Question

How much pure alcohol must be added to 400 ml of a 15% solution to make it's
Strength 32% ?

Answer 1

Initial amount of alcohol is

400 \times \frac{15}{100} = 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

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

Answer 2

Step-by-step explanation:

=400×15/100+X=400+X×32/100

=60+X=128+(8X/25)

=X-8X/25=128-60

=17X/25=68

=X=(68×25)/17

=X=1700/17

=X=100ml