My question is on the explanation on the bottom, how does he arrive at (1-X) * 0,25?
Thanks,
800guy wrote:
OA coming when some people have answered..
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) ¾
Let X be the fraction of 40% solution kept
Then we have
X * 0,4 + (1-X) * 0,25 = 0,35
<=> 0,4*X + 0,25 - 0,25*X = 0,35
<=> 0,15*X = 0,10
<=> X = 2/3
Mixture Problem
This topic has expert replies
- ssmiles08
- Master | Next Rank: 500 Posts
- Posts: 472
- Joined: Sun Mar 29, 2009 6:54 pm
- Thanked: 56 times
here is how I did it.
let total solution = 100.
40% of original solution = 40(100) = 40.
amount x is taken out. = - .4(x)
same amount x is added with different % = + .25(x)
net % of total solution is .35(100) = 35.
so 40 -.4x + .25x = 35
x = 100/3
so 100/3 was the amount taken out and added in with different percentages. Total solution is 100.
so (100/3) / 100 = 1/3
IMO B.
let total solution = 100.
40% of original solution = 40(100) = 40.
amount x is taken out. = - .4(x)
same amount x is added with different % = + .25(x)
net % of total solution is .35(100) = 35.
so 40 -.4x + .25x = 35
x = 100/3
so 100/3 was the amount taken out and added in with different percentages. Total solution is 100.
so (100/3) / 100 = 1/3
IMO B.
You got a dream... You gotta protect it. People can't do somethin' themselves, they wanna tell you you can't do it. If you want somethin', go get it. Period.