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Every other one seemed straightforward to solve except this. It looks simple but I still can't seem to solve it.

A, B and C are mixed according to the ratios A:B = 1:7 and B:C = 13:9. Find A, B and C.

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Without additional detail, the only thing one can conclude is the ration across all 3:

A:B = 1:7 = 13:91 B:C = 13:9 = 91:63

Thus, A:B:C = 13:91:63.

However, without more detail, there are many many solutions as A,B,C are in that ratio so $(A,B,C) = (13x,91x,63x)$ for any natural number $x$ which leads to more than a few solutions.


A:B = 2:5 and B:C = 10:11 Find A, B, C

In this case, the easy solution is to just double the A:B ratio so that the B value is the same. Thus, in this case, A:B = 4:10 and B:C = 10:11, so A:B:C = 4:10:11


In general, if A:B: = w:x and B:C = y:z then A:B:C = wy:xy:xz if you want the formula I'd used. Now, in the second case this would leave you with A:B:C = 20:50:55 where one could divide 5 out of all 3 terms to get 4:10:11.

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  • $\begingroup$ That was great..bt cud you apply the same method to this..it's the same except for the values. This could make me understand better - A:B = 2:5 and B:C = 10:11 Find A, B, C $\endgroup$ – patrickojeh May 8 '14 at 6:07
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Make the ratio 13 times larger such that A : B = 13 : 91. Do the similar action to the second ratio. Then the three can be compared.

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