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

It is currently 16 Dec 2018, 17:23

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
  • FREE Quant Workshop by e-GMAT!

     December 16, 2018

     December 16, 2018

     07:00 AM PST

     09:00 AM PST

    Get personalized insights on how to achieve your Target Quant Score.
  • Free GMAT Prep Hour

     December 16, 2018

     December 16, 2018

     03:00 PM EST

     04:00 PM EST

    Strategies and techniques for approaching featured GMAT topics

Two liquids are mixed in the ratio 3:2 and the vendor gains 10%

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

Hide Tags

Intern
Intern
avatar
Joined: 08 Oct 2016
Posts: 12
Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

Show Tags

New post 26 Dec 2016, 20:12
1
24
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

43% (02:51) correct 57% (02:34) wrong based on 371 sessions

HideShow timer Statistics

Two liquids are mixed in the ratio 3:2 and the vendor gains 10% by selling the mixture at $11/liter. If the first liquid costs $2 more than the second, find the cost price of first liquid.

A. $8
B. $10.8
C. $6
D. $8.8
E. $4

Is there any way to do this in weighted average method?
Most Helpful Community Reply
Director
Director
User avatar
S
Affiliations: CrackVerbal
Joined: 03 Oct 2013
Posts: 533
Location: India
GMAT 1: 780 Q51 V46
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

Show Tags

New post 27 Dec 2016, 12:49
4
1
Top Contributor
4
Hi voccubd,

Before I jump in to solve this question, let me provide you a brief explanation of how to approach weighted average questions on the GMAT. Weighted average questions can be easily solved by making use of the alligation/mixture diagram given below.

Attachment:
Mixtures 1.png
Mixtures 1.png [ 8.12 KiB | Viewed 10440 times ]


Putting in values in the alligation/mixture diagram and subtracting along the diagonals gives us a ratio in which two quantities are mixed. This ratio can now be used to find out what specific amounts of two quantities need to be mixed to obtain a particular mixture.

The only thing that you need to keep in mind here is that the values you need to use, that is the higher value, lower value and mean value have to be values which are associated with the word 'per' (percents, average, per km, per kg etc.).

The alligation/mixture diagram proves useful not only when mixing solutions or combining solids but also to explain the weighted average concept (the word average is also associated with the word per i.e. if the average marks of the class is 80, then it can be understood as 80 marks per student). Say if we have a class A where the average marks is 80 and another class B where the average marks is 70 and the combined average of both class A and B is 74, then we can definitely comment upon which class has the greater number of students. If we represent the average values in the mixture diagram, the ratio of students of Class A and Class B will be 2 : 3. This clearly indicates that class B has the greater number of students.

Now this alligation/mixture diagram can also be used in the above question, since we are mixing two prices (associated with the word 'per') and the ratio of mixing the two prices are given to us.

Let us keep the price of the second liquid to be 'x$' per liter. The price of the first liquid will be '(x+2)$' per liter. Since the vendor gains 10% on the cost price by selling the mixture at $11, the actual cost price of the mixture needs to be $10. We have also been given the ratio of 3 : 2. Creating the alligation/mixture diagram for the cost prices of the two liquids we get

Attachment:
Mixtures.png
Mixtures.png [ 6.3 KiB | Viewed 10425 times ]


Subtracting along the diagonals we get

x - 8 = 2
10 - x = 3. Cross multiplying we get

3x - 24 = 20 - 2x -----> 5x = 44 -----> x = 8.8

Since the question asks us the value of x + 2 -------> 8.8 + 2 = 10.8

OA : B

Hope this helps!

CrackVerbal Academics Team
_________________

For more info on GMAT and MBA, follow us on @AskCrackVerbal

Register for the Free GMAT Video Training Course : https://crackverbal.com/MBA-Through-GMAT-2019-Registration

General Discussion
Intern
Intern
avatar
B
Joined: 08 Aug 2016
Posts: 18
GMAT ToolKit User Premium Member
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

Show Tags

New post 26 Dec 2016, 20:45
vendor gains 10% by selling the mixture at $11/litre i.e he sells the mixture at 110% cost,
so Cost of mixture = 11/(110%)= $10/ litre
Let x be the cost of first liquid, so cost of second liquid= x-2
Two liquids are mixed in the ratio 3:2 i.e. 60:40 ratio
so Cost of mixture= x*60%+ (x-2)*40% =10
0.6x+0.4x-0.8=10
x=10.8
_________________

कर्मणयेवाधिकारस्ते मा फलेषु कदाचन|

Intern
Intern
avatar
Joined: 08 Oct 2016
Posts: 12
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

Show Tags

New post 27 Dec 2016, 03:55
Thank you. But I was wondering if any shortcut like weighted average method can be applied here.
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 7108
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

Show Tags

New post 27 Dec 2016, 05:42
2
1
voccubd wrote:
Two liquids are mixed in the ratio 3:2 and the vendor gains 10% by selling the mixture at $11/liter. If the first liquid costs $2 more than the second, find the cost price of first liquid.

