Question

how much pure alcohol should be added to 400 ml of a 15% solution to make its strength 32%

Answer 1

Let the required number of pure alcohol should be added be x ml.

15% of 400 = $\frac{15}{100} * 400$

                   = 15 * 4

                   = 60ml.


Given that strength is 32%.

$\frac{60+x}{400+x} = 32% [tex] \frac{60+x}{400+x} = \frac{32}{100}$

(60 + x) * 100 = (400 + x) * 32

6000 + 100x = 12800 + 32x

100x = 32x + 12800 - 6000

100x = 32x + 6800

100x - 32x = 6800

68x = 6800

x = 6800/68

x = 100.


Therefore 100ml of pure alcohol should be added to make its strength 32%.


Hope this helps!

Answer 2

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