How Many Liters Of 24% Solution Do I Need To Add To A 40% Solution To Get 30% Solution
Math Expert
Joined: 02 Sep 2009
Posts: 86392
How many liters of a 40% iodine solution need to be mixed with 35 light [#permalink] 
09 Jun 2016, 01:31
00:00
Question Stats:
79% (01:45) right 21% (02:07) incorrect based on 335 sessions
Hide Show timer Statistics
How many liters of a twoscore% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?
A. 35
B. 49
C. 100
D. 105
E. 140
_________________
GMAT Guild Legend
Joined: 11 Sep 2015
Posts: 6683
Location: Canada
Re: How many liters of a 40% iodine solution need to be mixed with 35 lite [#permalink] 29 Mar 2017, 05:36
Bunuel wrote:
How many liters of a 40% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?
A. 35
B. 49
C. 100
D. 105
East. 140
One approach is to sketch the solutions with their components separated.
Let's start with the 35 liters of twenty% iodine solution:
20% of 35 liters = 7 liters, so the original solution contains seven liters of iodine.
To this solution, we'll add 10 liters of the solution that'due south 40% iodine.
40% of x liters = 0.4x liters. So, we're adding 0.4x liters of iodine.
When we combine the two solutions, it's easy to see how the resulting mixture looks, since we need just add the volumes of iodine in each solution.
Equally you can see, our resulting mixture has a TOTAL volume of 35 + x liters.
Nosotros can besides see that the resulting mixture contains 7 + 0.4x liters of iodine.
Since the resulting mixture must past 35% (aka 35/100) iodine, we tin write the following equation:
(7 + 0.4x)/(35 + x liters) = 35/100
Offset simplify 35/100 to get: (seven + 0.4x)/(35 + x liters) = 7/twenty
Cantankerous multiply to get: 20(7 + 0.4x) = vii(35 + 10 liters)
Expand: 140 + 8x = 245 + 7x
Solve: x = 105
Answer: D
RELATED VIDEO
_________________
Brent Hanneson – Creator of gmatprepnow.com
Math Expert
Joined: 02 Aug 2009
Posts: 10494
Re: How many liters of a xl% iodine solution need to be mixed with 35 lite [#permalink] 09 Jun 2016, 01:51
Bunuel wrote:
How many liters of a 40% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?
A. 35
B. 49
C. 100
D. 105
E. 140
These Qs are best done with weighted Average method....
Two solutions - 10 of 40% and 35 of 20% results in a solution of 35% solution..
And then ratio = \(\frac{QTY- of- 40}{Qty- of- twenty%} =\frac{35-20}{40-35} = \frac{15}{5}.....\)
and so \(\frac{x}{35} = \frac{15}{v} = 3..................x = 35*3 = 105\)
D
_________________
SVP
Joined: 06 Nov 2014
Posts: 1824
Re: How many liters of a 40% iodine solution need to be mixed with 35 low-cal [#permalink] 09 Jun 2016, 01:44
Bunuel wrote:
How many liters of a xl% iodine solution demand to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?
A. 35
B. 49
C. 100
D. 105
Due east. 140
Solution 1:
Assume the iodine solution to be mixed = x lts.
Iodine = 0.4x lts, Water = 0.6x lts.
Solution 2: 35 liters of a 20% iodine solution
Iodine = 7 lts, H2o = 28 lts.
Total iodine = 0.4x + seven
Total water = 0.6x + 28
The resultant is a 35% idoine solution.
Hence (0.4x + 7) / (x + 35) = 35/100
40x + 700 = 35x + 1225
5x = 525
10 = 105 lts
Correct Choice: D
Board of Directors
Joined: 01 Sep 2010
Posts: 3591
Re: How many liters of a 40% iodine solution need to be mixed with 35 lite [#permalink] 09 Jun 2016, 02:33
A more straight arroyo
\(0.4X + 0.two Y = 0.35X + 0.35 Y\)
\(0.05X = 0.15Y\) AND our \(Y = 35\)
\(0.05X = 5.25\)
\(X = \frac{5.25}{0.05} = 105\)
D is the i
_________________
Electric current Student
Joined: 18 Oct 2014
Posts: 727
Location: United States
GPA: 3.98
Re: How many liters of a 40% iodine solution need to be mixed with 35 calorie-free [#permalink] 09 Jun 2016, 08:32
Bunuel wrote:
How many liters of a 40% iodine solution need to exist mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?
