The CAT exam is known to test its test-takers with a lot of questions based on Mixtures and Averages, as it belongs to the Arithmetic part of the Quantitative Ability section in the CAT Question Paper. This makes it all the more important for CAT aspirants to practise questions from CAT Previous Year Paper. Try your hands on this problem from CAT previous Year Paper 2018 Slot 2 and assess where you stand in your CAT online preparation.
Question 3 : 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
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 \\frac{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 \\frac{13}{20})th of the overall mixture. From here on, it is weighted averages.
So, \\frac{\frac{18}{5} × 3 + 𝑥 × 4}{7}) = \\frac{13}{20})
\\frac{54}{25}) + 4𝑥 = \\frac{91}{20})
4𝑥 = \\frac{91}{20}) - \\frac{54}{25})
4𝑥 = \\frac{455 - 216}{100}) = \\frac{239}{100})
𝑥 = \\frac{239}{400})
In drum 2, A is \\frac{239}{400}) so B should be the remaining. Therefore, A : B = 239 : 161
The question is "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 "
Choice B is the correct answer.
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