Mixture Problem

mynn · Posted: Tue Jul 22, 2008 12:36 am Post subject: Mixture Problem

Hi,
I always have a problem with a mixture problem..I really do not have any idea how to solve this question.

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

Please explain. TQ

sudhir3127 · Posted: Tue Jul 22, 2008 3:03 am Post subject:

let x be the fraction replaced. if x is the fraction then the total is 1.

(1-x)*40 + x*25 = 35
40 - 40x + 25x = 35
5 = 15x
x = 1/3

Xins · Posted: Fri Oct 31, 2008 12:19 pm Post subject: Mixture problem

Guess the soultion above is the simplest but here my way to it

Let X be the total quantity
Let Y be the replaced quantity

Fraction n = Y/X . Hence the Y = nX

The problem translates to this equation

0.4X- n(0.4X) + 0.25nX = 0.35X

If you solve this equation n = 1/3.

Hope this helps.

tpcool · Posted: Thu Nov 27, 2008 5:28 pm Post subject:

Alternative explanation and approach for the same.

No. of liters of original solution = x
No. of liters replaced = y

*** NOTE that the solution is REPLACED, which means that the quantity of the resulting mixture is still same as the original solution. In this case, the quantity of resulting mixture will still be x ****

put in the equation

40% of (x - y) + 25 % y = 35 % of x

2/5 * (x - y) + y/4 = 7x/20

solving, you will get y/x = 1/3

logitech · Posted: Thu Nov 27, 2008 8:19 pm Post subject: Re: Mixture Problem

40%.....35%.......25 %

40-35=5
35-25=10

replacement : 5/15