Question

how much pure alcohol must be added to 400 ml of a 15% solution to make its 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

Given that,

Solution = 400 ml

Strength = 32 %

Let x ml of alcohol be added

We need to calculate the quantity of alcohol

Using formula for quantity

Quantity of alcohol in 400 ml solution $Q =\dfrac{15}{100}\times400$

$Q=60\ ml$

We need to calculate the value of x

Using formula for strength

$\dfrac{60+x}{400+x}=\dfrac{32}{100}$

$100\times(60+x)=32(400+x)$

$6000+100x=12800+32x$

$100x-32x=12800-6000$

$68x=6800$

$x=\dfrac{6800}{68}$

$x=100\ ml$

Hence, 100 ml pure alcohol must be added .