The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is
Started 2 weeks ago by Shashank in
Explanatory Answer
Given that the strength of the salt solution is p% if 100 ml of the solution contains p grams of salt
It is also given that three salt solutions A , B , C are mixed in the proportion 1 : 2 : 3, then the
resulting solution has strength 20%.
So \({A+2B+3C\over6}\) = 0.2 ⟹ A + 2B + 3C = 1.2 ------(1)
If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%.
So, \({3A+2B+C\over6}\) = 0.3 ⟹ 3A +2B + C = 1.8 -------(2)
It is given that 4 th solution D is produced by mixing B and C in the ratio 2 : 7
So D = \({2B+7C\over9}\)
We have to find the ratio of the strength of D : A
Subtracting equations 1 and 2, we get
2A – 2C = 0.6 or A – C = 0.3
Since we could not find anything from the above methods, we can eliminate the number part and
get the ratio going
A + 2B + 3C = 1.2
3A + 2B + C = 1.8
So let us multiply eqn 1 and 2 with 3 and 2,
3A + 6B + 9C = 6A + 4B + 2C
2B + 7C = 3A
It is given that D = \({2B+7C\over9}\)
\({2B+7C\over9}\) = \(3A\over9\) ⟹ 2 \({2B+7C\over9}\) = A/3
Hence D = A/3
Therefore the ratio D : A = 1 : 3
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