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In what ratio must a person mix three kinds of wheat costing him 1.20,1.44 and 1.74 dollars per kg., so that the mixture formed is worth 1.41 dollars per kg? a)11:77:7 b)12:7:7 c)ratio other than a and b d)no solution

Is alligation valid here?

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  • $\begingroup$ There are infinitely many solutions. To answer the question, one can check by substitution whether a) or b) work. $\endgroup$ – André Nicolas Mar 3 '16 at 0:19
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You should be able to answer c without doing any computation at all. As $1.41$ is between the upper and lower cost elements there is some mix that will produce it, so d is out. If you had only two things you could mix you could find a unique mix that would produce that cost. Given that you have three, there is a one dimensional set of solutions, so there will be some other than a and b. If you trust the problem setter to have made a unique solution, neither a nor b should yield the correct cost. You could check if you want.

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  • $\begingroup$ Thanks for the answer. What would have been the solution if there were no options? $\endgroup$ – Priyaranjan Mar 3 '16 at 0:24
  • $\begingroup$ As I said, there are an infinite number of them. You could find one using just the $1.20$ and $1.74$ wheat. If fraction $x$ of the mix is $1.20$ you have $1.20x+1.74(1-x)=1.41$ For the mix using all three, if $x$ is the fraction of $1.20$ wheat and $y$ is the fraction of $1.44$, you have $1.20x+1.44y+1.74(1-x-y)=1.41$ One equation, two unknowns. $\endgroup$ – Ross Millikan Mar 3 '16 at 0:27
  • $\begingroup$ I think the word you are trying to use is allocation, but it doesn't really fit here. $\endgroup$ – Ross Millikan Mar 3 '16 at 0:49

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