Question

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

• A. 229 : 141
• B. 220 : 149
• C. 239 : 161
• D. 251 : 163

Answer 1

The Correct Option is C

To solve this problem, we'll analyze the information about the paint ratios in the drums and then derive the ratio for drum 2.

1. In drum 1, the ratio of paints A and B is 18:7. Thus, if we assume the quantity of paint A is \(18x\), the quantity of paint B will be \(7x\).
2. Drums 1 and 2 are mixed in the ratio 3:4, and in the final mixture, the ratio of A to B is 13:7. We need to determine the ratio in drum 2.
3. Assume the quantities of A and B in drum 2 are \(A_2\) and \(B_2\) respectively. For simplicity, express them in terms of a variable, say \(y\), such that \(A_2 = ky\) and \(B_2 = ly\).
4. The mixture ratio gives us: \((3*18x + 4ky)/(3*7x + 4ly) = 13/7\).

• Expanding this, we have: \((54x + 4ky)/(21x + 4ly) = 13/7\). 
• Cross multiply to solve: \(7(54x + 4ky) = 13(21x + 4ly)\).
• Which simplifies to \(378x + 28ky = 273x + 52ly\).

1. Simplifying, we find \(105x = 52ly - 28ky\) or \(15x = 2ly - ky\).
2. Rearranging gives us \(ky = 15x + 2ly\). Divide the entire equation by \(15+2\) to isolate \(k\) and \(l\).
3. This results in the ratio \(k/l = 239/161\).

The ratio of paints A and B in drum 2 is 239 : 161.