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

It is currently 18 Dec 2018, 23:34

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
Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
  • Happy Christmas 20% Sale! Math Revolution All-In-One Products!

     December 20, 2018

     December 20, 2018

     10:00 PM PST

     11:00 PM PST

    This is the most inexpensive and attractive price in the market. Get the course now!
  • Key Strategies to Master GMAT SC

     December 22, 2018

     December 22, 2018

     07:00 AM PST

     09:00 AM PST

    Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.

A chemistry student is working with solution M, which has a 20% calciu

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

Hide Tags

Manager
Manager
User avatar
G
Joined: 08 Sep 2017
Posts: 77
GMAT 1: 710 Q49 V39
GMAT ToolKit User Premium Member Reviews Badge CAT Tests
A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

Show Tags

New post 12 Jul 2018, 19:17
1
12
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

53% (02:25) correct 47% (03:04) wrong based on 148 sessions

HideShow timer Statistics

A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove three-fourths of this solution and then add a new solution, solution N, to create a third solution, solution P, with double the volume of solution M. What must the calcium chloride concentration of solution N be to create a calcium chloride concentration of 37.5% in solution P?

a) 32%
b) 35%
c) 40%
d) 44%
e) 55%

_________________

Kudos please if you liked my post


Thanks!

Intern
Intern
avatar
Joined: 11 Jul 2018
Posts: 20
A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

Show Tags

New post 12 Jul 2018, 19:37
Assume you have 1000 ml M out of which 200 ml is calcium chloride leading to 20% concentration.
Remove 3/4 of M, which will mean now we have (1000/4) = 250 ml M out of which (200/4) = 50 ml is calcium chloride leading to 20% concentration.
To have 37.5% concentration in 2000 ml we should have 750 ml in 2000 ml.

2000 ml is taken as question says we need twice the volume of solution M in final solution P.
(One small assumption here that twice the volume of P meant twice from original quantity of M and not after removing 3/4th)

In this remove remaining quantities from solution M => 750 - 50 = 700 ml & 2000 - 250 = 1750 ml
These quantities need to come from solution N.

Concentration of calcium chloride in solution N = (700/1750)*100 = 40
Answer C
Manager
Manager
User avatar
S
Joined: 19 Nov 2017
Posts: 174
Location: India
Schools: ISB
GMAT 1: 670 Q49 V32
GPA: 4
Re: A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

Show Tags

New post 12 Jul 2018, 19:41
3
Consider Solution M ha a volume of 100.
calcium chloride = 20% = 20.

Solution M is reduced by \frac{3}{4}.
New volume of M = 25
calcium chloride = 5

Solution P = N + 25
We know P has double the volume of M.
Therefore, 200 = N + 25
N = 175.

Concentration of calcium chloride required in P = \(37.5%\) or \(75\) (200*37.5%)
We already have 5 units of calcium chloride from M, we need 70 more.

\(X*\frac{175}{100} = 70\)
\(X*\frac{7}{4} = 70\)
\(X = 40\)

Option C is the answer
_________________

Regards,

Vaibhav



Sky is the limit. 800 is the limit.

~GMAC

Manager
Manager
User avatar
S
Joined: 21 Dec 2014
Posts: 66
Location: Viet Nam
Concentration: Entrepreneurship, General Management
Schools: Duke '21
GMAT 1: 710 Q49 V37
GPA: 3.8
WE: Other (Other)
GMAT ToolKit User Reviews Badge
Re: A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

Show Tags

New post 12 Jul 2018, 19:51
1
Carlgrc wrote:
A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove three-fourths of this solution and then add a new solution, solution N, to create a third solution, solution P, with double the volume of solution M. What must the calcium chloride concentration of solution N be to create a calcium chloride concentration of 37.5% in solution P?

a) 32%
b) 35%
c) 40%
d) 44%
e) 55%


X=Concentration of Ca in N
M=Volume of solution M
N=volume of solution N
P=volume of solution P
Remove three-fourths of M and then add N to create P, P=2M
=> 1/4 M +N=2M=>N=7/4 M
Concentration Of Ca in P=0.375
1/4 M*0.2 + X *7/4M =3/8 *2M=>x=0.4

Anser is C
_________________

Kudo is nothing but encouragement!

Manager
Manager
User avatar
G
Joined: 08 Sep 2017
Posts: 77
GMAT 1: 710 Q49 V39
GMAT ToolKit User Premium Member Reviews Badge CAT Tests
A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

Show Tags

New post 13 Jul 2018, 17:22
1
1
The easiest way to solve this mixture problem is with the general formula (see @VeritasPrepKarishma Mixtures):

The formula is: \(\frac{w1}{w2}=\frac{A2-A}{A-A1}\)

Concentration of Ca in M = 20% = A1
Concentration of Ca in P = 37,5% = A
Concentration of Ca in N =? = A2 = N

If V is the initial volume of M:

Volume of M = w1 = 1/4 V
Volume of P = 2V
Volume of N = w2 --> 2V = 1/4 V + Volume of N --> Volume of N = 7/4 V = w2

\(\frac{1}{4} * \frac{4}{7}=\frac{N-37,5}{37,5-20}\)

\(N = \frac{375}{10} + \frac{175}{10} * \frac{1}{7}\)

N = 40

Answer C
_________________

Kudos please if you liked my post


Thanks!

Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 7113
Re: A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

Show Tags

New post 13 Jul 2018, 17:57
Carlgrc wrote:
A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove three-fourths of this solution and then add a new solution, solution N, to create a third solution, solution P, with double the volume of solution M. What must the calcium chloride concentration of solution N be to create a calcium chloride concentration of 37.5% in solution P?

a) 32%
b) 35%
c) 40%
d) 44%
e) 55%



weighted average method
Volume of initial solution M is 1/4 and concentration is 20%.But since final volume is 2 times, the volume of remaining M as a part of P is 1/4*1/2=1/8 so 1 part
Volume of solution P is 1 or8 parts and concentration is 37.5%
Volume of N is 1-1/8=7/8 , that is 7 parts and volume be v.

M...................P......N
1...................8.......7
20%..........37.5%...v
So \(v=P + \frac{(37.5-20)}{7}=37.5+\frac{17.5}{7}=37.5+2.5=40%\)

C
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


GMAT online Tutor

Manager
Manager
avatar
S
Joined: 07 Feb 2017
Posts: 188
A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

Show Tags

New post 13 Jul 2018, 18:20
Bak 2 chem

M
x volume
.2x cacl
.25x volume
.05x cacl

N
2x-.25x volume
nx cacl

(.05x+nx)/(2x)=.375

n=.7

n/1.75 = .4
Answer C
VP
VP
avatar
P
Joined: 07 Dec 2014
Posts: 1130
Re: A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

Show Tags

New post 14 Jul 2018, 11:56
Carlgrc wrote:
A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove three-fourths of this solution and then add a new solution, solution N, to create a third solution, solution P, with double the volume of solution M. What must the calcium chloride concentration of solution N be to create a calcium chloride concentration of 37.5% in solution P?

a) 32%
b) 35%
c) 40%
d) 44%
e) 55%


let x=calcium chloride concentration of solution N
.2*(1/4)+x*(7/4)=.375*2
x=.4=40%
C
Intern
Intern
User avatar
B
Joined: 05 Sep 2016
Posts: 29
Location: India
Concentration: General Management, Operations
WE: Engineering (Energy and Utilities)
Re: A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

Show Tags

New post 30 Jul 2018, 05:17
chetan2u wrote:
Carlgrc wrote:
A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove three-fourths of this solution and then add a new solution, solution N, to create a third solution, solution P, with double the volume of solution M. What must the calcium chloride concentration of solution N be to create a calcium chloride concentration of 37.5% in solution P?

a) 32%
b) 35%
c) 40%
d) 44%
e) 55%



weighted average method
Volume of initial solution M is 1/4 and concentration is 20%.But since final volume is 2 times, the volume of remaining M as a part of P is 1/4*1/2=1/8 so 1 part
Volume of solution P is 1 or8 parts and concentration is 37.5%
Volume of N is 1-1/8=7/8 , that is 7 parts and volume be v.

M...................P......N
1...................8.......7
20%..........37.5%...v
So \(v=P + \frac{(37.5-20)}{7}=37.5+\frac{17.5}{7}=37.5+2.5=40%\)

C


Hi Chetan,

It will really help if you can explain the weighted average method and/or maybe paste links pointing to more questions and similar theories. :-) I am struggling to understand, being thick headed!! :(
Target Test Prep Representative
User avatar
P
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4328
Location: United States (CA)
Re: A chemistry student is working with solution M, which has a 20% calciu  [#permalink]

Show Tags

New post 10 Aug 2018, 17:48
dimmak wrote:
A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove three-fourths of this solution and then add a new solution, solution N, to create a third solution, solution P, with double the volume of solution M. What must the calcium chloride concentration of solution N be to create a calcium chloride concentration of 37.5% in solution P?

a) 32%
b) 35%
c) 40%
d) 44%
e) 55%


We can let amount of solution M = 20 liters, so it has 0.2 x 20 = 4 liters of calcium chloride. We also remove ¾ x 20 = 15 liters of M, which includes ¾ x 4 = 3 liters of calcium chloride. Thus, we have 5 liters of M left (which includes 1 liter of calcium chloride), and we have to add 35 liters of solution N to make 40 liters of solution P (recall that solution P is double the amount of solution M).

We can x = the percent of calcium chloride in solution N and create the equation:

(1 + x/100 * 35)/40 = 0.375

1 + 35x/100 = 15

35x/100 = 14

35x = 1400

x = 40

Alternate Solution:

Assume we start with 4 liters of 20% solution. We will remove ¾ of it, which is 3 liters of 20% solution. Now we add 7 liters of unknown concentration x, and the result is 8 liters of 37.5% solution. Note that we need to end with twice the volume of what we started with, so we end with 8 liters of solution. We can create the following equation:

4(0.20) - 3(0.20) + 7x = 8(0.375)

0.8 - 0.6 + 7x = 3

7x = 2.8

x = 0.40, or 40%

Answer: C
_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

GMAT Club Bot
Re: A chemistry student is working with solution M, which has a 20% calciu &nbs [#permalink] 10 Aug 2018, 17:48
Display posts from previous: Sort by

A chemistry student is working with solution M, which has a 20% calciu

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


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.