Question

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

Answer 1

Let the alcohol must be added be x
intial consecration of pure palcohol=400x15/100
pure alcohol = 60%
total solution = 400+x
A/C to question
=> 60+x/400+x =32/100
=> 100(60+x) = 32(400+x)
=>6000+100x = 12800+32x
=> 100x-32x = 12800-6000
=> 68x = 6800
=> x =6800/68
=> x = 100
Hence the alcohol must be added is x =100

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.