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

It is currently 15 Oct 2019, 19:12

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Some part of a 50% solution of acid was replaced with an equal amount

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Manager
Manager
avatar
Joined: 23 May 2005
Posts: 227
Location: Sing/ HK
Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post Updated on: 22 Jun 2019, 02:13
2
1
10
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

82% (01:23) correct 18% (01:59) wrong based on 307 sessions

HideShow timer Statistics

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. \(\frac{4}{5}\)


M25Q30

_________________
Impossible is nothing

Originally posted by Hermione on 18 Nov 2006, 20:56.
Last edited by Bunuel on 22 Jun 2019, 02:13, edited 1 time in total.
Renamed the topic and edited the question.
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58340
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 16 Jun 2013, 23:07
2
6
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.

Answer: C.

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

Answer: C.
_________________
Most Helpful Community Reply
Manager
Manager
avatar
Joined: 21 Mar 2005
Posts: 95
Location: Basel
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 19 Nov 2006, 07:50
6
1
perhaps the fastest way is :
1/2*x + 3/10* x = 0.4
->x= 1/2
General Discussion
VP
VP
avatar
Joined: 17 Jun 2008
Posts: 1174
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 01 Dec 2008, 07:11
4
2
Look at it this way.

Total part is 1. Out of this, x part was replaced.

Hence, (1-x)*50 (percent) + x*30 = 1*40
or, 50 - 20x = 40
or, 10 = 20x
or, x = 1/2....or, 50% of original solution was replaced.
Intern
Intern
avatar
Joined: 18 Oct 2009
Posts: 5
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 30 Dec 2009, 15:40
1
1
C. 1/2

In the table below, this refers to the original solution, not the acid concentration. Sorry about the formatting...it should be a 3X4 table with one expression in each box.

Original Removed Added Result
Concentration 0.50 0.50 0.70 0.60
Amount 1 x x 1
Multiply 0.50 0.50x 0.70x 0.60

0.50 (Original amount) - 0.50x (Removed) + 0.70x (Added) = 0.60 (End Result)
0.20x = 0.10
x = 1/2

Explanation:
Because the question asks for the original solution, you can change the numbers to reflect that. If 50% of the solution is acid, then 50% of the solution is the original solution. Then you're removing 50% concentration of acid (which is also 50% original solution) and then adding the same amount back but with 30% concentration of acid (also 70% concentration of original solution). What I mean by that is if you remove 50% of the acid, then you are also removing 50% concentration of the original solution. If you are adding an amount with 30% solution of acid, then you are also adding 70% of the original solution. The end result should be 40% acid---60% original solution, which is the focus since the question asks for the original amount of the solution.

Looking at the table, the original amount is 50% of let's say, 1 unit. The original solution is 0.50. You remove 50% concentration of original solution of some amount x and then you add back the same amount x but with 70% of original solution concentration. The end result is 60% original solution. Multiply down, then remove and add terms and solve for x. The amount you replaced, or x, is 1/2.

Try this problem: mixture-problem-78345.html and see how you do.

I had trouble with mixture problems, too, but I came across the user Economist's link, http://www.onlinemathlearning.com/mixture-problems.html, and this is where I got the above strategy from. You should check it out.
Senior Manager
Senior Manager
User avatar
Joined: 22 Dec 2009
Posts: 253
GMAT ToolKit User
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 01 Jan 2010, 10:35
1
Another way to look at this problem could be this.

Let the total solution be T and the part removed be P.

Now T has 50% acid
And P is removed from T.

Therefore left over acid (by quantity) would be:
\(\frac{50}{100} * (T-P) = \frac{T-P}{2}\)

to this P was added which had 30% acid in it. Hence added acid quanity is:
\(\frac{30}{100} * P\)

The resultant mixture had 40% acid in it. Hence the resultant acid quantity is:
\(\frac{40}{100} * T = \frac{T-P}{2} + \frac{30}{100}* P\)


Solving for P, we get \(P = \frac{T}{2}\) which gives the part as \(\frac{1}{2}\) of Original

Therefore answer is C

Hope this helps!

Cheers!
JT
_________________
Cheers!
JT...........
If u like my post..... payback in Kudos!! :beer

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|


~~Better Burn Out... Than Fade Away~~
Intern
Intern
avatar
Joined: 17 Jan 2010
Posts: 21
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 23 Jan 2010, 03:00
2
since the replaced amount is equal, let x be the amount replaced.

so, A - xA + xB

A - xA + xB = 0.4

so,

0.5 - 0.5x + 0.3x = 0.4
-0.5x + 0.3x = -0.1
-0.2x = -0.1
\(x = \frac{1}{2}\)

IMO, it's C
Manager
Manager
User avatar
B
Joined: 15 Sep 2009
Posts: 186
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 17 Sep 2012, 05:14
3
Alligative Method is such a time-saving method.

Here we go:

30%-----40%-----50%
Distances 1:1

This means that both contribute an equal amount of the final solution.

Straight C: 1/2

Cheers,
Der alte Fritz.
_________________
+1 Kudos me - I'm half Irish, half Prussian.
Senior Manager
Senior Manager
avatar
Joined: 15 Sep 2011
Posts: 310
Location: United States
WE: Corporate Finance (Manufacturing)
GMAT ToolKit User
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 16 Jun 2013, 16:55
4
2
If 50% solution becomes 40% solution, then 40% solution must be equally weighted between 50% solution and 30% solution - i.e. \(\frac{1}{2}\) of 50% solution and \(\frac{1}{2}\) of 30% solution. If the solution must be equally weighted, then 30% solution must replace \(\frac{1}{2}\) of 50% solution (assuming original solution was 50% solution of acid).

