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Mixture A is 15 percent alcohol, and mixture B is 50 percent

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Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 14 Jan 2011, 06:18
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Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?

A. 1.5
B. 1.7
C. 2.3
D. 2.5
E. 3.0
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Re: Mixture Problem (What is the best way to approach this?)  [#permalink]

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New post 14 Jan 2011, 20:01
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Another option is to use the scale method discussed here:
http://gmatclub.com/forum/tough-ds-105651.html#p828579

Attachment:
Ques1.jpg
Ques1.jpg [ 3.9 KiB | Viewed 12286 times ]

Now we see that the two solutions A:B will be in the ratio 4:3.
So in a 4 gallon mixture, volume of solution A = (4/7)*4 = 2.3 gallon
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Re: Mixture Problem (What is the best way to approach this?)  [#permalink]

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New post 14 Jan 2011, 07:21
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Nobody wrote:
The Question:
Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?[/color]
A. 1.5
B. 1.7
C. 2.3
D. 2.5
E. 3.0


You can treat this question as wighted average problem: \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(0.3=\frac{0.15*A+0.5(4-A)}{4}\) --> \(A\approx{2.3}\).

Answer: C.

Also you can notice that as wighted average (alcohol share) of final mixture (30%) is closer to that of mixture A (15%) than to that of mixture B (50%) then there should be more of mixture A in final solution of 4 gallons than of mixture B, so answer choices A and B are out right away. Plus, if in final mixture there were equal amounts of mixtures A and B then the final solution would have (15%+50%)/2=32.5% of alcohol, and as 32.5% is a little more than 30% (actual concentration) then there should be a little more of mixture A than mixture B in 4 gallons, answer choice C fits best.

Hope it's clear.
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Re: Mixture Problem (What is the best way to approach this?)  [#permalink]

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New post 14 Jan 2011, 07:37
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Hi,
It is always better to approach this problem as a simple word problem rather than a Problem of Mixtures. Word problems should be framed in to simple equations. Here let us assume mixture A is x gallons and mixture B is Y gallons.

Number of gallons of alcohol in A = 0.15x (Mixture A is 15 percent alcohol)
Number of gallons of alcohol in B = 0.5Y(mixture B is 50 percent alcohol)
Number of gallons of alcohol in a mixture formed by mixing A and B i.e in a mixture of x+y gallons =0.3(x+y)

Hence equating the alcohol content we get the equation
0.15x+0.5y=0.3(x+y)
We also have x+y=4(two Mixtures are poured together to create a 4-gallon mixture)
Here you can solve the two equations to get X, else substitute the choices. While substitution it is always preferable to start substituting from a median of options(Here 2.3 is median among the set of answers 1.5,1.7,2.3,2.5,3.0). This approach helps us to eliminate other two options based on the result we get by substituting median value, by either moving up or down.Here 2.3 satisfies our equation. Hope this helps.

Thank and Regards,
Sashikanth.
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 13 Jun 2013, 03:09
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 13 Jul 2013, 23:45
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using Table format :

x+y = 4 ----(1) volume equation

0.15x+0.5y = 0.3*4 ------(2) concentrate equation

solving (1) & (2) :
0.15x+0.5y= 1.2
15x+50y=120
15x+50(4-x) =120 ---(from (1))

15x+200-50x=120
35x = 80
x= 80/35= 16/7 = 2.28
ans C.
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 14 Jul 2013, 00:19
Nobody wrote:
Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?[/color]

A. 1.5
B. 1.7
C. 2.3
D. 2.5
E. 3.0


All the good methods are already outlined above: Another way of doing this problem, without the need to write down anything:

Assume that both Mixture A and Mixture B are in equal proportion \(\to\) The total amount of alcohol would be 50 % of 2 gallons+15% of 2 gallons \(\to\)1+0.3 = 1.3 gallons. As per the problem, we need only 1.2 gallons. Thus, to account for this extra 0.1 gallons, and we can safely assume that the mixture which has more percentage of alcohol would be lesser, and Mixture A will be a little more than 2 gallons

Now, if we are confused between 2.3 or 2.5, we can always back calculate :

For eg, assuming that 2.5 is the correct answer, amount of Mixture B = 4-2.5 = 1.5 gallons \(\to\) 50% of this is 0.75 gallons.

