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

- 251 : 163
- 239 : 161
- 220 : 149
- 229 : 141

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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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