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There are two drums, each containing a mixture of paints $\text{A}$ and $\text{B}$. In drum $1, \text{A}$ and $\text{B}$ 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, $\text{A}$ and $\text{B}$ are in the ratio $13 :7$. In drum $2$, then $\text{A}$ and $\text{B}$ were in the ratio

1. $229:141$
2. $220:149$
3. $239: 161$
4. $251: 163$

### 1 comment

option (C)

Given that,

DIAGRAM

In drum$1,$ $A$ and $B$ are in the ratio of $18 : 7.$

So, the quantity of $A = \frac{18}{25}$

The quantity of $B = \frac{7}{25}$

In final mixture, $A$ and $B$ are in the ratio of $13 : 7.$

So, the quantity of $A = \frac{13}{20}$

The quantity of $B = \frac{7}{20}$

In drum$2,$ let us assume the proportion of $B$ with respect to the overall mixture is $x.$

So, $\frac{ \frac{7}{25} \times 3 + x \times 4}{7} = \frac{7}{20}$

$\Rightarrow \frac{21}{25} + 4x = \frac{49}{20}$

$\Rightarrow 20 (21 + 100x) = 25 \times 49$

$\Rightarrow 420 + 2000x = 1225$

$\Rightarrow 2000x = 805$

$x = \frac{805}{2000}$

$\boxed{x = \frac{161}{400}}$

In drum$2, \; B$ is $\frac{161}{400}$

So, $A$ should be the remaining.

Thus, $A$ is $\frac{239}{400}.$

$\therefore$ In drum$2, \; A$ and $B$ are in the ratio of $239 : 161.$

Correct Answer$: \text{C}$
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