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question in percentage

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question in percentage

by Guidetti » Tue Aug 11, 2009 11:15 pm
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


Please explain in the details how to solve it
Thank you
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by tohellandback » Tue Aug 11, 2009 11:31 pm
IMO B
let X is the fraction of solution removed

X*.25 + (1-x) * .4=.35
X=1/3.


Easier way.
if you remove half the solution the resultion concentration will be (40+25)/2=32.5
so you should remove a little less than 50 percent
answer B
The powers of two are bloody impolite!!
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by [email protected] » Wed Aug 12, 2009 10:00 pm
As per Formula:

(Mean% - % of Lower Concentration) / (% of Higher Concentration-Mean)= (Ratio of qty of Lower Conc [x] / (Ratio of qty of Higher Conc [y]

=> (35%-25%)/(40%-35%) = x/y
=> 10/5 = x/y
=> x+y =15

Therfore, Quantity of Higher conc. replaced = (x/x+y)=5/15=1/3

Ans 1/3
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by ket » Thu Aug 13, 2009 1:49 pm
Guidetti

Because you asked for a detailed explanation here is my version:

• Let’s say the total quantity of the liquid (let's say its water) is A.

• The problem tells us that 40% of A is some sort of substance (let’s say salt.)

• Then they tell us part of water A was replaced with another substance (let’s say another type of water) but this substance is 25% solution (only 25% of it is salt).

• Let’s say the removed part in quantity is B. (Notice that problem asks what part is B of A i.e. we need to find B/A)

• So look what happened: After they removed B amount of water we were left with (A-B) amount right? And hence the amount of salt in water was reduced to (A-B) X40%.

• But we know after the removal they added B quantity of another type of water by doing this it's clear that they added BX25% amount of salt (remember B contained only 25% of salt)

• Notice that because they just replaced B quantity of existing water with another type of water still the total quantity of water is A.

• But what has changes is the % of salt in A, Now the amount of salt is :
(A-B)X40%+BX25% and from the problem we know that this is equal to the 35% of A. „the new concentration is 35%“

• Therefore the equation is (A-B)X40%+BX25% =AX35%
• A5%=B15% and B/A = 1/3
• Hence the answer is B

Hope this helps.
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by ghacker » Thu Aug 13, 2009 7:15 pm
This is easy

Look at the relative positions of the mixtures , the final mixture is 35% and the initial mixture is 40% but ur adding a 25% mixture

35 is closer to 40 than to 25 so the the amount that was taken out from the 40% solution should not be that big

10 5

________25 ___________35 __________40

Hence the ratio is 2:1 so it should be 1/3

Answer is 1/3
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