Solution/Mixture problem

[manjeshmurthy]

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

On general i have difficulty in solving this kind of solution/mixture problems.
i was trying to refer Manhattan strategy guide for some hints to solve this. I found 3 places related to this topic - weight average on page 107(word translations) or percentage strategy page 54/page 96 & all talk about different approaches.
Im kinda confused here..
pls guide me here with some tips to attack this kind of problem

Appreciate your help

[ChrisB]

Hi,

What's the source of this problem? It's worded pretty poorly so my assumption is that this problem is a "home brew" GMAT problem crafted by someone on another GMAT forum.

Please provide a source and we'll get after this problem. Alternatively, if you can find a mixture problem from a "good source" such as our CAT, GMAT Prep, or our strategy guides I'd be more than happy to help!

Thanks,
Chris

[manjeshmurthy]

Hello Chris,

Thanks for jumping in to help.
the source is crack-gmat, free practice test.

[mschwrtz]

Although, as Chris notes, this question is horribly written, it does have one very interesting feature: you are asked what portion of the stronger solution is "replaced" by the weaker solution, while the typical mixture problem would ask what portion of a mixture consists of the weaker solution. Now, those are exactly the same question mathematically, but that might not be plain even to someone practiced with mixture problems.

A few notes:

1) The various methods you've found confusing are all consistent with one another. Choose the one you find most congenial and run with that one.

2) Notice that the concentration of the mixture is much closer to 40% than to 25%, and so the mixture must have much more of the 40% solution than of the 25% solution.

3) In fact, the concentration of the mixture is twice as close to 40% as to 25%, so the mixture must have twice as much of the strong solution as of the weak.

4) That's the same reasoning as you'll find with that picture of a see-saw/number line in the WT guide.

5) The point of the mixture-problem table in the WT guide is to produce two equations with the same two variables, which you can then solve algebraically. Those equations are the same as the equations produced by the more explicitly algebraically approach you've seen in the FDP book.

6) For this problem, the equations would be

x+y=100
.4x+.25y=35
solve for y/(x+y)

[manjeshmurthy]

I couldnt get your name, but thank you for the explanation.

good thing is on solving the equation you mentioned i got the right answer.
x/(x+y) = 100/3 / 100

100/300 = 1/3

now on the explanation, as im still trying to understand

.4x+ .25y = 35

1) I assume x & y are the quantities of different concentration(25% & 40%) making upto 100.
here you are adding to quantities when question says its "replacing" stronger by weaker concentration

2) should it not be equation written as .4x + 2.5y = .35(x+y)
or FDP way(page 97) using mixture chart to get equation
40 + .25y = .35(100+x)
both these equations lead to same answer.

3) since the question talks about "replace"
can i write equation as (100-n)40 + n25 = 35(100) considering we have solution of total 100 & n is quantity replaced.

as i finish writing i think im understanding whats going on here
but your inputs will help to connect dots

[tim]

aside from some minor typographical errors, it looks like you totally understand this one, and the two alternatives you proposed are equivalent to each other and to Michael's solution. nice job..

[emailnaik]

I have a hard time understanding mixtures and allegations; this problem baffles me further.

Can anyone please detail out a step by step methodology for solving the above problem that is easy to comprehend.

Thanks.

[jnelson0612]

5) The point of the mixture-problem table in the WT guide is to produce two equations with the same two variables, which you can then solve algebraically. Those equations are the same as the equations produced by the more explicitly algebraically approach you've seen in the FDP book.

6) For this problem, the equations would be

x+y=100
.4x+.25y=35
solve for y/(x+y)

emailnaik,
Go back and take a look at Michael's excellent answer, particularly his last two points. We want to obtain two equations using the same two variables, so they can be combined to solve for the variables.

In this case, x is the percentage amount of the old mixture and y is the percentage amount of the new mixture. x + y will equal 100, or 100% of the new mixture.

Looking at the second equation, we know that x has a 40% solution and y has a 25% solution. We know that combining the two into the new mixture has a result of a 35% solution. Thus, .40x + .25y = 35.

Now, we can say that if x+y=100, x=100-y and plus that into our second equation, solving for x. Once we have a value for x, we can plug it back in and easily obtain a value for y.

Finally, we know that the new solution's percentage of y is the amount of y divided by the total of the entire solution (amount of x+y). There's our answer.

I hope this helps. There was another excellent point made about x contributing 40% and y contributing 25%, and the resulting 35% obviously coming from a larger portion of x than y. You can often ballpark the problem very successfully using this approach.

[emailnaik]

Thanks Jamie; I got the underlying basis on how to approach the problem and deduce the solution. Although Michael's approach was excellent I was getting lost in the methods outlined. Now I've better clarity than before. Appreciate your help.

[jnelson0612]

My pleasure emailnaik. Best wishes!

[Angus.I.Hsu]

PS 223 in the 12th edition gives a real mixture problem.

One comment though. It was not an easy question but why did it warrant a 700-800 difficulty in the companion guide? I guesstimated and reasoned it out in about 1 minute 30 seconds- getting the answer correct (33 1/3 percent).

[RonPurewal]

PS 223 in the 12th edition gives a real mixture problem.

One comment though. It was not an easy question but why did it warrant a 700-800 difficulty in the companion guide? I guesstimated and reasoned it out in about 1 minute 30 seconds- getting the answer correct (33 1/3 percent).

this is not something that you should bother to think about.
read here:
http://www.manhattangmat.com/blog/index ... 02/05/446/

[rohilla_sandeep]

Although it has been explained nicely, I am giving one more approach :
Let the fraction replaced is x, therefore,
40/100(1-x) + 25/100 x = 35/100
Solving it we get: x = 1/3

[jnelson0612]

Although it has been explained nicely, I am giving one more approach :
Let the fraction replaced is x, therefore,
40/100(1-x) + 25/100 x = 35/100
Solving it we get: x = 1/3

That works too, rohilla. Thank you.

[abhinav.hsci]

mix n solutions probs can be solved with some simple tactics

in this case:

original mix- 40 %
replacing mix- 25%
resulting mix- 35%

now, 40%-35%= 5 (gives no. of parts of weaker mix)
&, 35%-25%= 10 (gives no. of parts of stronger mix)

thus, 10:5= 2:1
so to get a resultant mix of 35% we need to mix 40% mix and 25% mix in ratio of 2:1

thus the ans 2/3 of 40% mix + 1/3 of 25% mix = 35% mix

thus replaced amount= 1/3 (ans)

all mixture problems can be solved with this method.

-Abhi