Maths Homework

[Clarry]

Hi folks, my daughter asked me for help with her maths homework but a few of the questions stumped me. I never was very good at maths myself, but I do know there are some very knowledgeable people on this forum so I'm asking for some help.
Q1. (Place Value) Suppose we were allowed to put more than a single digit in each place column. What are the different numbers that could be represented by 2134?

Q2. (Perimeter Possibilities) Using just whole numbers, how many rectangles can you draw that have a perimeter of 36 units? Is there a short way of finding out the number of rectangles you can draw with any even perimeter?

Q3. (The liquid Problem) Suppose you have a large vat of liquid.
(a) You have a 3lt measure and a 5lt measure. How can you take 1 litre from the vat?
(b) You have a 10lt measure and a 7lt measure. How can you take 9 litres of liquid from the vat?

Any help is greatly appreciated.

[MrB]

Q3 (a) Fill the 3L jug, tip into the 5L jug.. Fill the 3L jug again and tip into 5L jug untill it is full... 1L remains in the 3L jug.

[Clarry]

Of coarse, this is why I'm a chef and not a rocket scientist. Cheers.

[OICURMT]

Q1) N = n!
Q2) N = C/4

I think this is correct...

[MrB]

Bah, (a) came to me in seconds, had to have some dinner and a smoke to nut-out Q3 (b) haha

Fill the 10L jug and tip it into 7L jug untill full.
3L remains in the 10L jug.
Empty the 7L jug back into the vat.
Transfer the 3L that remains in the 10L jug into the 7L jug.

Fill the 10L jug again, tip it onto the 3L in the 7L jug untill full.
6L remains in the 10L jug.
Empty the 7L jug into the vat again.
Transfer the 6L that remains in the 10L jug into the 7L jug.

Fill the 10L jug yet again, tip onto the 6L in the 7L jug untill full .
9L remains in the 10L jug.

[Clarry]

Thanks MrB, this is no simple equation but you got it. Both parts would be easier if you just had a 1 lt jug to start with.

Sorry OICURMT but those answers leave me more confused than the question.

[MrB]

Equations? hmmm
Best I can come up with are:
(a) 3-(5-3) = 1
(b) 10-(7-(10-(7-(10-7)))) = 9

But they are both very ugly and don't realy correlate with what is happening with the measures(jugs).. I'd go with the descriptions.

[OICURMT]

Sorry clarification...


Q1) N = n! (factorial) = 1*2*3*4 = 24 permutations...
Q2) N = C/4 = 36/4 = 9 possible unique combination's limited to integers

[sheeny]

Q1 = 10, 22, 37, 64, 217, 334, 2134 = 7 (eg 10 = 2+1+3+4, 334 = (2+1)x100+3x10+4)

Q2 = 2x16, 3x15, 4x14, 5x13, 6x12, 7x11, 8x10 = 7 = 36/4-2 = n/4-2 (eg perimeter divided by number of sides misus the 2 solutions where the rectangle is a square or a straight line). n/4-1 if you allow a square as a rectangle.

Q3a =

Pour 3L into the 5L measure. Fill the 3L measure and use this to fill the 5L measure. 1L is left in the 3L measure. Pour the 5L back in the vat.

Q3b =
Pour 7L into the 10L measure. Fill the 7L measure and use this to fill the 10L measure leaving 4L in the 7L measure. Pour the 10L measure back in the vat. Pour the 4L into the 10L measure and fill the 7L measure. Fill the 10L meaasure from the 7L measure leaving 1L in the 7L measure. Pour the 10L measure back in the vat.

[Clarry]

Thanks folks, I knew this was the place to come for answers. Now if you can just tell me the meaning of life I can rest easy.

[wasyoungonce]

I was faced with similar issues a few years back..went back and did VCE (Vic year 11/12) maths methods, spec maths, physics, chemistry etc..etc by distance education.

