A) \[33\frac{1}{3}\]
B) \[16\frac{2}{3}\]
C) \[25\]
D) \[12\frac{1}{2}\]
Correct Answer: B
Solution :
[b] Amount of alcohol in 20 L of mixture \[=20%\]of \[20\,L=\frac{20\times 20}{100}=4\,L\] \[\therefore \] Water in the mixture \[=20-4=16\,L\] Now, 4 L of water is further added \[\therefore \] Amount of water in\[16+4=20\,L\] \[\therefore \] Percentage of alcohol in new mixture \[\text{=}\frac{\text{Amount}\,\,\text{of}\,\,\text{alcohol}}{\text{Total}\,\,\text{mixture}}\text{ }\!\!\times\!\!\text{ 100 }\!\!%\!\!\text{ }\] \[=\frac{4}{24}\times 100=\frac{100}{6}%=\frac{50}{3}%=16\frac{2}{3}%\] |
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