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(A) Let the cost price of the second liquid be $x$ per litre.Then,the cost price of the first liquid is $(x + 2)$ per litre.The selling price of the m

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(A) Let the cost price of the second liquid be $x$ per litre.Then,the cost price of the first liquid is $(x + 2)$ per litre.The selling price of the mixture is $Rs. 11$ per litre at a profit of $10 \%$.Therefore,the cost price $(CP)$ of the mixture is:$CP = 11 	imes \frac{100}{110} = Rs. 10$ per litre.Using the method of alligation:Ratio of first liquid to second liquid = $(10 - x) : (x + 2 - 10) = 3 : 2$$\frac{10 - x}{x - 8} = \frac{3}{2}$$2(10 - x) = 3(x - 8)$$20 - 2x = 3x - 24$$5x = 44$$x = 8.8$Thus,the cost per litre of the second liquid is $Rs. 8.80$.

Two liquids are mixed in the ratio of $3: 2$ and the mixture is sold at $Rs. 11$ per litre at a $10 \%$ profit. If the $1$st liquid costs $Rs. 2$ more per litre than the second,what is the cost per litre of the second liquid (in $.80$)? (in $Rs.$)

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