This was so I could astound and assist my Kids with my knowledge.;)

My master plan all went to pieces...kids would have a bar of me helping them...all ended in tears!:shrug:

Good kids gone bad I say!:sadeyes:

[renormalised]

42:)

[DavidU]

I wish IIS was around when I was at school !
Nice work fellas.

[Allan_L]

Hey Sheeny
What about 1x17?
(a rectangle 1 metre by 17 metres is not a straight line).
[the straight line only comes into contention where "units" equals pixels]

And a Square is ALWAYS a rectangle. Opposite sides equal and all angles 90 degrees. (I doubled checked definitions before going into print on this one)

So the formula is n/4 (and ... always rounded DOWN)

and I don't understand Q1 at all ???

[wasyoungonce]

Permutations. If you had a 4 digit number and wrote that number into 4 blank boxes (one at a time, reducing the selections left by not re-using the same digit) how many way can it be written.

ie the number is 2134. We can write combinations or permutations of this as:
Any of the 4 digits x any of the 3 digits left x any of the 2 digits left x any of the 1 digit left = 4 x 3 x 2 x 1 = 24 ways of writing it.

It really doesn't matter what the digits are they could be any as the number of ways of writing permutations of it is the same.

In maths it's written 4!

[AstralTraveller]

For Q1 no one has explained the logic. If I understand the question correctly you can put any digit in the first column you fill (left, right, other - it doesn't matter). So, you have 4 options. For the second colunm you fill you only have three options since you cannot re-use the digit you put in the first column. Similarly you have 2 options for the third column and only 1 for the fourth. So the total combinations = 4 x 3 x 2 x 1 = 4! (four factorial).

For Q2, finding all the possible rectangles that have a perimeter of 36 is the same as finding all the combinations of two numbers that add up to 18 (the second two sides are just a repeat of the first two). So you have 1+17, 2+16 ...... 9+9 or nine possibilities = 36/4 or n/4 for the general case. In the case where the original number is not evenly divisible by 4 it is Rounddown(n/4) - work it out yourself for n=38.

[TrevorW]

Q1. To find the number of different arrangements of the set we select a first choice; there are 4 possible choices.

Now we take a second choice; there are 3 choices. Now pick a third choice; there are 2 choices.

Finally, there is 1 choice for the last selection. Thus, there are 4 * 3 * 2 * 1 or 24

The total number of permutations of a set of n objects is given by n!

Q3. take out 3 lots of 10 put back in 3 lots of 7 you are left with 9L (the question does not state using only the containers you have)

[MrB]

I got tricked at first too, the question asked for 9L

[TrevorW]

Q3b. take out 3 lots of 10 put back in 3 lots of 7 you are left with 9L (the question does not state using only the containers you have)

[sheeny]

:rolleyes:

I knew I shouldn't have posted a quick answer while getting ready for work...



Absolutely correct on the rectangles!:thumbsup:

Well, now that I've read it some more, there are some rules missing to make the question unambiguous.;) I assumed that the digits sharing the columns would be added together... we all know about ASSuME! The question doesn't say whether they are added, multiplied, or whether normal place column rules apply.:rolleyes:



I don't believe the question is about permutations or combinations. they are clearly refering to place columns i.e. the RH column is units, the 2nd from the right is tens, etc. The question is what real numbers are possible if you are allowed to have more than one digit in each place column. I assumed the digits in each place column would be added together (e.g. 1 & 3 in column 2 = (1+3)x10 = 40). But since this problem is playing with some fundamental maths rules, the behaviour you are expected to use for these digits needs to be spelled out to make it unambiguous. For example, the 1 & 3 in column 2 could be interpreted as 13, hence 13 x 10 = 130):shrug:.



Oh yeah...:rolleyes: ... what did my old school teachers always tell us??? Read the question!

:lol:

Al.

[OICURMT]

http://en.wikipedia.org/wiki/Permutation

[Clarry]

I'm going to have to give you good folk more of my daughters maths homework. You all seem to enjoy it so much.

[toc]

Jez, I have a son that has just started high school - im not looking forward to this at all :) Guess I can always hire a tutor :D