Question

How much pure alcohol must be added to 400 mL of a 15% solution to make its
strength 32%?
​

Answer 1

Step-by-step explanation:

Initial amount of alcohol is

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

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

400+x

60+x

×100=32

60+x=128+0.32x

0.68x=68

x=100ml

Answer 2

Answer:

Initial amount of alcohol is

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

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

400+x

60+x

×100=32

60+x=128+0.32x

0.68x=68

x=100ml

Step-by-step explanation:

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