Answer:
A) 20+0.6![x](https://tex.z-dn.net/?f=x)
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 = ![\frac{40}{100} \times 50 = 20\ litre](https://tex.z-dn.net/?f=%5Cfrac%7B40%7D%7B100%7D%20%5Ctimes%2050%20%3D%2020%5C%20litre)
litres are removed.
Amount of acid removed = 40% of
litre.
Now, remaining acid in the container = (20 - 40% of
) litre
Now, replaced with 100% acid.
So, final acid in the container = (20 - 40% of
+ 100% of
) litre
Amount of acid in the final mixture:
![20 - \dfrac{40}{100} \times x + \dfrac{100}{100} \times x\\\Rightarrow 20 +\dfrac{100-40}{100}x\\\Rightarrow 20 +\dfrac{60}{100}x](https://tex.z-dn.net/?f=20%20-%20%5Cdfrac%7B40%7D%7B100%7D%20%5Ctimes%20x%20%2B%20%5Cdfrac%7B100%7D%7B100%7D%20%5Ctimes%20x%5C%5C%5CRightarrow%2020%20%2B%5Cdfrac%7B100-40%7D%7B100%7Dx%5C%5C%5CRightarrow%2020%20%2B%5Cdfrac%7B60%7D%7B100%7Dx)
Answer A) Amount of acid in the final mixture = 20+0.6
Answer B)
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:
![20+0.6x =25\\\Rightarrow 0.6x =5\\\Rightarrow \bold{x =8.33\ litres}](https://tex.z-dn.net/?f=20%2B0.6x%20%3D25%5C%5C%5CRightarrow%200.6x%20%3D5%5C%5C%5CRightarrow%20%5Cbold%7Bx%20%3D8.33%5C%20litres%7D)