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# Anagram and others

Fri 7 Dec, 2007 04:24 am
1. A large number of children have had their Christmas lunch at school. 40 kids had mince pies; 50 had ice cream; and 60 had Christmas cake. of these number, 11 had mince pies AND ice cream; 11 had mince pies AND Christmas cake; and 11 had ice cream AND Christmas cake. hard to believe, but 9 little gluttons had all THREE. so how many children were there in total?
2. ACCORDANCE LIMITS SHARKS is an anagram of which well known novel AND the surname of its author?
3. What is the longest English word that can be spelled using the letters from CHRISTMAS DAY? You can have two As and two Ss because there are two, but not, say, two Cs or two Ds.)
Many, many thanks for any help.
TUFDEVIL! meant to say 3 months not 3 weeks! A long haul but enjoyable.
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Type: Discussion • Score: 1 • Views: 1,675 • Replies: 16
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fresco

1
Fri 7 Dec, 2007 11:35 am
First one...126 Children.

Working on the others.
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fresco

1
Fri 7 Dec, 2007 11:47 am
Second one... "A CHRISTMAS CAROL" DICKENS
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upandrunning

1
Fri 7 Dec, 2007 11:59 am
Many thanks fresco, very much appreciated.
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Tufdevil

1
Fri 7 Dec, 2007 02:51 pm
Don't like arguing but I believe that the answer to 1 is 99 kids.

9 kids had all 3 and 33 kids had two of the three, so the first 42 kids shared out equally 31 of each.

That leaves 9 who just had mince pies, 19 who just had ice cream and 29 who just had christmas cake

Therefore 33+9+9+19+29 = 99

Do you agree, fresco?

I'll think about the anagram after the pub, Upandrunning. Mind you, I might come up with a very strange word!

Tufdevil
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Tufdevil

1
Fri 7 Dec, 2007 02:52 pm
I have CHARISMA to be going on with, that uses 8 of the 12 letters but I am sure there is something better.
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lmur

1
Fri 7 Dec, 2007 04:06 pm
DRAMATICS has 9.
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fresco

1
Fri 7 Dec, 2007 05:18 pm
Tufdevil,

This is a typical Venn Diagram question with three intersecting circles (One for each food),

9 goes in the middle leaving 2 in each of the three central "petals" to give the 11's. By Subtraction you are left with 27,37,and 47 who have one food only, The sum of all the areas is 27+37+47 +2+2+2+9 =126.
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fresco

1
Fri 7 Dec, 2007 05:26 pm
Most google "anagram solvers" yield several 9 letter words for item 3.
The implication of the question is that there may be a word with more than 9.
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Tufdevil

1
Fri 7 Dec, 2007 06:22 pm
Evening Fresco

I seem to remember, sometime in the dim and distant past, getting A levels in both Maths and Further Maths, but sometimes I think that Arithmetic is the answer. This is one of those times. I didn't use Venn diagrams or arithmetical formulae, I used a spreadsheet and basic logic and I am sure that the answer is no more than 99. I didn't do Venn diagrams in my days, so, if you have gone wrong somewhere, I don't know where, but I stand by my answer. Upandrunning, whichever answer you go with, please tell us who was right, or what the answer was if neither of us is right.

Tufdevil
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fresco

1
Fri 7 Dec, 2007 09:57 pm
Tufdevil,

Forget A-level ! They are usually harder than that on current Manchester Grammar (11+) entry exams. ! :wink:

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upandrunning

1
Sat 8 Dec, 2007 02:20 am
Many thanks fresco, tufdevil and lmur. I agree with fresco regarding the anagram, I think the answer may include most of the letters, even maybe then all, still working on it.
Still trying the maths one too. I have two answers, which don't match yours tufdevil and fresco, so back to pencil and paper! Many thanks for taking the time. Answers not expected until the new year, but will keep in touch.
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Tufdevil

1
Sat 8 Dec, 2007 11:08 am
Afternoon Upandrunning.

Bit more sober now so I can think more clearly. As I said it is a simple arithmetic problem. Assuming no child had 2 of the same item, we have 40 mince pies, 50 ice creams and 60 Christmas cakes, which is 150 items altogether. Once we've established that, we can forget about the items and just look at the numbers:

9 kids had 3 items = 27 items = 9 kids
33 kids had 2 items = 66 items = 33 kids

That makes 93 items shared by 42 kids. If there are 150 items altogether, then the remaining 57 items must have been eaten by 57 kids. Add the other 42 kids and you get 99 kids.

So I still get the same answer when I'm sober!

I am sure that the anagram must be at least 10 letters. Otherwise, you could just have CHRISTMAS. It gave me something to occuopy my mind while hanging around in a hospital waiting room this morning, but I still haven't come up with an answer.

Good luck

Tufdevil
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fresco

1
Sat 8 Dec, 2007 04:44 pm
Tufdevil,

You obviously havn't tried the link then !
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fresco

1
Sat 8 Dec, 2007 05:12 pm
Further to item three there seems to be a ten letter version DYSCHRASIA of the more normal (ha ha !) DYSCRASIA, meaning blood disorder, as in.....

Quote:
Multiple myeloma is a severe plasma dyschrasia with no known treatment or cure, even bone marrow transplantation.
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Tufdevil

1
Sat 8 Dec, 2007 07:50 pm
OK Fresco, I see where we are at odds. It is in the interpretation of the question, and I don't know whose interpretation is correct (they may both be which would make it a bad question).

In your interpretation, the 11 who had mince pies and ice cream includes the 9 who had all three, etc. In my interpretation, the 9 who had all 3 are in addition to the 11 who only had two.

So, Upandrunning, I am now sure that the answer you are looking for is...

either 99 or 126. Depends whose interpretation you prefer. Over to you!
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fresco

1
Sat 8 Dec, 2007 09:34 pm
Tufdevil,

I did hundreds of these when I used to teach !!+. I assure you that there is only one interpretation.

Note that Venn Diagrams and Boolean algebra (involving logical operators AND OR NOT etc) have now largely replaced geometry as an exercise in logic. Venn Diagrams abound in the "Bond Assessment Papers" which are standard texts for UK primary schools.
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