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Rodrick mixes a martini that has a volume of 'n' ounces havi [#permalink]
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28 Dec 2009, 13:18
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Rodrick mixes a martini that has a volume of 'n' ounces having 40% Vermouth and 60% Gin by volume. He wants to change it so that the martini is 25% Vermouth by volume. How many ounces of Gin must he add? A) n/6 B) n/3 C) 3n/5 D) 5n/6 E) 8n/5
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Re: Method of Allegation in Percentages [#permalink]
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28 Dec 2009, 13:25
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Re: Method of Allegation in Percentages [#permalink]
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28 Dec 2009, 15:35
Bunuel, thanks for the solution but as i said, i already know the "equations" method. The problem is that I cant solve it using the Allegation method. Any help on that would be nice.



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Re: Method of Allegation in Percentages [#permalink]
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29 Dec 2009, 08:50
Any response as to how to solve the problem using Allegation method would be useful. Please help!!



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Re: Method of Allegation in Percentages [#permalink]
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01 Jan 2010, 20:25
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Not sure if this is the method you're looking for but:
Verm Gin 40% 0%
25%
25 15
Which reduces to 5:3



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Re: Method of Allegation in Percentages [#permalink]
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13 Jul 2010, 05:48
I want to respond to the following question appearing in this forum: _____ Rodrick mixes a martini that has a volume of 'n' ounces having 40% Vermouth and 60% Gin by volume. He wants to change it so that the martini is 25% Vermouth by volume. How many ounces of Gin must he add?
A) n/6 B) n/3 C) 3n/5 D) 5n/6 E) 8n/5 _____
The answer C) 3n/5 for Gin is obviously not correct for the following reason. Gin's concentration is 60% which is much higher than that of Vermouth (40%). Obviously, you need less of Gin than Vermouth to make n quantity of Martin. Consequently, the answer we should be looking for must be below 1/2 of the quantity of the total (n). The correct proportional volume of Gin is 3n/10 using the Alligation Method. Make a nine cell table like the one below (Unfortunately, I can't draw the table here. I have added starred lines to space the figures to create a Table image). Obtain diagonal differences and enter them as shown in top right and bottom right boxes(always positive numbers). Add right hand column (15+35=50). Calculate amount of Gin proportion as (n/50)*15 =3/10 n. Calculate amount of Vermouth proportion as (n/50)*35 =7/10 n
60(Gin)****************************15(4025) *******************25(Martin) 40(Vermouth)***********************35(6025)
Mathew



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Re: Method of Allegation in Percentages [#permalink]
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22 Jul 2011, 05:02
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Dear Friends,
Using allegations we will view the action that Rodrick performs as below:
"He will Mix Martini [60 % gin by volume ] with Gin [ 100 % gin by volume] to yield a martini [75 % gin by volume]."
By the method of allegations : If we have 2 quantities to mix ,one with higher percentage H and other with lower percentage L ,to get a desired solution R %we proceed as Percentage of Lower in final Result = HR Percentage of Higher in final Result = RL.
Therefore using the allegations method we have %volume of Pure Gin required = 75 60 = 15 %volume of Martini (60 % gin)required = 100 75 =25
Which reduces to Pure Gin to Martini =3 :5. Which can be inferred as for every 5 ml of martini if we should add 5 ml of pure gin to get a drink that is 75 % gin and 25 % Vermouth by volume.
So , we have : For 5 ml Martini we need 3 ml Gin For 1 ml martini we need 3/5 ml of Gin For 'n' ml of martini we need 3n/5 ml of Gin.
I hope the solution helps as a good reference.
Cheers, Rohan



