| 20 L of a mixture contains 20% alcohol and the rest water. If 4 L of water be mixed in it, the percentage of alcohol in the new mixture, will be [SSC (CGL) Mains 2014] |
A) \[33\frac{1}{3}\]%
B) \[16\frac{2}{3}\]%
C) \[25\]%
D) \[12\frac{1}{2}\]%
Correct Answer: B
Solution :
| Amount of alcohol in 20 L of mixture |
| \[=20%\,\,\text{of 20}\,\,\text{L=}\frac{20\times 20}{100}=4\,\,\text{L}\] |
| \[\therefore \]Water in the mixture \[=20-4=16\,\,\text{L}\] |
| Now, 4 L of water is further added |
| \[\therefore \]Amount of water \[=16+4=20\,\,\text{L}\] |
| \[\therefore \]Percentage of alcohol in new mixture |
| \[=\frac{\text{Amount}\,\,\text{of}\,\,\text{alcohol}}{\text{Total}\,\,\text{mixture}}\times 100%\] |
| \[=\frac{4}{24}\times 100=\frac{100}{6}\]%\[=\frac{50}{3}\]%\[=16\frac{2}{3}\]% |
You need to login to perform this action.
You will be redirected in
3 sec
