Answer:
A) 20+0.6[tex]x[/tex]
B) range is [0, 50] (i.e. both inclusive)
C) 8.33 litres
Step-by-step explanation:
Given that concentration of acid in 50 litre container is 40%.
Amount of acid in the container = 40% of 50 litres
Amount of acid in the container = [tex]\frac{40}{100} \times 50 = 20\ litre[/tex]
[tex]x[/tex] litres are removed.
Amount of acid removed = 40% of [tex]x[/tex] litre.
Now, remaining acid in the container = (20 - 40% of [tex]x[/tex]) litre
Now, replaced with 100% acid.
So, final acid in the container = (20 - 40% of [tex]x[/tex] + 100% of [tex]x[/tex] ) litre
Amount of acid in the final mixture:
[tex]20 - \dfrac{40}{100} \times x + \dfrac{100}{100} \times x\\\Rightarrow 20 +\dfrac{100-40}{100}x\\\Rightarrow 20 +\dfrac{60}{100}x[/tex]
Answer A) Amount of acid in the final mixture = 20+0.6[tex]x[/tex]
Answer B) [tex]x[/tex] can not be greater than 50 litres (initial volume of container) and can not be lesser than 0 litres.
so, range is [0, 50] (i.e. both inclusive)
Answer C)
Given that final mixture is 50% acid.
amount of acid = 50% of 50 litres = 25 litres
Using the equation:
[tex]20+0.6x =25\\\Rightarrow 0.6x =5\\\Rightarrow \bold{x =8.33\ litres}[/tex]
