Question

A can contains a mixture of two liquids A and B in the ratio 7:5. When a liter of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many litres of liquid A was contained by the can initially?​

Answer 1

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Answer 2

$\underline{ \underline{ \bf Question}}$

A can contains a mixture of two liquids A and B in the ratio 7 : 5. When a 9 litre of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?

$\mapsto \sf \: To \: Find$

How many litres of liquid A was contained by the can initially?

$\star \underline{\rm \: Answer}$

$\colorbox{cyan}{ \bf21 \: L } \: \bf\: of \: \: Liquid( A)$

Solution

Suppose the can initially contains 7x and 5x of mixture A & B respectively.

$\odot \boxed{\rm\:Quantity \: of \: A \: in \: mixture \: left}$

$\hookrightarrow \sf(7x - \frac{7}{12} \times 9) l = (7x - \frac{21}{4}) l$

$\odot \boxed{\rm\:Quantity \: of \: B \: in \: mixture \: left}$

$\hookrightarrow \sf(5x - \frac{5}{12} \times 9)l = (5x - \frac{15}{4})l$

According to the question

$\implies \dfrac{7x - \dfrac{21}{4} }{5x - \dfrac{15}{4} } = \dfrac{7}{9}$

$\implies \dfrac{28x - 21}{20x + 21} = \dfrac{7}{9}$

$\leadsto \rm252x - 189 = 140x + 147 \\ \rightarrow \rm112x = 336 \\ \rm \:x = \dfrac{ \cancel{336}}{ \cancel{112}} = 3$

Here , we have supposed 7x of mixture A to be present in can now we get the value of x to be 3

Now ,

$\boxed{ \red{7x = 7 \times 3 = 21}}$

Conclusion

The can contains 21 L of mixture A .

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