A) 10 L
B) 20 L
C) 21 L
D) 25 L
Correct Answer: C
Solution :
Initially, let the can hold \[7x\] L and \[5x\] L of A and B respectively. |
Quantity of A in remaining mixture |
\[=\left( 7x-\frac{7}{12}\times 9 \right)\,\,\text{L}\] |
\[=\left( 7x-\frac{21}{4} \right)\,\,\text{L}\] |
Quantity of B in remaining mixture |
\[=\left( 5x-\frac{5}{12}\times 9 \right)\,\,\text{L}\] |
\[=\left( 5x-\frac{15}{4} \right)\,\,\text{L}\] |
\[\therefore \] \[\frac{\left( 7x-\frac{21}{4} \right)}{\left( 5x-\frac{15}{4} \right)+9}=\frac{7}{9}\]\[\Rightarrow \]\[\frac{28x-21}{20x+21}=\frac{7}{9}\] |
\[\Rightarrow \] \[252x-189=140x+147\] |
\[\Rightarrow \] \[112x=336\]\[\Rightarrow \]\[x=3\] |
Initially, the can contained \[(7\times 3)\,\,\text{L}=21\,\,\text{L}\] |
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