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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?

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?

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.

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

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 = H-R Percentage of Higher in final Result = R-L.

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.

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?

Re: Method of Allegation in Percentages [#permalink]

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18 Sep 2011, 09:02

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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?

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?

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

Re: Rodrick mixes a martini that has a volume of 'n' ounces havi [#permalink]

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09 Sep 2013, 01:30

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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..

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?

Re: Rodrick mixes a martini that has a volume of 'n' ounces havi [#permalink]

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13 Jan 2015, 07:50

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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. (40-25)/(25) = 3/5 i.e. Martini should be diluted 3/5 times

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?

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

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.
_________________

Rodrick mixes a martini that has a volume of 'n' ounces havi [#permalink]

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14 Jan 2016, 23:10

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

This is how I did it. It took me 1m 10 S: For simplicity, start by assuming the gin has a volume of 100 units.

Current amount of vermouth: 40 units

I rephrased the question to say: With the numerator being 40, what should the denominator be, so that the fraction becomes 1/4 (25% Vermouth)? For what value of x will the ratio 40/x become 1/4?

A quick mental math will tell you x should be 160 (since 40*4=160)

Thus, we have created a condition where the question's requirement is satisfied.

Thus the amount of gin in the new mix will be =160-40=120 Of this 120, 60 was already present. Thus he needs to add another 60 units of gin to make the mixture 25% Vermouth

However, the answer choices are in terms of n, the volume of the whole mix.

So lets convert 60 to terms of n. We had started by assuming that the mix had 100 units for simplicity. So what % of 100 is 60? easy its 60%

60% in fraction is 3/5. Thus the answer should be 3n/5