Question

Solution A is 50% hydrochloric acid, while solution B is 75% hydrochloric acid.
How many liters of each solution should be used to make 100 liters of a solution
which is 60% hydrochloric acid?



Please give a step by step answer.

Answer 1

Answer:

X ml taken 50% solution

y ml taken 75% solution

y ml = (100-x) ml since total sol is 100ml

now, 50% of x +75% of (100-x)= 60% of 100

1/2x + 3/4×100 - 3/4x=60

1/2x - 3/4x= 60-75

2-3/4x = -15

-x= -15×4

x=60ml

y= 100-60 = 40ml

Answer 2

The quantity of A and B solutions are 60 liters and 40 liters, respectively.

Step-by-step explanation:

Given: 50% hydrochloric acid is solution A.

75% hydrochloric acid is solution B.

60% hydrochloric acid should be in 100 liters.

To Find: The quantity of A and B solutions that should use to make 100 liters of a solution of 60% hydrochloric acid.

Solution: As given- 50% hydrochloric acid is solution A.

Since solution A contains 50% hydrochloric acid, one liter of solution A contains 0.5 liters of hydrochloric acid.

As given- 75% hydrochloric acid is solution B.

Since solution B contains 75% hydrochloric acid, one liter of solution B contains 0.75 liters of hydrochloric acid.

As given - 60% hydrochloric acid should be in a 100 liters solution.

Let Solution A =A and Solution B =B.

Since the total amount of solution is 100 liters,$A+B=100$ --------- equation no.01.

Since the 100 liters solution is to be 60%, $0.5A+0.75B=0.6\times100 = 60.$                            

$0.5 A+0.75 B =60$   ----------- equation no.2.

Multiply equation no. 01 by 0.5 and subtract with equation 2.

Then $0.25B= 10$        $B=40$ liters.

Putting the value of B in equation 01.

Then $A+40=100$         A $= 60$ liters.

Thus, the quantity of A and B solutions are 60 liters and 40 liters, respectively.