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 = 5a
b = 19  c b = 19  (5a) b=14+a
d = 11a
so we have...
a, a+14 on one coin
5a, 11a 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
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