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Maths question
Posted: Thu Jan 24, 2008 12:23 am
by kosar.irshad
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.
Posted: Thu Jan 24, 2008 8:05 am
by Guest55
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
Posted: Thu Jan 24, 2008 6:58 pm
by dadofkent
Only one of the numbers is greater than 10 though.
Posted: Thu Jan 24, 2008 7:10 pm
by kosar.irshad
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??
Posted: Thu Jan 24, 2008 7:13 pm
by Guest55
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
Posted: Thu Jan 24, 2008 8:43 pm
by SunlampVexesEel
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
Posted: Thu Jan 24, 2008 8:46 pm
by Mike
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
Posted: Thu Jan 24, 2008 8:51 pm
by dadofkent
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.