Question

How much pure alcohol has to be added to 400ml of a 15 solution to make its strength 32%?

Answer 1

Answer:

Since we have given that

Quantity of solution = 400 ml

Let quantity of pure alcohol be 'x'.

According to question,

\dfrac{15}{100}\times 400+x=(400+x)\times \dfrac{32}{100}

So, we will solve it for the value of 'x':

60+x=(400+x)0.32\\\\60+x=128+0.32x\\\\x-0.32x=128-60\\\\0.68x=68\\\\x=\dfrac{68}{0.68}\\\\x=100\ ml

So, we will solve it for the value of 'x':

Hence, 100 ml of pure alcohol must be added to 400 ml of a 15% solution.

Explanation:

Answer 2

Answer:

Hi mate,

Solution :

=> Volume of alcohol in 400 ml solution

➾ 0.15 × 400 ml

=> Volume of alcohol in 400 ml solution

➾ 60 ml

=> Volume of others liquid

➾ 340 ml

=> Let x ml of alcohol is added to make it 32% strong

=> Then, volume of alcohol

➾ 60+x ml

=> Volume of other liquid

➾ 340 ml

=> Volume of solution

➾ 400 + x ml

➾ 0.32 × (400 + x) = 60 + x

➾ 128 + 0.32x = 60 + x

➾ 0.68x = 68

➾ 68x = 6800

➾ x = 100 ml ...

Answer : 100 ml pure alcohol must be added to 400ml of a 15% solution to make it's strength 32%

i hope it helps you