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Liono4ka [1.6K]
3 years ago
5

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

Mathematics
1 answer:
insens350 [35]3 years ago
8 0

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 <em>6 L</em>.

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