Maths question
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Maths question
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.
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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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
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
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.
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.