Mixture Problem

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mynn: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Tue Apr 15, 2008 12:55 am; Location: Malaysia

Mixture Problem

by mynn » Tue Jul 22, 2008 12:36 am

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

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Source: Beat The GMAT — Problem Solving | Tue Jul 22, 2008 12:36 am

sudhir3127: Legendary Member; Posts: 829; Joined: Mon Jul 07, 2008 10:09 pm; Location: INDIA; Thanked: 84 times; Followed by:3 members

by sudhir3127 » Tue Jul 22, 2008 3:03 am

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

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Xins: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Mon Oct 20, 2008 7:11 am; GMAT Score:590

Mixture problem

by Xins » Fri Oct 31, 2008 12:19 pm

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.

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tpcool: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Tue Jul 01, 2008 7:15 am

by tpcool » Thu Nov 27, 2008 5:28 pm

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

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logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pm; Thanked: 237 times; Followed by:25 members; GMAT Score:730