Asolution of 5% acid is mixed with a solution of 6.5% acid to make 200 ml of 6% acid. how many mls of each of the original solutions are used?

66.67 ml of 5% solution and 133.33 ml of 6.5% solution are mixed.

Step-by-step explanation:

Let 200 ml of 6% solution is formed by mixing x ml of 5% solution and y ml of 6.5% solution.

We can write the equation x + y = 200 ........ (1)

Now, x ml of 5% solution will contain ml of acid  

Again, y ml of 6.5% solution will contain ml of acid.

So, mixing them together will give (0.05x + 0.065y) ml of acid in 200 ml of solution.

So, from the condition given we can write  

⇒ 0.05x + 0.065y = 12  

⇒ x + 1.3 y = 240 ........... (2) {Dividing both sides by 0.05}

Now, solving equations (1) and (2), we get

0.3y = (240 - 200) = 40

⇒ y = 133.33 ml  

From equation (1) we get, x = 200 - 133.33 = 66.67 ml

Therefore, 66.67 ml of 5% solution and 133.33 ml of 6.5% solution are mixed. (Answer)

4x and 6y

im not sure though

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step-by-step explanation: