 = 1570
7 Girls = 1570 * 7 = 10990
............. eto po dapat ang wala n kc n multiply n s 7 ung bag dun p lang s pangalawang line.

skit s ulo he he he.
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Post by rica Sun Jun 21, 2009 7:58 am

at last im back!!!hehehe
grabeh ntgalan ako bgo nkapagpost ulit, ngka problem kc sa pc ei..anyway
e2 na po ang susunod na problem..hehe

There are Five masters and their five dogs that needs to cross a river. There is a
boat that fits three beings at a time to cross it. Each dog has very
little training and will bite another human if its master is not
with it. So you cannot have a dog around other humans without its
master. But you can have the dog around other dogs without its master.
In order to cross the river you need to have one human in the boat to
row, because dogs cannot row. Except for Wilbert the wonder dog. He is
a special dog and can row the boat. How do you get all the dogs and
masters across the river without anyone getting bitten?


hope ull solve and enjoy..hehehe
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Post by corpsegrinder Sun Jun 21, 2009 8:52 pm

rica wrote:at last im back!!!hehehe
grabeh ntgalan ako bgo nkapagpost ulit, ngka problem kc sa pc ei..anyway
e2 na po ang susunod na problem..hehe

There are Five masters and their five dogs that needs to cross a river. There is a
boat that fits three beings at a time to cross it. Each dog has very
little training and will bite another human if its master is not
with it. So you cannot have a dog around other humans without its
master. But you can have the dog around other dogs without its master.
In order to cross the river you need to have one human in the boat to
row, because dogs cannot row. Except for Wilbert the wonder dog. He is
a special dog and can row the boat. How do you get all the dogs and
masters across the river without anyone getting bitten?


hope ull solve and enjoy..hehehe

Pwd true or false nlng?
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Post by pixelburn Mon Jun 22, 2009 4:17 pm

rica wrote:at last im back!!!hehehe
grabeh ntgalan ako bgo nkapagpost ulit, ngka problem kc sa pc ei..anyway
e2 na po ang susunod na problem..hehe

There are Five masters and their five dogs that needs to cross a river. There is a
boat that fits three beings at a time to cross it. Each dog has very
little training and will bite another human if its master is not
with it. So you cannot have a dog around other humans without its
master. But you can have the dog around other dogs without its master.
In order to cross the river you need to have one human in the boat to
row, because dogs cannot row. Except for Wilbert the wonder dog. He is
a special dog and can row the boat. How do you get all the dogs and
masters across the river without anyone getting bitten?


hope ull solve and enjoy..hehehe

Given;

Aa, Bb, Cc, Dd, Ww

(capital letter ay "amo",,,,, small letter ay "aso")

w = ay si wilbert
***********************************************************************************

Where;

P1 = "pinanggalingang pangpang"

P2 = "patutunguhang pangpang"

***********************************************************************************

Solution

1. ihahatid ni wilbert ung tatlong aso sa P2 (dalawang byahe ang gagawin ni wibert pabalikbalik)

A, B, C, Dd, Ww .......>>>> a, b , c ,

---------------------------------------------------------------------------------------------------------------------

2. tapos ung tatlong mayari nung aso na nasa P1, ang babyahe naman papunta P2 .

Ww, Dd .......>>>>>> Aa, Bb, Cc

---------------------------------------------------------------------------------------------------------------------

3. tapos ung isang pares ng magamong aso na nasa P2 ay babalik sa P1 .

Ww, Dd , Aa <<<<<........Bb, Cc

-------------------------------------------------------------------------------------------------------------- ------

4. babyahe si wilbert at ang kanyang amo papunta sa P2.

Dd, Aa ..........>>>> Ww , Bb , Cc

--------------------------------------------------------------------------------------------------------------------

5. babayahe naman ang isang pares ng mag-among aso pabalik ng P1.

Dd , Aa , Bb <<<<<......... Ww , Cc

--------------------------------------------------------------------------------------------------------------------

6. babayahe naman ang tatlong amo papuntang P2.

d , a , b , .............>>>>>>> D , A , B, Ww , Cc ,

--------------------------------------------------------------------------------------------------------------------

7. at susunduin naman ni wilbert ang natitirang aso papuntang P2. (dalawang byahe ang gagawin nya pabalik balik)

d, .............>>>> w , a , b , >>>>.......... D , A , B, W, Cc


Last edited by pixelburn on Mon Jun 22, 2009 5:12 pm; edited 1 time in total
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Post by Guest Mon Jun 22, 2009 4:53 pm

rica wrote:at last im back!!!hehehe
grabeh ntgalan ako bgo nkapagpost ulit, ngka problem kc sa pc ei..anyway
e2 na po ang susunod na problem..hehe

There are Five masters and their five dogs that needs to cross a river. There is a
boat that fits three beings at a time to cross it. Each dog has very
little training and will bite another human if its master is not
with it. So you cannot have a dog around other humans without its
master. But you can have the dog around other dogs without its master.
In order to cross the river you need to have one human in the boat to
row, because dogs cannot row. Except for Wilbert the wonder dog. He is
a special dog and can row the boat. How do you get all the dogs and
masters across the river without anyone getting bitten?


hope ull solve and enjoy..hehehe

nabasa ko na to kay Dr. Math dati hehehe abang mode na lang ako sa sasagot Very Happy

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Post by rica Sat Jun 27, 2009 9:13 am

hehehe...uu nga sir kietsmark..dun ko tlga nkuha yan ei..hehehe

well, to sir pixelburn....parang alam ko kng san mo knuha ung answer mu...hehe
anyway..thanks po sa pagpost ng answer mu ha..galing!!! ;-)
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Post by JAKE Sat Jun 27, 2009 11:22 am

wala na ba? ung medyo mahirap naman! peace man
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Post by rica Sun Jun 28, 2009 7:15 am

JAKE wrote:wala na ba? ung medyo mahirap naman! peace man

wait lng po sir..nagreresearch pa aku ei..hehehe peace man
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Post by bakugan Sun Jun 28, 2009 7:49 am

fourteen legs (14) po yata. hehe
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Post by pixelburn Mon Jun 29, 2009 2:08 am

rica wrote:
JAKE wrote:wala na ba? ung medyo mahirap naman! peace man

wait lng po sir..nagreresearch pa aku ei..hehehe peace man

abangan po namin mam rica... bounce
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