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Thread ID: 95362	2008-12-03 09:15:00	Maths problem	bk T (215)	PC World Chat

Post ID	Timestamp	Content	User
725452	2008-12-03 10:49:00	How do I solve this problem: Greg puts his pet cats in a box to take them to his auntie's house. How many cats are there in the box: a) If there were 21 more legs than tails? b) If there were 14 more legs than eyes? My daughter brought this home from school expecting me to help her out. But I can't solve it other than by trial and error. I know the answer is 7 for both the questions but don't really know how to show her the work out. Can some wise person here explain the ways to solve problems of this nature? The most effective way to approach these types of problems is by simultaneous equations. For both parts, let c = the number of cats. For (a), you have 4c = the number of legs from the cats, and c = the number of tails. Thus, 4c - c = 21 3c = 21 c = 7. For (b), similarly 4c = the number of legs, and 2c = the number of eyes. Thus, 4c - 2c = 14 2c = 14 c = 7. Metla's suggestion to simply draw a picture is also another useful way to approach a problem like this. Since I can't be ****ed actually drawing a picture, I'll just use slashes to represent legs and hyphens to represent tails. In order to solve this problem the approach is similar to that provided by James, but easier to follow in my opinion. All you do is draw cats until the difference is that which is required. /\ /\ - (this is a cat btw, 4 legs and a tail) /\ /\ - 1 cat gives a difference of 3, /\ /\ - /\ /\ - 2 cats gives a difference of 6, /\ /\ - /\ /\ - /\ /\ - 3 cats gives a difference of 9, ... /\ /\ - /\ /\ - /\ /\ - /\ /\ - /\ /\ - /\ /\ - /\ /\ - 7 cats gives the required difference of 21. Hope this helps /maths teacher post.	roddy_boy (4115)
725453	2008-12-03 10:54:00	Just to add, the approach one would use for this type of problem in school seems to change with age. Younger children probably haven't been introduced to the concept of simultaneous equations, so using this approach might not be the best method. If your daughter is older though this would probably be the best way to teach it. If she is older than 16, can I have her number?	roddy_boy (4115)
725454	2008-12-03 11:01:00	I've just realised it doesn't look like I've used simultaneous equations here. I skipped steps which just seem obvious but are essential to leave in if you're not familiar with forming equations to fit a problem. I'll redo the beginning of part (a) for clarification. Let c = the number of cats, Let t = the number of tails, Let l = the number of legs. From the problem we can form the equation l - t = 21 However, from common knowledge we know that l = 4c (there are 4 times as many legs as there are cats - 4 legs per cat) and t = c (each cat has only 1 tail so they are equivalent for this problem) substituting l for 4c and t for c gives 4c - c = 21 and the rest follows as above.	roddy_boy (4115)
725455	2008-12-03 11:03:00	^ Or you could just ask Greg's Aunty...	--Wolf-- (128)
725456	2008-12-03 21:00:00	I thank you all for the interesting inputs. My brain is quite rusty after leaving school for so many donkey years. Without all your assistance I certainly will have a hard time facing my daughter! Cheers	bk T (215)
725457	2008-12-03 21:02:00	Good lord, I didn't know roddy_boy could be so mature!	ubergeek85 (131)
725458	2008-12-03 22:05:00	Good lord, I didn't know roddy_boy could be so mature! Me neither. :lol:	jwil1 (65)
725459	2008-12-03 22:14:00	Good lord, I didn't know roddy_boy could be so mature! Someone (not 'somebody') has hijacked his login....	johcar (6283)
725460	2008-12-04 00:15:00	So if I make a useful post around here I still get ****? There's just no winning with you guys.	roddy_boy (4115)
725461	2008-12-04 04:34:00	So if I make a useful post around here I still get ****? There's just no winning with you guys. Is that an admission that many of his previous posts were not useful?:horrified	Roscoe (6288)
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