Author 
Message 
TAGS:

Hide Tags

Manager
Status: GMAT in 4 weeks
Joined: 28 Mar 2010
Posts: 147
GPA: 3.89

Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
Show Tags
Updated on: 27 Mar 2012, 05:31
Question Stats:
29% (02:34) correct 71% (02:35) wrong based on 438 sessions
HideShow timer Statistics
Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of A. 18% B. 20% C. 21% D. 23% E. Cannot be determined
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by hussi9 on 20 May 2011, 00:31.
Last edited by Bunuel on 27 Mar 2012, 05:31, edited 3 times in total.
Edited the question




Math Expert
Joined: 02 Sep 2009
Posts: 60778

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
Show Tags
13 Sep 2013, 02:01
ishanbhat455 wrote: Hi Bunuel,
I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?
Thanks, Ishan Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit ofA. 18% B. 20% C. 21% D. 23% E. Cannot be determined The profit on the first kind of vodka = x%; The profit on the second kind of vodka = y%. When they are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%: (x + 2y)/3 = 10. When they are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%: (2x + y)/3 = 20. Solving gives: x = 30% and y = 0%. After the individual profit percent on them are increased by 4/3 and 5/3 times respectively the profit becomes 40% and 0%, on the first and te second kinds of vodka, respectively. If they are mixed in equal ratio (1:1), then the mixture will fetch the profit of (40 + 0)/2 = 20%. Answer: B. Hope it's clear.
_________________




Director
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 779

Re: Mixing Vodka ;)
[#permalink]
Show Tags
20 May 2011, 02:59
Mixture 1 and 2 having individual profit % of 10 and 20 respectively. individual profits of vodka A for mixture 1 = 1/3 * 10 = 3.33 individual profits of vodka B for mixture 1 = 2/3 * 10 = 6.77
Individual profits for A and B are increased 4/3 and 5/3 times.
Means profits of mixtures 1 and 2 will increase too.
Increased profits of Vodka A for mixture 1 = 4/3 * 3.33 = 4.44 Increased profits of Vodka B for mixture 2 = 5/3 * 6.77 = 10.55 total increased profits for mixture 1 = 14.99
Similarly total increased profits for mixture 2 = 24.42
using allegation formula
Va: Vb = 1:1 = [24.42  (profit of mixture)] / [(profit of mixture) 14.99]
hence profit of mixture = 19.705 approx = 20%
B



Manager
Status: GMAT in 4 weeks
Joined: 28 Mar 2010
Posts: 147
GPA: 3.89

Re: Mixing Vodka ;)
[#permalink]
Show Tags
20 May 2011, 03:59
Solution:
Let the CP of two vodkas be Rs 100 and Rs 100x and individual profit in Rs on them being A and B. => (A+2B)/3 = 10/100*(100+200x)/3 and (2A+B)/3 = 20/100*(200 + 100x)/3.
solving we get A = (70+20x)/3 and B = (20x20)/3 => profit percentages on each is (70+20x)/3 and (20x20)/3x.
When they are increased to 4/3 and 5/3 times respectively and mixed in the ratio 1:1 we get total profit % as (4/3*100*(70+20x)/3 + 5/3*100x*(20x20)/3x)/(100+100x) = 100*(20x+20)/(100+100x) = 20
=> choice (b) is the right answer.



Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 631

Re: Mixing Vodka ;)
[#permalink]
Show Tags
20 May 2011, 20:25
I did not do any calculation on this one. There are two ingredients in V1 and V2 and we know from the stem that the first ingredient is expensive than the second. Case I  V1 profit is 10% when mix is 1:2 => profit from second ingredient is more Case II  V2 profit is 20% when mix is 2:1 => profit from first ingredient is more. Decreasing the first ingredient results in less profit  Case I. Hence first ingredient is more expensive. If V1 and V2 are mixed in equal ratio then the profit will be closer to 20%. Even though we have equal quantity of each ingredient but first ingredient is more expensive. Now the trick. When the profit on V1 is increased by 4/3 and V2 is increased by 5/3. This is equivalent to increasing the quantity of V2 as 1.33 < 1.67. It means the weighted mean will move even closer to 20% Hence the final mean will be almost 20. hussi9 wrote: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by \(\frac{4}{3}\) and \(\frac{5}{3}\) times respectively, then the mixture will fetch the profit of
(a) 18% (b) 20% (c) 21 % (d) 23% (e) Cannot be determined