Now, we know that 10% of 2.5 gallons can be calculated in a jiffy and it equals 0.25. Thus, 15% would be 10%+5% \(\to\) 0.25+0.125 = 0.375 and the net alcohol content stands at 1.125 which is less than 1.2 gallons.Hence, we have to increase the alcohol content a little, and this can be done by increasing amount of Mixture B. Thus, the correct answer is C.
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 24 Oct 2013, 23:48
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Please refer screenshot below:
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 22 Sep 2014, 04:10
0.15a+0.5b=0.3*4=1.2
a+b=4

b=4-a
0.15a+0.5(4-a)=1.2 ===> a=2.3

C
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 24 Feb 2015, 04:34
I kind of used a variation of the unknown multiplier approach.

I started with the representation of the percentages in the final solution:
15x+50y = 30 (x+y)
20y = 15x
y/x = 15/20
y/x = 3/4

Then, I again used the unknown multiplier to figure out the amounts:
3x+4x = 4
7x = 4
x = 4/7

For A 3*(4/7) = 2.3 (about). So, ANS C

Does it make sense?
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 05 Jul 2015, 05:22
Bunuel wrote:
Nobody wrote:
The Question:
Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?[/color]
A. 1.5
B. 1.7
C. 2.3
D. 2.5
E. 3.0


You can treat this question as wighted average problem: \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(0.3=\frac{0.15*A+0.5(4-A)}{4}\) --> \(A\approx{2.3}\).

Answer: C.

Also you can notice that as wighted average (alcohol share) of final mixture (30%) is closer to that of mixture A (15%) than to that of mixture B (50%) then there should be more of mixture A in final solution of 4 gallons than of mixture B, so answer choices A and B are out right away. Plus, if in final mixture there were equal amounts of mixtures A and B then the final solution would have (15%+50%)/2=32.5% of alcohol, and as 32.5% is a little more than 30% (actual concentration) then there should be a little more of mixture A than mixture B in 4 gallons, answer choice C fits best.

Hope it's clear.


Why can't we use this formula :

0,15A + 0,5B = 0,3
A + B = 4

and then solve for A ?

I have difficulties understanding when to use: 0,15A + 0,5B = 0,3 (A+B) and when to use the formula above.
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 05 Jul 2015, 05:33
LaxAvenger wrote:
Bunuel wrote:
Nobody wrote:
The Question:
Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?[/color]
A. 1.5
B. 1.7
C. 2.3
D. 2.5
E. 3.0


You can treat this question as wighted average problem: \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(0.3=\frac{0.15*A+0.5(4-A)}{4}\) --> \(A\approx{2.3}\).

Answer: C.

Also you can notice that as wighted average (alcohol share) of final mixture (30%) is closer to that of mixture A (15%) than to that of mixture B (50%) then there should be more of mixture A in final solution of 4 gallons than of mixture B, so answer choices A and B are out right away. Plus, if in final mixture there were equal amounts of mixtures A and B then the final solution would have (15%+50%)/2=32.5% of alcohol, and as 32.5% is a little more than 30% (actual concentration) then there should be a little more of mixture A than mixture B in 4 gallons, answer choice C fits best.

Hope it's clear.


Why can't we use this formula :

0,15A + 0,5B = 0,3
A + B = 4

and then solve for A ?

I have difficulties understanding when to use: 0,15A + 0,5B = 0,3 (A+B) and when to use the formula above.


Hi lax,
it is this very formula that has been used earlier too..
your two eqs are:
\(0.15A + 0.5B = 0.3(A+B)\)
A + B = 4 or B=4-A....
substitue value of B and you will get
\(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(0.3=\frac{0.15*A+0.5(4-A)}{4}\)
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 05 Jul 2015, 06:10
Why can't we use this formula :

0,15A + 0,5B = 0,3
A + B = 4

and then solve for A ?