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Re: Method of Allegation in Percentages [#permalink]
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18 Sep 2011, 01:45
Bunuel wrote: gaurav05nov wrote: Guys, please help me out. I want to know how to solve this question using the Allegation method. I know the other way(using equations) but I need to understand the Allegation method. Rodrick mixes a martini that has a volume of 'n' ounces having 40% Vermouth and 60% Gin by volume. He wants to change it so that the martini is 25% Vermouth by volume. How many ounces of Gin must he add? A) n/6 B) n/3 C) 3n/5 D) 5n/6 E) 8n/5 Spoiler : Note that after we add pure Gin, the volume of Vermouth will remain the same. Based on this set the equation: 0.4n=0.25(n+g) > g=3n/5 Answer: C. Fluke/Bunuel, can you please explain this to me? I don't understand the equation 0.4n=0.25(n+g).
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Re: Method of Allegation in Percentages [#permalink]
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18 Sep 2011, 04:08
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petrifiedbutstanding wrote: Fluke/Bunuel, can you please explain this to me? I don't understand the equation 0.4n=0.25(n+g).
I will try. Volume of martini given = n Volume of vermouth in martini = 40% of n = 0.4n Let "g" be the volume of gin we add. New Volume of martini = n+g Also, we did not add any vermouth. New Volume of vermouth = 0.4n (still same as old) New volume of vermouth= 25 % New martini (n+g) 0.4n=0.25(n+g) n=3n/5 Pick number and see. It will be clear.
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Re: Method of Allegation in Percentages [#permalink]
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total V G 1 ounce 0.4 0.6 n ounce 0.4n 0.6n initial expression lets say g ounces of gin is added to this mixture n+g 0.4n 0.6n+g final expression given that after adding g ounces of gin , V should become 25% of the total volume. =>Volume of V/total volume = 25/100 => 0.4n /n+g = 1/4 => 1.6n = n+g => g = 3n/5 Answer is C. petrifiedbutstanding wrote: Bunuel wrote: gaurav05nov wrote: Guys, please help me out. I want to know how to solve this question using the Allegation method. I know the other way(using equations) but I need to understand the Allegation method. Rodrick mixes a martini that has a volume of 'n' ounces having 40% Vermouth and 60% Gin by volume. He wants to change it so that the martini is 25% Vermouth by volume. How many ounces of Gin must he add? A) n/6 B) n/3 C) 3n/5 D) 5n/6 E) 8n/5 Spoiler : Note that after we add pure Gin, the volume of Vermouth will remain the same. Based on this set the equation: 0.4n=0.25(n+g) > g=3n/5 Answer: C. Fluke/Bunuel, can you please explain this to me? I don't understand the equation 0.4n=0.25(n+g).



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Re: Method of Allegation in Percentages [#permalink]
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18 Sep 2011, 09:07
jamifahad wrote: petrifiedbutstanding wrote: Fluke/Bunuel, can you please explain this to me? I don't understand the equation 0.4n=0.25(n+g).
I will try. Volume of martini given = n Volume of vermouth in martini = 40% of n = 0.4n Let "g" be the volume of gin we add. New Volume of martini = n+g Also, we did not add any vermouth. New Volume of vermouth = 0.4n (still same as old) New volume of vermouth= 25 % New martini (n+g) 0.4n=0.25(n+g) n=3n/5 Pick number and see. It will be clear. Thanks! This does help!
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Re: Method of Allegation in Percentages [#permalink]
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gaurav05nov wrote: Guys, please help me out. I want to know how to solve this question using the Allegation method. I know the other way(using equations) but I need to understand the Allegation method. Rodrick mixes a martini that has a volume of 'n' ounces having 40% Vermouth and 60% Gin by volume. He wants to change it so that the martini is 25% Vermouth by volume. How many ounces of Gin must he add? A) n/6 B) n/3 C) 3n/5 D) 5n/6 E) 8n/5 Spoiler : He wants to mix a martini of 60% Gin with a liquid that is 100% Gin to give a mixture with 75% Gin (because 25% is Vermouth) Using the mixtures formula discussed here ( http://www.veritasprep.com/blog/2011/03 ... averages/), we get, w1/w2 = (100  75)/(75  60) = 5:3 Martini:Pure Gin must be added in the ratio 5:3. So if martini is n, pure gin must be (3/5)n A very similar question is discussed in the 'Responses' of this post (at the bottom): http://www.veritasprep.com/blog/2011/04 ... mixtures/
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Re: Method of Allegation in Percentages [#permalink]
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mixture already has 60% gin, and we need 75%. \(\frac{0.6n + x}{n + x} = 0.75\) \(x = \frac{3}{5}n\)
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Re: Rodrick mixes a martini that has a volume of 'n' ounces havi [#permalink]
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gaurav05nov wrote: Rodrick mixes a martini that has a volume of 'n' ounces having 40% Vermouth and 60% Gin by volume. He wants to change it so that the martini is 25% Vermouth by volume. How many ounces of Gin must he add?
A) n/6 B) n/3 C) 3n/5 D) 5n/6 E) 8n/5 Initially Martini = V : G = 4 : 6 = 2 : 3 Ratio required is = 25 : 75 = 1 : 3 Since V is same hence ratio should be 2 : 6 Hence 3 parts of gin should be added to the initial mixture i.e. same as what was in the initial mixture i.e. 60n/100 = 3n/5 The mixture related problems if dealt with ratios becomes much easier..



