A) 60 mL
B) 100 mL
C) 128 mL
D) 68 mL
Correct Answer: B
Solution :
Quantity of alcohol in mixture \[=15%\,\,\text{of}\,\,400\] \[\frac{15\times 400}{100}\] \[=60\,mL\] Let x mL of alcohol should be added to the mixture \[\frac{60+x}{400+x}=\frac{32}{100}\] \[\Rightarrow \] \[25(60+x)=8(400+x)\] \[\Rightarrow \] \[17x=1700\] \[\Rightarrow \] \[x=100\] \[\therefore \] 100 mL of alcohol should be added to the mixture.You need to login to perform this action.
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