Manager
Status: GMAT in 4 weeks
Joined: 28 Mar 2010
Posts: 147
GPA: 3.89

Re: Mixing Vodka ;)
[#permalink]
Show Tags
20 May 2011, 23:04
Your logic is right but when options are as close as 20 and 21 its not safe to assume 20. Only if it were 20, 30 , 40 then its safe to assume 20. gmat1220 wrote: I did not do any calculation on this one. There are two ingredients in V1 and V2 and we know from the stem that the first ingredient is expensive than the second. Case I  V1 profit is 10% when mix is 1:2 => profit from second ingredient is more Case II  V2 profit is 20% when mix is 2:1 => profit from first ingredient is more. Decreasing the first ingredient results in less profit  Case I. Hence first ingredient is more expensive. If V1 and V2 are mixed in equal ratio then the profit will be closer to 20%. Even though we have equal quantity of each ingredient but first ingredient is more expensive. Now the trick. When the profit on V1 is increased by 4/3 and V2 is increased by 5/3. This is equivalent to increasing the quantity of V2 as 1.33 < 1.67. It means the weighted mean will move even closer to 20% Hence the final mean will be almost 20. hussi9 wrote: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by \(\frac{4}{3}\) and \(\frac{5}{3}\) times respectively, then the mixture will fetch the profit of
(a) 18% (b) 20% (c) 21 % (d) 23% (e) Cannot be determined



Intern
Joined: 01 Feb 2013
Posts: 12
Location: India

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
Show Tags
10 Sep 2013, 10:57
Hi Bunuel,
I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?
Thanks, Ishan



Intern
Joined: 01 Feb 2013
Posts: 12
Location: India

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
Show Tags
13 Sep 2013, 05:47
Bunuel wrote: ishanbhat455 wrote: Hi Bunuel,
I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?
Thanks, Ishan Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit ofA. 18% B. 20% C. 21% D. 23% E. Cannot be determined The profit on the first kind of vodka = x%; The profit on the second kind of vodka = y%. When they are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%: (x + 2y)/3 = 10. When they are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%: (2x + 2y)/3 = 20. Solving gives: x = 30% and y = 0%. After the individual profit percent on them are increased by 4/3 and 5/3 times respectively the profit becomes 40% and 0%, on the first and te second kinds of vodka, respectively. If they are mixed in equal ratio (1:1), then the mixture will fetch the profit of (40 + 0)/2 = 20%. Answer: B. Hope it's clear. Thanks Bunuel for this explanation. It's quite clear this way. Kudos!



Manager
Joined: 29 Aug 2013
Posts: 69
Location: United States
Concentration: Finance, International Business
GMAT 1: 590 Q41 V29 GMAT 2: 540 Q44 V20
GPA: 3.5
WE: Programming (Computer Software)

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
Show Tags
15 Sep 2013, 06:38
Bunuel wrote: ishanbhat455 wrote: Hi Bunuel,
I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?
Thanks, Ishan Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit ofA. 18% B. 20% C. 21% D. 23% E. Cannot be determined The profit on the first kind of vodka = x%; The profit on the second kind of vodka = y%. When they are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%: (x + 2y)/3 = 10. When they are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%: (2x + y)/3 = 20. Solving gives: x = 30% and y = 0%. After the individual profit percent on them are increased by 4/3 and 5/3 times respectively the profit becomes 40% and 0%, on the first and te second kinds of vodka, respectively. If they are mixed in equal ratio (1:1), then the mixture will fetch the profit of (40 + 0)/2 = 20%. Answer: B. Hope it's clear. Hi Bunuel, When we say Increased By  Don't we add the values i.e. previous + current... in this case if the profit increased by 4/3 times the previous value shouldn't we add 40 + 30.. i.e. increased by 40% means 70%.. If the question had been increased to 4/3 times then we could have taken 40% i.e. from 30% it has changed to 40%. Please help!!!!



Manager
Joined: 06 Mar 2014
Posts: 220
Location: India
GMAT Date: 04302015

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
Show Tags
22 Apr 2015, 01:00
Bunuel wrote: ishanbhat455 wrote: Hi Bunuel,
I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?
Thanks, Ishan Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them areincreased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit ofA. 18% B. 20% C. 21% D. 23% E. Cannot be determined The profit on the first kind of vodka = x%; The profit on the second kind of vodka = y%. When they are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%: (x + 2y)/3 = 10. When they are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%: (2x + y)/3 = 20. Solving gives: x = 30% and y = 0%. After the individual profit percent on them are increased by 4/3 and 5/3 times respectively the profit becomes 40% and 0%, on the first and te second kinds of vodka, respectively. If they are mixed in equal ratio (1:1), then the mixture will fetch the profit of (40 + 0)/2 = 20%. Answer: B. Hope it's clear. I have a doubt here. If x is the profit and it is increased by 4/3 times, then the new profit will be 4x/3 or 7x/3 ?



