Turkish
A certain fraction of a 15% salt solution by weight was replaced by a 25% salt solution by weight. The resulting solution has 21 % salt by weight. What fraction of the original solution was replaced?
A. 1/2
B. 3/5
C. 2/3
D. 7/10
E. 3/4
Weighted average with one variable approach.
A is the original solution
B is the resultant solution
x = the amount (volume) replaced:
1) x is a volume of
salt, both removed and replaced. The ingredient is the same. You don't need a second variable;
2) Volume removed = volume replaced;
3) However, the concentration (percentage) of salt is different in the volume removed and the volume replaced. Same volume, different concentrations of salt;
4) x, the volume removed
and replaced is a part of the whole, A. Part x of whole A = fraction of A. Answers are all 0 < x < 1. Assume A's volume = 1
5) Volume of A and volume of B are equal.
Original, (salt concentration)(volume): (.15)(1)
Removed: (.15)(x)
Replaced: (.25)(x)
Resultant solution, (salt concen.)(volume): (.21)(1)
Original - removed + replaced = Resultant solution
(.15)(1) - .15(x) + .25(x) = .21(1)
Multiply all by 100 to make arithmetic easier.
(15)(1) - 15x + 25x = 21(1)
10x = 6
x = \(\frac{6}{10}=\frac{3}{5}\)
Answer B
--10x = 6 might seem odd. Fractional replacements are hard. Think in volumes first, then in fractions.
--Amount x = a volume. A has a volume of 1. x is the unknown volume removed and replaced. And it is not all of A.
--Fractions: all of A did not go. Only part of A's volume got removed and replaced. (Part) x of (whole) A got removed and replaced. \(\frac{part}{whole}\) = fraction.