How much pure alcohol should be added to $400 m l$ of strength $15 %$ to make its strength $32 %$ ?

Question

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Answer 1

The amount of alcohol in $400 m l$ solution of $15 %$ strength
$= 400 \times \frac{1 5}{1 0 0} m l = 60 m l$

Let the required amount of alcohol to be added $= x$ ml.
It will increase the amount of the solution by $x$ ml.

$∴$ The amount of alcohol now $= (60 + x) m l$

and the amount of the solution $= (400 + x) m l$ .
Then, the strength $= 32 % = \frac{6 0 + x}{4 0 0 + x} = \frac{3 2}{1 0 0}$

$\Rightarrow 25 x + 1500 = 3200 + 8 x \Rightarrow 17 x = 1700 \Rightarrow x = 100$ .
So, the required amount of pure alcohol to be added $= 100 m l$ .

Answer 2

The amount of alcohol in

400 m l

solution of

15 %

strength

= 400 \times \frac{1 5}{1 0 0} m l = 60 m l

Let the required amount of alcohol to be added

= x

ml.

It will increase the amount of the solution by

x

ml.

∴

The amount of alcohol now

= (60 + x) m l

and the amount of the solution

= (400 + x) m l

.

Then, the strength

= 32 % = \frac{6 0 + x}{4 0 0 + x} = \frac{3 2}{1 0 0}

\Rightarrow 25 x + 1500 = 3200 + 8 x \Rightarrow 17 x = 1700 \Rightarrow x = 100

.

So, the required amount of pure alcohol to be added

= 100 m l

.