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