A container contains a mixture of two liquids A and B in the ratio 7: 5. When 9 liters of mixture are drawn off and the container is filled with B, the ratio of A and B becomes 7: 9. How many liters of liquid A was contained in the container initially. (Mixture)

Question

A container contains a mixture of two liquids A and B in the ratio 7: 5. When 9 liters of mixture are drawn off and the container is filled with B, the ratio of A and B becomes 7: 9. How many liters of liquid A was contained in the container initially. (Mixture)

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2025-01-04 04:42

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Shohan · Accepted Answer · 2024-12-14T13:31:11+06:00

Let, the container primarily contains 7x and 5x of mixtures A and B respetively.

Now, after drawing off 9 litres of mixtures

The quantity of A left in the mixture = $7 x - \frac{7 \times 9}{5 + 7} = 7 x - \frac{63}{12} = 7 x - \frac{21}{4}$ litres.

And the quantity of B left in the mixture = $5 x - \frac{5 \times 9}{5 + 7} = 5 x - \frac{45}{12} - \frac{15}{4} l i t r e s .$

According to question, $\frac{7 x - \frac{21}{4}}{5 x - \frac{15}{4} + 9} = \frac{7}{9}$

$= \frac{\frac{28 x - 21}{4}}{\frac{20 x - 15 + 36}{4}} = \frac{7}{9} = \frac{28 x - 21}{4} \times \frac{4}{20 x + 21} = \frac{7}{9} = \frac{28 x - 21}{20 x + 21} = \frac{7}{9} = 252 x - 189 = 140 x + 147 = 112 x = 336 ∴ x = \frac{336}{112} = 3 ∴ T h e c o n t a i n e r c o n t a i n e d = 7 \times 3 = 21 l i t r e s o f A .$