Manager
Joined: 10 Jun 2015
Posts: 110

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
Show Tags
07 Aug 2015, 07:22
hussi9 wrote: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of
A. 18% B. 20% C. 21% D. 23% E. Cannot be determined since we do not know in what ratio the first 1:1 and second 1:1 mixtures are mixed we will not be able to determine the resultant profit. Hence, the option is (E)



Senior Manager
Joined: 03 Apr 2013
Posts: 263
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
Show Tags
28 Jun 2016, 00:23
hussi9 wrote: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of
A. 18% B. 20% C. 21% D. 23% E. Cannot be determined I got the individual profits as 30% and 0% and then I did it all wrong..why? because the question has wording problem..it says that the profits increase BY 4/3..that mean that the profit becomes 7/4th of itself..which is wrong...instead..the question should say that the profits increase TO 4/3 of itself and same for the other ingredient..great question..bad wording spoiled it!



Manager
Joined: 06 Oct 2015
Posts: 86
Location: Bangladesh
Concentration: Accounting, Leadership

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
Show Tags
02 Sep 2016, 10:57
Bunuel wrote: ishanbhat455 wrote: Hi Bunuel,
I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?
Thanks, Ishan Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit ofA. 18% B. 20% C. 21% D. 23% E. Cannot be determined The profit on the first kind of vodka = x%; The profit on the second kind of vodka = y%. When they are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%: (x + 2y)/3 = 10. When they are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%: (2x + y)/3 = 20. Solving gives: x = 30% and y = 0%. After the individual profit percent on them are increased by 4/3 and 5/3 times respectively the profit becomes 40% and 0%, on the first and te second kinds of vodka, respectively. If they are mixed in equal ratio (1:1), then the mixture will fetch the profit of (40 + 0)/2 = 20%. Answer: B. Hope it's clear. Hi Bunuel, I got your method except how you got 30%. Will you explain, please, how you have solved and got 30%? Thanks in advance.



Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3552

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
Show Tags
02 Sep 2016, 11:50
NaeemHasan wrote: Bunuel wrote: ishanbhat455 wrote: Hi Bunuel,
I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?
Thanks, Ishan Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit ofA. 18% B. 20% C. 21% D. 23% E. Cannot be determined The profit on the first kind of vodka = x%; The profit on the second kind of vodka = y%. When they are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%: (x + 2y)/3 = 10. When they are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%: (2x + y)/3 = 20. Solving gives: x = 30% and y = 0%. After the individual profit percent on them are increased by 4/3 and 5/3 times respectively the profit becomes 40% and 0%, on the first and te second kinds of vodka, respectively. If they are mixed in equal ratio (1:1), then the mixture will fetch the profit of (40 + 0)/2 = 20%. Answer: B. Hope it's clear. Hi Bunuel, I got your method except how you got 30%. Will you explain, please, how you have solved and got 30%? Thanks in advance. We have x+ 2y = 30 and 2x+y = 60 Multiplying 2nd equation by 2, we can say 2*( 2x+y = 60) = 4x +2y=120 Now subtract the 1st equation from this new equation, we will have 4x + 2y=120  x + 2y=30  3x = 90 or x=30. I Hope its clear now.
_________________
My LinkedIn abhimahna.My GMAT Story: From V21 to V40My MBA Journey: My 10 years long MBA DreamMy Secret Hacks: Best way to use GMATClub  Importance of an Error Log!Verbal Resources: All SC Resources at one place  All CR Resources at one placeBlog: Subscribe to Question of the Day BlogGMAT Club Inbuilt Error Log Functionality  View More. New Visa Forum  Ask all your Visa Related Questions  here. New! Best Reply Functionality on GMAT Club!Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for freeCheck our new About Us Page here.



Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3552

Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
Show Tags
03 Sep 2016, 21:58
NaeemHasan wrote: earnit wrote: Bunuel wrote: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them areincreased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of A. 18% B. 20% C. 21% D. 23% E. Cannot be determined
The profit on the first kind of vodka = x%; The profit on the second kind of vodka = y%.
When they are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%: (x + 2y)/3 = 10. When they are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%: (2x + y)/3 = 20.
Solving gives: x = 30% and y = 0%.
After the individual profit percent on them are increased by 4/3 and 5/3 times respectively the profit becomes 40% and 0%, on the first and te second kinds of vodka, respectively.
If they are mixed in equal ratio (1:1), then the mixture will fetch the profit of (40 + 0)/2 = 20%.
Answer: B.
Hope it's clear. I have a doubt here. If x is the profit and it is increased by 4/3 times, then the new profit will be 4x/3 or 7x/3 ? I agree with you. Bunuel, I believe there is a wording problem with the question. It should be Increased TO and not BY. Kindly confirm.
_________________
My LinkedIn abhimahna.My GMAT Story: From V21 to V40My MBA Journey: My 10 years long MBA DreamMy Secret Hacks: Best way to use GMATClub  Importance of an Error Log!Verbal Resources: All SC Resources at one place  All CR Resources at one placeBlog: Subscribe to Question of the Day BlogGMAT Club Inbuilt Error Log Functionality  View More. New Visa Forum  Ask all your Visa Related Questions  here. New! Best Reply Functionality on GMAT Club!Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for freeCheck our new About Us Page here.