A. 35
B. 49
C. 100
D. 105
E. 140
threescore% water solution mixed with eighty% h2o solution to get 65% water solution.
Solution will be mixed in the ratio::-
x/35= 80-65/65-60
ten/35= 15/v
x= 3*35= 105
D is the answer
_________________
I welcome disquisitional assay of my mail!! That volition aid me reach 700+
Intern
Joined: 11 Aug 2013
Posts: 44
Re: How many liters of a twoscore% iodine solution need to be mixed with 35 low-cal [#permalink] 09 Jun 2016, 08:44
D it will be
weighted avg concept :-
20 --------------------35-------------------40
qyantity of 20%/quantity of 40% = xl- 35/35-20
35/x = 5/xv
ten = 105
Senior Manager
Joined: 21 Mar 2016
Posts: 464
Re: How many liters of a 40% iodine solution need to be mixed with 35 lite [#permalink] ten Jun 2016, 05:07
Divyadisha wrote:
Bunuel wrote:
How many liters of a 40% iodine solution need to be mixed with 35 liters of a xx% iodine solution to create a 35% iodine solution?
A. 35
B. 49
C. 100
D. 105
East. 140
sixty% water solution mixed with 80% water solution to get 65% h2o solution.
Solution will be mixed in the ratio::-
x/35= 80-65/65-60
x/35= 15/five
x= 3*35= 105
D is the answer
Hi, Divyadisha ,, may i knw how did u get that formula
Senior Manager
Joined: 21 Mar 2016
Posts: 464
Re: How many liters of a 40% iodine solution need to be mixed with 35 lite [#permalink] 10 Jun 2016, 05:08
chetan2u wrote:
Bunuel wrote:
How many liters of a 40% iodine solution need to exist mixed with 35 liters of a xx% iodine solution to create a 35% iodine solution?
A. 35
B. 49
C. 100
D. 105
Due east. 140
These Qs are best done with weighted Boilerplate method....
Two solutions - x of 40% and 35 of twenty% results in a solution of 35% solution..
SO ratio = \(\frac{QTY- of- 40}{Qty- of- 20%} =\frac{35-xx}{xl-35} = \frac{15}{5}.....\)
so \(\frac{x}{35} = \frac{15}{v} = iii..................ten = 35*3 = 105\)
D
hi chetan2u tin can u plz explain tat formula,
Manager
Joined: 16 Mar 2016
Posts: 115
Location: French republic
GPA: 3.25
Re: How many liters of a 40% iodine solution need to be mixed with 35 calorie-free [#permalink] x Jun 2016, 08:34
If x is the number of liter of the twoscore% iodine solution, we take :
0.4 *x + 0.ii * 35 = 0.35 ( 35 + x)
0.four *10 + 0.ii * 35 = 0.35 * 35 + 0.35 * x
0.05 * x = 35 (0.35 - 0.2)
0.05 * x = 35 * 0.15
x = 35 * three = 105
VP
Joined: 07 Dec 2014
Posts: 1142
Re: How many liters of a xl% iodine solution demand to be mixed with 35 lite [#permalink] x Jun 2016, 19:17
let x=liters of 40% iodine solution needed
.4x+(.two)(35)=.35(x+35)
x=105 liters
Board of Directors
Joined: xi Jun 2011
Status:QA & VA Forum Moderator
Posts: 5673
Location: Republic of india
GPA: 3.5
WE:Business concern Evolution (Commercial Banking)
Re: How many liters of a 40% iodine solution need to be mixed with 35 low-cal [#permalink] 11 Jun 2016, 04:35
Bunuel wrote:
How many liters of a 40% iodine solution demand to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?
