Question

how much pure alcohol must be added to 400m.l of a 15 % soulation to make it strength 32%​

Answer 1

Answer:

Initial amount of alcohol is

\begin{lgathered}400 \times \frac{15}{100} = 60 \\\end{lgathered}

400×

100

15

=60

Let x ml of pure alcohol is to be added.

Then,

Amount of alcohol becomes 60 + x

Total amount of solution becomes 400 + x

\begin{lgathered}\frac{60 + x}{400 + x} \times 100 = 32 \\ 60 + x = 128 + 0.32x \\ 0.68x = 68 \\ x = 100 \: ml\end{lgathered}

400+x

60+x

×100=32

60+x=128+0.32x

0.68x=68

x=100ml