Intern
Joined: 12 Dec 2016
Posts: 1

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
Show Tags
01 May 2017, 19:10
Bunuel wrote: ishanbhat455 wrote: Hi Bunuel,
I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?
Thanks, Ishan Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit ofA. 18% B. 20% C. 21% D. 23% E. Cannot be determined The profit on the first kind of vodka = x%; The profit on the second kind of vodka = y%. When they are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%: (x + 2y)/3 = 10. When they are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%: (2x + y)/3 = 20. Solving gives: x = 30% and y = 0%. After the individual profit percent on them are increased by 4/3 and 5/3 times respectively the profit becomes 40% and 0%, on the first and te second kinds of vodka, respectively. If they are mixed in equal ratio (1:1), then the mixture will fetch the profit of (40 + 0)/2 = 20%. Answer: B. Hope it's clear. Hi Bunuel, I think the question's wording should be changed from profit increased [b]by[/b] to profit increased to. Increased by would imply that the new profit is (1+4/3)x %



Intern
Joined: 24 Apr 2016
Posts: 30

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
Show Tags
28 Nov 2018, 22:29
amit2k9 wrote: Mixture 1 and 2 having individual profit % of 10 and 20 respectively. individual profits of vodka A for mixture 1 = 1/3 * 10 = 3.33 individual profits of vodka B for mixture 1 = 2/3 * 10 = 6.77
Individual profits for A and B are increased 4/3 and 5/3 times.
Means profits of mixtures 1 and 2 will increase too.
Increased profits of Vodka A for mixture 1 = 4/3 * 3.33 = 4.44 Increased profits of Vodka B for mixture 2 = 5/3 * 6.77 = 10.55 total increased profits for mixture 1 = 14.99
Similarly total increased profits for mixture 2 = 24.42
using allegation formula
Va: Vb = 1:1 = [24.42  (profit of mixture)] / [(profit of mixture) 14.99]
hence profit of mixture = 19.705 approx = 20%
B I think the above solution is wrong. The method here is assuming that profit is the same for both Vodka A and Vodka B. If you want to do allegation you can do the following: 1. Allegation says that \(\frac{\text{cheaper quantity}}{\text{greater quantity}} = \frac{\text{profit of greater  average profit}}{\text{average profit  profit of cheaper}}\) Let the vodkas be A and B respectively. Since we know 2:1 has a greater profit than 1:2, we know that A must be the greater quantity. 1:2 => profit 10% gives \(\frac{x  10}{10y} = \frac{2a}{1a}\) where \(a\) is an integer. You can break this down further into the two equations\(x10 = 2a\) and \(10y = 1a\). 2:1 => profit 20% gives \(\frac{x  20}{20y} = \frac{1a}{2a}\). You can also break this down further into the two equations \(x20=a\) and \(20y=2a\). Now you have 3 variables and 4 equations so you can solve for a,x, and y. You get that x = 30 and y = 0. 2. since vodkas A and B are increased by 4/3 and 5/3 respectively, you get that A' = 40 and B' = 0. 3. Using allegation once again, you have \(\frac{1}{1} = \frac{40x}{x0} \implies x = 40x \implies x = 20.\). Hence B.



Intern
Joined: 22 Oct 2017
Posts: 19

Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
Show Tags
23 Feb 2019, 10:19
Bunuel wrote: ishanbhat455 wrote: Hi Bunuel,
I am not quite familiar with the allegation formula amit2k9 has used to solve this problem. Could you please show a simpler way of solving this problem?
Thanks, Ishan Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit ofA. 18% B. 20% C. 21% D. 23% E. Cannot be determined The profit on the first kind of vodka = x%; The profit on the second kind of vodka = y%.
When they are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%: (x + 2y)/3 = 10. When they are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%: (2x + y)/3 = 20.
Solving gives: x = 30% and y = 0%. After the individual profit percent on them are increased by 4/3 and 5/3 times respectively the profit becomes 40% and 0%, on the first and te second kinds of vodka, respectively. If they are mixed in equal ratio (1:1), then the mixture will fetch the profit of (40 + 0)/2 = 20%. Answer: B. Hope it's clear. Sorry guys but I don't get how the red part is formed, I would like to know the intuition behind it, maybe with an example, kudos guaranteed




Re: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and
[#permalink]
23 Feb 2019, 10:19






