Question

How many L of A 90% saline solution must be mixed with 4 L of a 50% saline solution to make a 74% solution ?

Answer 1

Answer:

6 L

Step-by-step explanation:

Given two solutions:

1st solution is 90% saline solution. Let the volume of this solution be $x$ litres.

2nd solution is 50% saline solution. Volume of this solution = 4 L

Resultant solution is 74% saline.

To find:

The volume of 1st solution = ?

Solution:

Total volume of the 74% saline mixture = (4+$x$) Litres

We can write equation here, as per the percentage of saline in the mixtures.

90% of $x$ L + 50% of 4 L = 74% of ($x$+4)

$\Rightarrow \dfrac{90x}{100} + \dfrac{50}{100}\times 4 = \dfrac{74}{100}\times (x+4)\\\Rightarrow 90x + 200 = 74x+296\\\Rightarrow 90x-74x = 296-200\\\Rightarrow 16x = 96\\\Rightarrow x = \bold{6\ L}$

Therefore, the volume must be 6 L.