Just another way of saying the same thing. Cheers
SVP
SVP
User avatar
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1750
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 10 Jul 2014, 02:09
3
Did using allegation method

Refer diagram below

Answer = 1/2

Bunuel, kindly update the OA.

Thanks in advance :)
Attachments

all.png
all.png [ 3.61 KiB | Viewed 7284 times ]


_________________
Kindly press "+1 Kudos" to appreciate :)
Veritas Prep GMAT Instructor
User avatar
V
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 20 Jul 2015, 06:11
4
chrissy28 wrote:
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. \(\frac{4}{5}\)

M25Q30


Responding to a pm:

It is worded to look like a replacement question but essentially, it is a simple mixtures problem only.

You have some 50% solution of acid and you remove some of it. Now to that 50% solution, you are adding 30% solution to get 40% solution. The ratio in which you mix the 50% solution and the 30% solution is given by w1/w2 = (50 - 40)/(40 - 30) = 1/1
Basically, you mix equal parts of 50% solution and 30% solution.
So initially you must have had 2 parts of 50% solution out of which you removed 1 part and replaced with 1 part of 30% solution. So you replaced half of the original solution.

The first question of this post discusses exactly this concept: http://www.veritasprep.com/blog/2012/01 ... -mixtures/
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Manager
Manager
User avatar
B
Joined: 11 May 2018
Posts: 99
Location: India
GMAT 1: 460 Q42 V14
GMAT ToolKit User
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 03 Sep 2018, 04:21
Hello chetan2u .

I was unable to interpret the question and the above solutions .
could you please explain
Thank you.
_________________
If you want to Thank me Give me a KUDOS
"I’ve spent months preparing for the day I’d face you. I’ve come a long way, GMAT"- SonGoku
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 7960
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 03 Sep 2018, 04:30
1
SonGoku wrote:
Hello chetan2u .

I was unable to interpret the question and the above solutions .
could you please explain
Thank you.


Hi..

The question means that initially a 50% solution of acid is there. 50%solution of acid means the percentage of acid is 50%, so of the solution is 2 litre , the acid is 50% or 1 litre.
Later some of it is changed with a 30% solution that is a solution which has 30% as acid, so if you took 2 litre, acid was 2000*30/100=600ml
And finally it became 40% acid...

These questions are best done through weighted average method..
So two solutions 50% and 30% are mixed together to get 40%..
Since 40% is half way between 30 and 50, we can say that we took half of each..
Otherwise the weight of 30% = (50-40)/(50-30)=1/2 ....
_________________
Target Test Prep Representative
User avatar
D
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8048
Location: United States (CA)
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 11 Sep 2018, 17:13
1
samrand wrote:
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. 1/5
B. 1/4
C. 1/2
D. 3/4


Let the original amount of the 50% solution of acid = 10 liters (so 5 liters are acid). Let x = the amount of solution that was replaced. So 0.5x acid was removed and replaced with 0.3x acid. We can create the equation:

(5 - 0.5x + 0.3x)/(10 - x + x) = 0.4

(5 - 0.2x)/10 = 4/10

5 - 0.2x = 4

1 = 0.2x

5 = x

Since 5/10 = 1/2, we see that 1/2 of the original solution was replaced.

Answer: C
_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
TTP - Target Test Prep Logo
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Intern
Intern
avatar
Joined: 30 Jan 2019
Posts: 1
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 12 Feb 2019, 00:07
my thought process was to use the process of elimination strategy, so I started with the middle answer of 1/2.

if half of the original solution was replaced that means I am left with:

50% of 50% acid solution + 50% of 30% solution = new acid solution (which happens to be 40% acid)

.5x.5 = .25
.5x.3 = .15

.25+.15 = .4. So, my first attempt was the answer.


Let's say the new acid solution was 35% instead of 40%, then after trying the 1/2 answer, I would realize that I need more of the 30% solution to bring down the amount of acid. From this point, I can eliminate A,B, and C.

I would then try 3/4 of the 30% solution.

.25*.5 + .75*.3 = new solution
.125+.225 = .35, which is the answer I was looking for.

bing bing bing
Director
Director
avatar
G
Joined: 09 Mar 2018
Posts: 996
Location: India
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 12 Feb 2019, 00:28
chrissy28 wrote:
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. \(\frac{4}{5}\)


M25Q30


so we have a mixture in which
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid.

40% solution of acid was obtained

By allegation method, we get the ratio
M1--------M2
50----------30
-------40-----
10-----------10

Mixture will be in the ratio of 1:1

what part of the original solution was replaced?

Total mixture = 2

1/2 was replaced

C
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.
Manager
Manager
User avatar
S
Joined: 25 Apr 2018
Posts: 76
Location: India
Concentration: Strategy, Finance
GMAT 1: 600 Q40 V33
Reviews Badge CAT Tests
Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

Show Tags

New post 04 Sep 2019, 10:50
We know this,

w1/w2=A2-Avg/Avg-A1

ie

w1/w2=50-40/40-30

w1/w2=1

originally 50% alcohol was 1
now it is 1/2

so reduction is 1/2
_________________
Procrastination kills ,Kudos doesn't 8-)
Press Kudos if my post helps you!
GMAT Club Bot
Re: Some part of a 50% solution of acid was replaced with an equal amount   [#permalink] 04 Sep 2019, 10:50
Display posts from previous: Sort by

Some part of a 50% solution of acid was replaced with an equal amount

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne