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 Post subject: Problem solving: heads and legsPosted: Sun Jan 16, 2011 4:44 pm

Joined: Thu Apr 08, 2010 10:28 pm
Posts: 95
Location: London
Please could someone help me on the following question.

In a farm there are 13 heads and 40 legs.

a) How many ducks are there ?

b) How many sheep's are there ?

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 Post subject: Re: Problem SolvingPosted: Sun Jan 16, 2011 4:52 pm

Joined: Fri Oct 12, 2007 12:42 pm
Posts: 3826
This is done most easily by trial and error.

Sheep have 4 legs so try a multiple of 4. 8 times 4 = 32.
40 - 32 = 8 legs for the ducks, which makes 4 ducks.
8+4 = 12 heads so wrong and try again.

As there were not enough heads you will need more ducks and fewer sheep.

7 times 4 = 28 legs (7 sheep with four legs)
40 - 28 = 12 legs for ducks, 6 ducks

7 sheep and 6 ducks = 13 heads.

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 Post subject: Re: Problem SolvingPosted: Sun Jan 16, 2011 5:30 pm

Joined: Fri Nov 17, 2006 8:54 pm
Posts: 1776
Location: caversham
Tulip wrote:
Please could someone help me on the following question.

In a farm there are 13 heads and 40 legs.

a) How many ducks are there ?

b) How many sheep's are there ?

I would construct a table of sheep, ducks, heads and legs!

There must be 13 animals as 13 heads.

Where to start well 13 can be split in half as 6 and 7 (no half animals!)

Looks like more sheep than ducks because lots of legs........this is a lucky guess.

7 sheep (4*7=28 legs) 6 ducks (6*2= legs), 13 heads, 40 legs!! Bingo.

I think that was more luck than judgement, expected to try a few variations of ducks and sheep. I expect the question was written so if an 11 year old tried they would get the answer.

For older kids you could use algebra and simultaneous equations.

S + D = 13.................................heads (1)
4S + 2D = 40...............................legs (2)

S=13-D rearaange (1) and substitue into (2)

4(13-D)+2D=40

52-4D+2D=40

52-2D=40

12=2D

D=6

then feed back into (1)

S+6=13

S=7

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 Post subject: Re: Problem SolvingPosted: Sun Jan 16, 2011 9:06 pm
I agree. I wouldn't go into algebra with a 10 year old but do it by trial and error. Too many parents try to teach their children the algebraic method of solving problems which is likely to be a) more error prone and b) more time consuming that a simple table of trial and error.

Try to remember they are only 10 and, even if they bomb through a test (which is not always the best idea), they probably will have, at most, 3 minutes to solve something like this.

So, rule of thumb, if a method is taking an ordinate time to work out, then it is probably not the smartest one.

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 Post subject: Re: Problem SolvingPosted: Mon Jan 17, 2011 1:56 pm

Joined: Thu Apr 08, 2010 10:28 pm
Posts: 95
Location: London

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 Post subject: Re: Problem SolvingPosted: Tue Jan 18, 2011 9:43 am

Joined: Sat Dec 06, 2008 1:41 pm
Posts: 154
This is how I think it can be done:

13 animals 40 legs. Some are ducks and some are sheep.

Ducks have 2 legs. Sheep have 4 legs.

1 duck has 2 legs. 1 sheep has 4 legs (1 x 4)

2 ducks have 4 legs. 2 sheep have 8 legs (2 x 4)

3 x 2 = 6 3 x 4 = 12

4 x 2 = 8 4 x 4 = 16

5 x 2 = 10 5 x 4 = 20

6 x 2 = 12 6 x 4 = 24

7 x 2 = 14 7 x 4 = 28

8 x 2 = 16 8 x 4 = 32

9 x 2 = 18 9 x 4 = 36

10 x 2 = 20 10 x 4 = 40

Since you could go on writing the times tables forever, use the number 40 legs to help you. You do not want to write timestables on both columns hich will add up to more than 40 because the number of animals on both sides which will eventually give you 40 legs are all you need (hope this makes sense).

Now find pairs of numbers from the sheep and duck columns that add up to 40, since you are looking for 40 legs.

The first pairing is 4 ducks and 8 sheep (12 legs from ducks and 32 legs from sheep = 40 legs) which will give you 12 animals. That is not the answer you want since the question says 13 animals.

You will also find 10 ducks and 5 sheep (20 legs from ducks and 20 legs from sheep = 40 legs) - that is still not the answer.

The pairing is 6 ducks and 7 sheep (12 legs from ducks and 28 legs from sheep = 40 legs). This is the answer because you get 13 animals.

Therefore the farmer has 6 ducks and 7 sheep.

Similar questions include:

Tammy has 33 straws and she makes rectangles and pentagons from these straws. She makes more rectangles than she does pentagons. How many of each does she make with her straws?

Or

Sir Dancealot is throwing a party. He can use round tables that will seat 5 guests on each table or rectangle tables that will seat 8 guests. What is the least number of tables Sir Dancealot will need to seat his 35 guests and himself. This question arose in the junior maths challenge one year.

Lengthy answer but I hope this helps.

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 Post subject: Re: Problem SolvingPosted: Tue Jan 18, 2011 3:20 pm

If they were all sheep, there would be 52 legs ( 4x`13).

This is 12 legs too many. You would get rid of 12 legs by swapping 6 sheep for 6 ducks.

This leaves you with 7 sheep and 6 ducks.

While not all children might grasp the above method, it is probably the quickest.

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 Post subject: Re: Problem SolvingPosted: Thu Jan 20, 2011 11:18 pm

Joined: Sat Aug 21, 2010 4:53 pm
Posts: 21
I believe equations with 2 variables works best, but to save time would go quicky scan 4 options from multiple choice answer, doing simple calculation in mind -> 4*sheep+2*duck would do the trick.. smartest/fastest and easiest way of doing it.

Regards
jmom

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