A. $8
B. $10.8
C. $6
D. $8.8
E. $4

Is there any way to do this in weighted average method?


Hi,
Weighted average method is best when there is Only one variable..
Here there are two, so if you want to use it , it may sound slightly different from a proper equation

Average price is 11*100/110=10..

1) AVG price 10...
The other two prices have a difference of 2..
So higher price will be closer to 10 as the quantity is more..
\(10 + 2*\frac{2}{(2+3)}=10+2*0.4=10.8\)

2)
So if the ratio is 3:2 the price will be 10(3+2)=50..

Let the price be x and x-2...
So 3*x + 2(x-2) =50...
3x+2x-4=50......5x=54.... x=54/5=10.8..

3)
AVG price is

Otherwise when you know average price is 10..
And the rates are different so one will be higher and the other lower than 10..
And we are looking for the higher value
Only 10.8 is higher than 10..
Ans is 10.8
_________________

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

Intern
Intern
User avatar
B
Joined: 19 Nov 2016
Posts: 16
Schools: Erasmus '18
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

Show Tags

New post 27 Dec 2016, 11:55
i still get confused with this question can anyone help? how to approach it in a different way?
Intern
Intern
avatar
Joined: 08 Oct 2016
Posts: 12
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

Show Tags

New post 28 Dec 2016, 08:46
Well I just googled and find a faster solution in a youtube video. In that solution the ratio was reversed as 2:3 and $2 was divided as $0.8 and $1.2. Then 10-1.2=8.8 and 10+0.8=10.8 was found. But I don't understand it.Can anyone please explain that method?
Intern
Intern
avatar
B
Joined: 11 Nov 2015
Posts: 20
Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

Show Tags

New post 11 Jan 2017, 10:32
5
C(1.10)=S ===> \(C=11/1.10=10\). The cost price of the mixture is 10$

We have 3 units bought at X$, and 2 units bought at (x-2)$, which gives us a mixture of 5units that costs 10$.

\(3(X)+2(X-2)=5(10)\)

x=54/5=10.8
VP
VP
avatar
P
Joined: 07 Dec 2014
Posts: 1129
Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

Show Tags

New post 11 Jan 2017, 15:47
voccubd wrote:
Two liquids are mixed in the ratio 3:2 and the vendor gains 10% by selling the mixture at $11/liter. If the first liquid costs $2 more than the second, find the cost price of first liquid.

A. $8
B. $10.8
C. $6
D. $8.8
E. $4

Is there any way to do this in weighted average method?


11/1.1=$10 cost of liter
3:2=6:4
let 6x=cost of first liquid
6x+4(x-2)=10
x=1.8
6x=$10.80
B
Target Test Prep Representative
User avatar
P
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4295
Location: United States (CA)
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

Show Tags

New post 16 Jan 2017, 16:58
1
voccubd wrote:
Two liquids are mixed in the ratio 3:2 and the vendor gains 10% by selling the mixture at $11/liter. If the first liquid costs $2 more than the second, find the cost price of first liquid.

A. $8
B. $10.8
C. $6
D. $8.8
E. $4


We are given that two liquids are mixed in the ratio 3:2. Thus, we can let 3x = weight of the first liquid and 2x = weight of the second liquid. Thus, the total weight = 5x.

We are also given that the vendor gains 10% by selling the mixture at $11/liter and the first liquid costs $2 more than the second. Thus we can let y = cost of the second liquid per liter and y + 2 = cost of the first liquid per liter. Furthermore, we can create a total cost equation and solve for y:

1.1[3x(y + 2) + 2x(y)] = 11(5x)

Divide both sides by 1.1:

3x(y + 2) + 2x(y) = 10(5x)

Divide both sides by x:

3y + 6 + 2y = 50

5y = 44

y = 44/5 = 8.8

Since y is the cost of the second liquid per liter and the cost of the first liquid per liter is $2 more, the cost of the first liquid per liter is 8.8 + 2 = $10.8.

