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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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14 Jan 2011, 05:18
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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
[Reveal] Spoiler: OA
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14 Jan 2011, 06:21
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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: $$weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}$$ --> $$0.3=\frac{0.15*A+0.5(4-A)}{4}$$ --> $$A\approx{2.3}$$.

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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14 Jan 2011, 06:37
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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 Problem (What is the best way to approach this?) [#permalink]

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14 Jan 2011, 19:01
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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 [ 3.9 KiB | Viewed 6589 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 A is 15 percent alcohol, and mixture B is 50 percent [#permalink]

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13 Jun 2013, 02:09
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent [#permalink]

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13 Jul 2013, 22:45
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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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13 Jul 2013, 23:19
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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24 Oct 2013, 22:48
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Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent [#permalink]

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22 Sep 2014, 03:10
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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24 Feb 2015, 03:34
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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05 Jul 2015, 04:22
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: $$weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}$$ --> $$0.3=\frac{0.15*A+0.5(4-A)}{4}$$ --> $$A\approx{2.3}$$.

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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05 Jul 2015, 04:33
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: $$weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}$$ --> $$0.3=\frac{0.15*A+0.5(4-A)}{4}$$ --> $$A\approx{2.3}$$.

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..
$$0.15A + 0.5B = 0.3(A+B)$$
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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05 Jul 2015, 05:10
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..
$$0.15A + 0.5B = 0.3(A+B)$$
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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05 Jul 2015, 05:50
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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05 Jul 2015, 05:52
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..
$$0.15A + 0.5B = 0.3(A+B)$$
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] 05 Jul 2015, 05:52
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