I have difficulties understanding when to use: 0,15A + 0,5B = 0,3 (A+B) and when to use the formula above.[/quote]

Hi lax,
it is this very formula that has been used earlier too..
your two eqs are:
\(0.15A + 0.5B = 0.3(A+B)\)
A + B = 4 or B=4-A....
substitue value of B and you will get
\(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(0.3=\frac{0.15*A+0.5(4-A)}{4}\)[/quote]


If I solve the eq 0,15A + 0,5B = 0,3
A + B = 4

I end up getting: -1.7 = -0.35A -> A = 4.8

so I do not understand why for some problems this formula gives correct results and for some we have to use another ..
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 05 Jul 2015, 06:50
hi,
0.15A+0.5B=0.3(A+B)
A + B = 4 or B=4-A......
so .15A+ .5(4-A)=.3*4...
.15A+2-.5A=1.2..
.35A=.8..
so A=.8/.35=2.3 nearly...
so the answer is coming correct..
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 05 Jul 2015, 06:52
LaxAvenger wrote:
Why can't we use this formula :

0,15A + 0,5B = 0,3
A + B = 4

and then solve for A ?

I have difficulties understanding when to use: 0,15A + 0,5B = 0,3 (A+B) and when to use the formula above.


Hi lax,
it is this very formula that has been used earlier too..
your two eqs are:
\(0.15A + 0.5B = 0.3(A+B)\)
A + B = 4 or B=4-A....
substitue value of B and you will get
\(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(0.3=\frac{0.15*A+0.5(4-A)}{4}\)[/quote]


If I solve the eq 0,15A + 0,5B = 0,3
A + B = 4

I end up getting: -1.7 = -0.35A -> A = 4.8

so I do not understand why for some problems this formula gives correct results and for some we have to use another ..[/quote]


Hi LaxAvenger,
where you are going wrong is in the formation of equation..
0,15A + 0,5B = 0,3....
0.3 of what?
0.3 of final mix that is equal to A+B..
so the eq will become
0,15A + 0,5B = 0,3(A+B)
now try and you will get the correct answer.. :) :)
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 12 Apr 2017, 10:10
Nobody wrote:
Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?

A. 1.5
B. 1.7
C. 2.3
D. 2.5
E. 3.0


.15A+.5B=.3(A+B)
A/B=4/3
A/(A+B)=4/7
4/7*4 gallons=2.3 gallons
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 16 Jan 2019, 05:35
Nobody wrote:
Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?

A. 1.5
B. 1.7
C. 2.3
D. 2.5
E. 3.0

Let´s see an "alternate" way to apply the ALLIGATION method:

Image

\(?\,\, = x\)

\({x \over 4} = {{50 - 30} \over {50 - 15}} = {4 \over 7}\,\,\,\,\, \Rightarrow \,\,\,\,\,7x = 16\,\,\,\,\, \Rightarrow x = {{14 + 2} \over 7} = 2{2 \over 7}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{C}} \right)\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent  [#permalink]

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New post 16 Jan 2019, 12:24
Hi All,

We're told that Mixture A is 15 percent alcohol and mixture B is 50 percent alcohol and when a certain amount of each is poured together to create a 4-gallon mixture, the result contains 30 percent alcohol. We're asked approximately how many gallons of mixture A are in the mixture. This "mixture" question can be approached in a number of different ways, including by TESTing THE ANSWERS.

To start, IF we had an EQUAL amount of each Mixture, then the average of the 4-gallon mixture would be (15 + 50)/2 = 65/2 = 32.5% alcohol. This result is TOO HIGH though; the actual mixture is 30%. This means that we'll need a little more of Mixture A... meaning that a little more than 2 gallons (of the 4-gallon total) will be Mixture A. Looking at the answer choices, it's highly likely that the correct answer is Answer C.

We can actually check to see if either Answer C is a match or if Answer D is "too small." We're looking for an answer that would lead to (4)(.3) = 1.2 gallons of alcohol. Answer D is a little 'easier' to work with, so let's TEST that Answer....

Answer D: 2.5 gallons

IF.... we have 2.5 gallons of Mixture A, then we have 4 - 2.5 = 1.5 gallons of Mixture B...
(2.5)(.15) + (1.5)(.5) =
.375 + .75 =
1.125 gallons of Alcohol. This answer is TOO SMALL (we need there to be 1.2 gallons of alcohol). To get more alcohol we need LESS of Mixture A than we have here. There's only one answer that fits...

Final Answer:

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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent   [#permalink] 16 Jan 2019, 12:24
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