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Re: Rodrick mixes a martini that has a volume of 'n' ounces havi [#permalink]
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16 Dec 2013, 06:52
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gaurav05nov wrote: Rodrick mixes a martini that has a volume of 'n' ounces having 40% Vermouth and 60% Gin by volume. He wants to change it so that the martini is 25% Vermouth by volume. How many ounces of Gin must he add?
A) n/6 B) n/3 C) 3n/5 D) 5n/6 E) 8n/5 Way I did it Smart Numbers n=100 So we'll have 60+x/100+x=3/4 x=60 So 60/100 = 3/5n Answer is C Hope it helps! Cheers! J



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Re: Rodrick mixes a martini that has a volume of 'n' ounces havi [#permalink]
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PROBLEM: Rodrick mixes a martini that has a volume of 'n' ounces having 40% Vermouth and 60% Gin by volume. He wants to change it so that the martini is 25% Vermouth by volume. How many ounces of Gin must he add?
A) n/6 B) n/3 C) 3n/5 D) 5n/6 E) 8n/5
SOLUTION:
A quick way to calculate without any algebra is:
(Old  New)/(New) i.e. (4025)/(25) = 3/5 i.e. Martini should be diluted 3/5 times
ANSWER: C



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Re: Rodrick mixes a martini that has a volume of 'n' ounces havi [#permalink]
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15 Jan 2015, 02:35
Place n = 1 in the question as well as in OA & say "x" volume of Gin is added. Equation setup would be as follows: \(\frac{25}{100}(1+x) = \frac{40}{100}\) \(x = \frac{3}{5}\) Answer = C
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Re: Rodrick mixes a martini that has a volume of 'n' ounces havi [#permalink]
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30 Dec 2015, 07:12
HI Karishma Is this approach also right (from one of your posts)? Volume (initial) x concentration (initial) = Volume (final) x Concentration (final)Since, the volume of Vermouth remains same (not Gin), this should be applied to Vermouth Therefore, n x 40% = Volume (final) x 25% Volume (final) = 8n/5 So, solution added = (8n/5)  n ==> 3n /5 VeritasPrepKarishma wrote: gaurav05nov wrote: Guys, please help me out. I want to know how to solve this question using the Allegation method. I know the other way(using equations) but I need to understand the Allegation method. Rodrick mixes a martini that has a volume of 'n' ounces having 40% Vermouth and 60% Gin by volume. He wants to change it so that the martini is 25% Vermouth by volume. How many ounces of Gin must he add? A) n/6 B) n/3 C) 3n/5 D) 5n/6 E) 8n/5 Spoiler : He wants to mix a martini of 60% Gin with a liquid that is 100% Gin to give a mixture with 75% Gin (because 25% is Vermouth) Using the mixtures formula discussed here ( http://www.veritasprep.com/blog/2011/03 ... averages/), we get, w1/w2 = (100  75)/(75  60) = 5:3 Martini:Pure Gin must be added in the ratio 5:3. So if martini is n, pure gin must be (3/5)n A very similar question is discussed in the 'Responses' of this post (at the bottom): http://www.veritasprep.com/blog/2011/04 ... mixtures/



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Re: Rodrick mixes a martini that has a volume of 'n' ounces havi [#permalink]
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30 Dec 2015, 23:26
madhusudhan237 wrote: HI Karishma
Is this approach also right (from one of your posts)?
Volume (initial) x concentration (initial) = Volume (final) x Concentration (final) Since, the volume of Vermouth remains same (not Gin), this should be applied to Vermouth Therefore, n x 40% = Volume (final) x 25% Volume (final) = 8n/5
So, solution added = (8n/5)  n ==> 3n /5
Yes, correct. Since amount of Vermouth stays the same in the two cases, Volume of solution*Concentration of Vermouth should be equal in the two cases.
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