Question

a chemist has one solution which is 40% acid and another which is 20% acid how much each Solutions should the chemist mix to get 10 litre of a solution which is 25℅ acid

Answer 1

Let the qty of 1st solution needed be x and that of 2nd be y.

For x ltr of solution, there will be 0.4x acid and rest 0.6x water

For y ltr of solution, there will be 0.2y acid and 0.8y rest water

x + y = 10....(1)

0.4x + 0.2y / 0.6x + 0.8y = 0.25/0.75 [ acid-water fraction]

4x + 2y / 6x + 8y = 1/3

2x + y / 3x + 4y = 1/3

3(2x + y) = 3x + 4y

6x + 3y = 3x + 4y

y = 3x...(2)

From (1) and (2) we get

3x + x = 10

4x = 10

x = 2.5 ltr

y = 7.5 ltr

Answer 2

Define x and y:

Let x be the amount of 40% acid

Let y be the amount of 20% acid

Form equation 1 :

Total amount is 10 litres:

x + y = 10  ----------------------- [ 1 ]

Form equation 2:

The final mixture is 25% acid

0.4x + 0.2y = 0.25 (10)

-0.4x + 0.2y = 2.5 ---------------------- [ 2 ]

Put the 2 equations together:

x + y = 10                ----------------------- [ 1 ]

0.4x + 0.2y = 2.5 ----------------------- [ 2 ]

From equation [ 1 ]:

x + y = 10

x = 10 - y  ----------------------- Sub into [ 2 ]

Solve y:

0.4(10 - y) + 0.2y = 2.5

4 - 0.4y + 0.2y = 2.5

0.2y = 1.5

y = 7.5 ----------------------- Sub into [ 1 ]

Solve x:

x + y = 10

x + 7.5 = 10

x = 2.5

Answer: Each solution has 2.5 litres of 40% acid and 7.5 litres of the 40% acid.