Question

how much pure alcohol must be added to 400ml of a 15 percent solution to make its strength 32 percent can u pls help me because I have my math exam tomorrow

Answer 1

Given 15% of 400ml should be added to 100% pure alcohol.

 i.e 15% of 400 = 15/100 * 400

                          = 15 *  4

                          = 60ml.


Let the quantity of alcohol to be added to make its strength 32% = x.

Then the required solution will be

60 + x/400 + x = 32%

60 + x/400 + x = 32/100

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

6000 + 100x = 12800 + 32x

6000 + 100x - 32x = 12800

6000 + 68x = 12800

68x = 6800

x = 6800/68

x = 100.


Therefore the Quantity of alcohol to be added is 100ml.


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