GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Feb 2019, 09:42

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT Algebra Webinar

February 17, 2019

February 17, 2019

07:00 AM PST

09:00 AM PST

Attend this Free Algebra Webinar and learn how to master Inequalities and Absolute Value problems on GMAT.
• ### Free GMAT Strategy Webinar

February 16, 2019

February 16, 2019

07:00 AM PST

09:00 AM PST

Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

# Some part of the 50% solution of acid was replaced with the

Author Message
Manager
Joined: 30 Sep 2010
Posts: 55
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

### Show Tags

07 Nov 2010, 09:10
If you notice average of 50 and 30 = 40
That means, taking the weighted average, we know 50% and 30% solution should be of the same quantity.

That mean 1/2 part of 50% solution was replaced by 30% solution
Manager
Joined: 14 May 2007
Posts: 180
Location: India
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

### Show Tags

08 Nov 2010, 07:10
Thanks a lot guys for your help.
Senior Manager
Status: Up again.
Joined: 31 Oct 2010
Posts: 489
Concentration: Strategy, Operations
GMAT 1: 710 Q48 V40
GMAT 2: 740 Q49 V42
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

### Show Tags

11 Feb 2011, 14:11
Wonderful stuff this!!

If anyone is looking for a tabular approach, here is a link you may find useful:

_________________

My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html

Retired Moderator
Joined: 20 Dec 2010
Posts: 1795
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

### Show Tags

11 Feb 2011, 23:55
One may solve it by alligation as well.

50 30
\ /
40
/ \
10 10

50 - Percent of acid in first solution
30 - Percent of acid in second solution
40 - Percent of acid in mixture of both
LHS 10 -> Percent of acid in mixture of both - Percent of acid in second solution = 40-30 -> Portion of the second in the mixture
RHS 10 -> Percent of acid in first solution - Percent of acid in mixture of both = 50-40 ->Portion of the first in the mixture

$$\frac{Portion \quad of \quad second}{Portion \quad of \quad first}=\frac{10}{10}=\frac{1}{1}$$
$$\frac{Portion \quad of \quad first}{Total \quad mixture}=\frac{1}{1+1}=\frac{1}{2}$$

If the new mixture contains 1/2 of the first solution; then half of the first solution must have been replaced with second solution.

Ans: "C"
_________________
Senior Manager
Joined: 08 Nov 2010
Posts: 331
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

### Show Tags

12 Feb 2011, 11:16
Karishmas system her works great.

the 50% "pulls" as hard as the 30% so if its going to 40% it means both "pulls" exactly the same.

so it means we have the same amount of 30% and 50% solution. so we have 1/2 replaced.
_________________
Manager
Joined: 19 Dec 2010
Posts: 99
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

### Show Tags

12 Mar 2011, 22:27

for replacement problems use the following formula.

Whole - part removed + part added = New total.
Assume 100 is the total here
Therefore, 50% (100) - 50% (x) + 30% (X) = 40% (100)

Solve...
Manager
Joined: 20 Dec 2010
Posts: 186
Schools: UNC Duke Kellogg
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

### Show Tags

26 Jun 2011, 16:48
Good problem this...took more more than 2 mins -- but C it is. Enjoyed reading the explanations -- which seem to use much more efficient methods.
Intern
Joined: 09 Jun 2011
Posts: 21
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

### Show Tags

28 Jun 2011, 11:19
Seems like it's easier just to solve it straight.
(30%+50%)/2=40%. So the amount replaced was 50-30 or 20%.
Or am I missing something?
Manager
Joined: 20 Dec 2010
Posts: 186
Schools: UNC Duke Kellogg
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

### Show Tags

28 Jun 2011, 13:05
imerial wrote:
Seems like it's easier just to solve it straight.
(30%+50%)/2=40%. So the amount replaced was 50-30 or 20%.
Or am I missing something?

I don't think this is correct.

You have a 50% solution and a 30% solution. You are replacing part of the 50% solution with the 30% solution to make a 40% solution.

There are lots of different approaches to solve the problem.

Intuitively -- you know that 40% is in between 30% and 50% -- so 1/2 of the 50% solution will need to be replaced.

You can use algebra and solve this or use the tables. You can also use the "allegation" method -- which is quite useful in these problems. I don't quite follow how you got 20%.
Intern
Joined: 09 Jun 2011
Posts: 21
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

### Show Tags

28 Jun 2011, 19:25
Sorry, the 20% was a typo. My thought process was:

1) We have a 50% solution.
2) We want a 40% solution.
3) We have at our disposal a 30% solution.
4) 40 is the average of 50 and 30, so the solution needs to be half 50 and half 30.
5) Therefore, we replace half the 50% solution with 30% solution.
6) Now instead of being all 50 we're half 50 half 30. That means we replaced half the solution.

Does that work or have I misunderstood? I saw all the complex math in the answers and got confused as to why it was necessary.
Manager
Joined: 20 Dec 2010
Posts: 186
Schools: UNC Duke Kellogg
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

### Show Tags

28 Jun 2011, 20:13
imerial wrote:
Sorry, the 20% was a typo. My thought process was:

1) We have a 50% solution.
2) We want a 40% solution.
3) We have at our disposal a 30% solution.
4) 40 is the average of 50 and 30, so the solution needs to be half 50 and half 30.
5) Therefore, we replace half the 50% solution with 30% solution.
6) Now instead of being all 50 we're half 50 half 30. That means we replaced half the solution.

Does that work or have I misunderstood? I saw all the complex math in the answers and got confused as to why it was necessary.

Yeah..that's perfect.
Manager
Joined: 25 May 2011
Posts: 116
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

### Show Tags

07 Nov 2011, 02:24
lonewolf wrote:
Hermione wrote:
Does anyone have a quick and easy way of solving this problem? I had to pick from the answer choices but I want to know how to solve it outright.

Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution f acid was obtained. what part of the original solution was replaced?

1/5
1/4
1/2
3/4
4/5

Let X = Part of acid solution replaced, then 1-X will be the parts not replaced.

(1-X)*0.5 + 0.3*x = 0.4
0.5 - 0.5X +0.3X = 0.4
-0.2X = -0.1
X=1/2

Good explanation
Thank you
Math Expert
Joined: 02 Sep 2009
Posts: 52902
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

### Show Tags

06 Sep 2014, 18:54
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?

A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
E. $$\frac{4}{5}$$

Since the concentration of acid in resulting solution (40%) is the average of 50% and 30% then exactly half of the original solution was replaced.

Alternately you can do the following, say $$x$$ part of the original solution was replaced then: $$(1-x)*0.5+x*0.3=1*0.4$$ --> $$x=\frac{1}{2}$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: some-part-of-a-50-solution-of-acid-was-replaced-with-an-68805.html
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 9835
Re: Some part of the 50% solution of acid was replaced with the  [#permalink]

### Show Tags

11 Oct 2018, 10:00
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Some part of the 50% solution of acid was replaced with the   [#permalink] 11 Oct 2018, 10:00

Go to page   Previous    1   2   [ 34 posts ]

Display posts from previous: Sort by