Question

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

Answer 1

let the required number of millimetres of alcohol be x ml
Given 15% of 400ml should be added with 100% pure alcohol
Hence 400×15% + x×100% = (400+x)×32%
or, 400×15/100+x = (400+x)×32/100
or,60+x= 128+ (8/25)x
or, x-(8/25)x= 128-60
or,(25x-8x)/25 = 68
or, 17x = 68×25
or, 17x = 1700
or, x= 100ml
Hence 100ml must be added

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