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 Post subject: Maths questionPosted: Wed Jan 23, 2008 11:23 pm

Joined: Fri Nov 16, 2007 10:20 pm
Posts: 41
Any help with the following question would be appreciated.
Two discs with numbers on each side are thrown. The numbers showing, add up to 5. This is repeated three more times and the total of the two numbers on each of these throws are 11 19 25.
What are the four numbers on each side of the discs?. Two of the numbers on the discs are greater than 10.

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 Post subject: Posted: Thu Jan 24, 2008 7:05 am

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 14005
The clue is that the sum of two of them is 11 - this must be made up off 10 + 1

25 must be the two biggest added up so if one is 10 the other must be 15

so the other small one must be 4 to get 5

11 = 10 + 1

25 = 10 + 15

5 = 1 + 4

check 19 = 15 + 4

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 Post subject: Posted: Thu Jan 24, 2008 5:58 pm

Joined: Tue Jan 01, 2008 1:05 pm
Posts: 515
Only one of the numbers is greater than 10 though.

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 Post subject: Posted: Thu Jan 24, 2008 6:10 pm

Joined: Fri Nov 16, 2007 10:20 pm
Posts: 41
Yes I agree that only one number is greater than ten in the above solution, but It does stipulate in the question that two of the numbers are greater than ten. Anyone any other ideas??

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 Post subject: Posted: Thu Jan 24, 2008 6:13 pm

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 14005
Sorry in my solution 1,4 10, 15 -

must be

11 = 11 + 0

25 = 11 + 14

19 = 14 + 5

5 = 5 + 0

so 0,5,11,14

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 Post subject: Posted: Thu Jan 24, 2008 7:43 pm

Joined: Fri Jul 06, 2007 8:31 pm
Posts: 1245
If the coins are... a on one side; b on the other
and c on one side; d on the other...

the combinations are therefore...

a+c
a+d
b+c
b+d

you note that (a+c)+(b+d) is the same as (a+b)+(c+d)...
given the answers are 5,11,19,25... you can see that 5+25=30 and 11+19=30.

Therefore we can safely say... a+b+c+d=30

Arbitarily assigning...
a+c=5 b+d=25 a+d=11 b+c=19

Clearly there aren't enough equations here to get a single solution so
let's rearrange using a.

c = 5-a
b = 19 - c b = 19 - (5-a) b=14+a
d = 11-a

so we have...

a, a+14 on one coin
5-a, 11-a on the other

so... combinations are...

0,14 & 5,11
1,15 & 4,10
2,16 & 3,9
3,17 & 2,8
4,18 & 1,7
5,19 & 0,6

other combinations have -ve numbers which is unlikely!

The only combination where two numbers are >10 is...
0,14 and 5,11.

Personally I don't like the zero... If the question is reworded to more than or equal to 10 then the answer would be..

1,15 & 4,10

Regards
SVE

_________________
Animis opibusque parati

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 Post subject: Posted: Thu Jan 24, 2008 7:46 pm

Joined: Thu Jan 12, 2006 3:29 pm
Posts: 625
Hi

You have to us "0" because there are two numbers higher than ten.

so first coin 0 and 14
second coin 5 and 11

Regards

Mike

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 Post subject: Posted: Thu Jan 24, 2008 7:51 pm

Joined: Tue Jan 01, 2008 1:05 pm
Posts: 515
2 numbers greater than 10.
Therefore the 2 smallest numbers must equal 5, i.e each 5 or less.
Therefore, 11 must include one of the two larger numbers, both of which are greater than 10.
Therefore 2 of the numbers are 11 and 0.
Therefore the second smaller nunber is 5.
Therefore remaining number is 14.

Alternatively the question is badly worded and Guest55's first answer is correct.

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