Answer: B
_________________

Scott Woodbury-Stewart
Founder and CEO

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

Intern
Intern
avatar
S
Joined: 21 May 2016
Posts: 19
Premium Member
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

Show Tags

New post 04 Feb 2017, 08:26
ScottTargetTestPrep wrote:
voccubd wrote:
Two liquids are mixed in the ratio 3:2 and the vendor gains 10% by selling the mixture at $11/liter. If the first liquid costs $2 more than the second, find the cost price of first liquid.

A. $8
B. $10.8
C. $6
D. $8.8
E. $4


We are given that two liquids are mixed in the ratio 3:2. Thus, we can let 3x = weight of the first liquid and 2x = weight of the second liquid. Thus, the total weight = 5x.

We are also given that the vendor gains 10% by selling the mixture at $11/liter and the first liquid costs $2 more than the second. Thus we can let y = cost of the second liquid per liter and y + 2 = cost of the first liquid per liter. Furthermore, we can create a total cost equation and solve for y:

1.1[3x(y + 2) + 2x(y)] = 11(5x)

Divide both sides by 1.1:

3x(y + 2) + 2x(y) = 10(5x)

Divide both sides by x:

3y + 6 + 2y = 50

5y = 44

y = 44/5 = 8.8

Since y is the cost of the second liquid per liter and the cost of the first liquid per liter is $2 more, the cost of the first liquid per liter is 8.8 + 2 = $10.8.

Answer: B


In this question, how can we be sure that 3x is the first liquid and 2x is the second liquid? Why cant 3x be the second liquid and 2x the first?

Also, I solved it backward through the choices. 4,6 and 8 get ruled out when we try and substitute say :

First substitute 4 in equation 3x+2y = mixture and we are given that x=y+2 or y=x+2 so: 3(4)+2(2)=16, second : 3(2)+2(4)=14 (since I dont know which is the first liquid - if it's 3x or 2x), similar workings for 6 and 8 leads me to answers that are not multiples of 10. We need a multiple of 10 since the cost/litre is 10. I'm now left with 10.8 and 8.8 and both give me multiples of 10 using this method. Finally I chose 10.8 since question is asking for y+2 or x+2, so I guessed the value due to paucity of time. Is there anything wrong with this method?
Director
Director
User avatar
P
Joined: 14 Dec 2017
Posts: 518
Location: India
Premium Member
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

Show Tags

New post 16 Aug 2018, 02:51
voccubd wrote:
Two liquids are mixed in the ratio 3:2 and the vendor gains 10% by selling the mixture at $11/liter. If the first liquid costs $2 more than the second, find the cost price of first liquid.

A. $8
B. $10.8
C. $6
D. $8.8
E. $4



Given the two liquids are mixed in the ratio 3:2, hence 3k litres of liquid 1 & 2k litres of liquid 2.

Total quantity of final solution = 5k litres

Selling Price = $11/liter, hence SP of solution = $55k

Profit is 10%, hence Cost price of solution = 55k/1.1 = $50k

Let x be price of liquid 1 per liter & y be price of liquid per liter, hence

Hence Cost price of Solution = 3k*x + 2k*y = 50k

given that y = x - 2

we get 3x + 2(x-2) = 50

Solving we get, x = $10.8


Answer B.



Thanks,
GyM
_________________

New to GMAT Club - https://gmatclub.com/forum/new-to-gmat-club-need-help-271131.html#p2098335

Intern
Intern
avatar
B
Joined: 03 Feb 2016
Posts: 11
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

Show Tags

New post 21 Aug 2018, 08:35
x=cost price of 1st liquid
y=cost price of 2nd liquid
so,
x-y=2...(1)
(x*3/5+y*2/5)*110/100=11.....(2)
equate the above 2 equations and get
y=44/5$
x=54/5=10.8$
Intern
Intern
avatar
Joined: 19 Sep 2018
Posts: 3
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10%  [#permalink]

Show Tags

New post 22 Sep 2018, 22:07
3/5(x+2)+2/5(x)=10
=> 3x+6+2x=10*5
=> x= 8.8
=> x+2= 8.8+2= 10.8
GMAT Club Bot
Re: Two liquids are mixed in the ratio 3:2 and the vendor gains 10% &nbs [#permalink] 22 Sep 2018, 22:07
Display posts from previous: Sort by

Two liquids are mixed in the ratio 3:2 and the vendor gains 10%

  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®.