There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio
Started 2 weeks ago by Shashank in
Explanatory Answer
Given that two drums each containing a mixture of paints A and B. In drum 1, A and B are in the
ratio 18 : 7. Drums 1 and 2 are mixed in the ratio 3 : 4 and in the final mixture A and B are in the
ratio 13 : 7. For these kinds of questions do not consider separately as A and B. Deal with either A
or B as a share of overall
Here A is 18/25 of overall
In D2, let us assume the proportion of A with respect to the overall mixture is x. 3 parts of D1 is
mixed with parts of D2 to give proportion of A of which is (13/20) th of the overall mixture. From
here on, it is weighted averages.
So, {(18/5)×3 + (x×4)} / 7 = 13/20
54/25 + 4𝑥 = 91/20
4𝑥 = 91/20 - 54/25
4𝑥 = (455−216)/100 = 239/100
𝑥 = 239/400
In drum 2, A is 239/400 so B should be the remaining. Therefore, A : B = 239 : 161
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