Question

How much pure alcohol has to be added to 400ml of a solution?

Answer 1

Let x = the required number of milliliters of
pure alcohol.
Then we have:
400(15%) + x(100%) = (400+x)(32%) Change
the percents to decimals.
400(0.15) + x(1) = (400+x)(0.32) Simplify
and solve for x.
60 + x = 128 + 0.32x Subtract 0.32x from
both sides.
60 + 0.68x = 128 Subtract 60 from both
sides.
0.68x = 68 Divide both sides by 0.68
x = 100 ml of pure alcohol is required.
Check:
400(0.15) + 100 = (400+100)(0.32)
60 + 100 = 500(0.32)
160 = 160 Check ok.

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