A. 35
B. 49
C. 100
D. 105
E. 140
Attachment:
Capture.PNG [ 4.45 KiB | Viewed 12679 times ]
35% ----> 7 + 0.4x
100%---->\(\frac{(7 + 0.4x)}{35} *100\) = \(\frac{(7 + 0.4x)}{7}*20\)
Or, \(\frac{(140 + 8x)}{vii}\) = 35 + x
Or, 140 + 8x = 245 + 7x
Or, x = 245 - 140
Or, x = 105
Hence answer volition be (D) 105
_________________
GMAT Society Legend
Joined: 11 Sep 2015
Posts: 6683
Location: Canada
Re: How many liters of a xl% iodine solution demand to be mixed with 35 lite [#permalink] 29 Mar 2017, 05:49
Bunuel wrote:
How many liters of a 40% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?
A. 35
B. 49
C. 100
D. 105
E. 140
Some other approach is to use Weighted Averages:
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B boilerplate) + (grouping C proportion)(group C average) + ...
In this case, we have:
35% iodine = (proportion of 20% solution)(20% iodine ) + (proportion of 40% solution)(40% iodine)
To determine the proportions, permit's let x = the volume of 40% solution needed
Nosotros already know we have 35 liters of 20% solution
This means the Full volume = 35 + 10
So, the proportion of 20% solution 35/(35 + 10)
And the proportion of forty% solution x/(35 + x)
Plug these values into formula to get: 35 = [35/(35 + x)](20) + [x/(35 + ten)](xl)
Multiply both sides by (35 + x) to get: 35(35 + x) = (35)(20) + (x)(xl)
Aggrandize: 1225 + 35x = 700 + 40x
Decrease 700 from both sides: 525 + 35x = 40x
Subtract 35x from both sides: 525 = 5x
Solve: 10 = 105
Answer: D
RELATED VIDEO
_________________
Brent Hanneson – Creator of gmatprepnow.com
VP
Joined: 27 May 2012
Posts: 1437
How many liters of a 40% iodine solution demand to be mixed with 35 light [#permalink] xv Sep 2018, 04:30
Bunuel wrote:
How many liters of a twoscore% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?
A. 35
B. 49
C. 100
D. 105
Eastward. 140
Dear Moderator,
Please alter tag from " work/rate" to "mixture Bug ". Thank yous.
_________________
Math Proficient
Joined: 02 Sep 2009
Posts: 86392
Re: How many liters of a 40% iodine solution need to exist mixed with 35 low-cal [#permalink] xv Sep 2018, 04:31
stne wrote:
Bunuel wrote:
How many liters of a 40% iodine solution need to exist mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?
A. 35
B. 49
C. 100
D. 105
E. 140
Beloved Moderator,
Please change tag from " work/charge per unit" to "mixture Problems ". Thank you.
_________________
Washed. Thanks.
_________________
Non-Man User
Joined: 09 Sep 2013
Posts: 22930
Re: How many liters of a 40% iodine solution need to be mixed with 35 light [#permalink] 19 May 2021, 05:05
Hello from the GMAT Club BumpBot!
Thanks to some other GMAT Lodge member, I have just discovered this valuable topic, withal it had no word for over a yr. I am at present bumping it upward - doing my chore. I retrieve you lot may notice 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 contour area as well as via email.
_________________
Re: How many liters of a 40% iodine solution demand to be mixed with 35 lite [#permalink]
xix May 2021, 05:05
Moderators:
Senior SC Moderator
5460 posts
Senior Moderator - Masters Forum
2903 posts
How Many Liters Of 24% Solution Do I Need To Add To A 40% Solution To Get 30% Solution,
Source: https://gmatclub.com/forum/how-many-liters-of-a-40-iodine-solution-need-to-be-mixed-with-35-lite-219911.html
Posted by: boltonhicing.blogspot.com
0 Response to "How Many Liters Of 24% Solution Do I Need To Add To A 40% Solution To Get 30% Solution"
Post a Comment