Joy Shaha wrote: ↑Fri Jul 15, 2016 5:42 am
Q.A certain quantity of 40% solution is replaced with 25% solution such that the new concentration is 35%. What is the fraction of the solution that was replaced?
(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) 3/4
When solving mixture questions, we can often get a much clearer picture by sketching the solutions with the components
separated
Since the question asks us to find a certain FRACTION, let's assign a nice value to the starting volume.
So let's start with 100 ml of a solution that's 40% salt (Why salt? Why not?).
If we separate the two components, we get the following:
Now let's REMOVE x ml of solution.
This means 100 - x = the resulting volume of the entire solution
Since 40% of this volume is salt, the resulting volume of salt = 40% of (100 - x) ml
In other words: the resulting volume of salt = 0.4(100 - x) ml
Expand to get: the resulting volume of salt =
40 - 0.4x ml
Now let's ADD x ml of the 25% solution.
The volume of salt in this solution = 25% of x
In other words: the volume of salt in this solution =
0.25x ml
To determine how much salt there is in the resulting mixture, simply add the volume of salt from the two mixtures we're combining
In other words, the TOTAL amount of salt = (
40 - 0.4x ml) +
0.25x ml
=
40 - 0.15x ml
We get:
Since we are told the resulting mixture is 35% salt, we can write:
(40 - 0.15x)/100 = 35/100
This means that: 40 - 0.15x = 35
Rearrange to get: 5 = 0.15x
Solve: x = 5/0.15 = 500/15 = 100/3 = 33 1/3
In other words, 33 1/3 mls were removed from the original 100 ml of solution.
In other words 1/3 of the solution was removed
Answer: B
Cheers,
Brent
