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Question

$$37.85\%$$ and $$92\%$$ alcoholic solutions are mixed to get $$35$$ liters of an $$89\%$$ alcoholic solution. How many liters of each solution are there in the new mixture?


A
10 L of the 1st, 25 L of the 2nd
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B
20 L of the 1st, 15 L of the 2nd
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C
15 L of the 1st, 20 L of the 2nd
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D
None
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Solution

The correct option is D None
Let $$x$$ litres of the $$37.85\%$$ alcoholic solution and $$(35 - x)$$ litres of $$92\%$$ alcoholic solution be required to get $$35$$ L of $$89\%$$ solution. 
Then $$37.85\%$$ of $$x + 92\%$$ of $$(35 - x) = 89\%$$ of $$35$$
$$\displaystyle \Rightarrow \frac{37.85x}{100}+\frac{92\times 35}{100}-\frac{92\times x}{100}=\frac{89}{100}\times 35$$
$$\displaystyle \Rightarrow 54.15x=105$$
$$\Rightarrow x=\dfrac{105}{54.15}=1.94\: L$$ (approx)

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