Question

a 400ml mixture contains 15%alcohol. how much alcohol much be added so that the concentration of alcohol will become 32%.

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

Percentage of concentration of alcohol in the mixture = ((volume of alcohol in ml)/(volume of mixture in ml)) *100 .let the volume of alcohol be 'y'

So, 15%=(y/400)*100, 15=y/4 and y=15*4=60 ml.

Let the extra amount of alcohol that should be added be 'x'.

Now, concentration of alcohol after adding 'x'is 32%

So, 32%=((60+x)/(400+x))*100.

12800+32x=6000+100x

12800-6000=100x-32x

6800=68x

Therefore, x=(6800/68)

=100 ml

So, 100 ml of alcohol should be added