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505-555 (Easy)|   Algebra|   Mixture Problems|                              
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cant we do this problem in allegation way if so how?
Thanks
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KarishmaB
Akgmat85
Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid, approximately how many liters of the 2 percent solution will be required?

A) 18
B) 20
C) 24
D) 36
E) 42

Use weighted average:

2% and 12% solutions mix to give 5% solution.

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (12 - 5)/(5 - 2) = 7/3

You need 7 parts of 2% solution and 3 parts of 12% solution to get 10 parts of 5% solution.
If total 5% solution is actually 60 litres, you need 7*6 = 42 litres of 2% solution and 3*6 = 18 litres of 12% solution.

Answer (E)

The formula and its application in mixtures are discussed in the following two posts:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... -averages/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/04 ... -mixtures/

KarishmaB the attached links aren't working. Please could you look into it :)
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Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
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Hoozan
KarishmaB
Akgmat85
Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid, approximately how many liters of the 2 percent solution will be required?

A) 18
B) 20
C) 24
D) 36
E) 42

Use weighted average:

2% and 12% solutions mix to give 5% solution.

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (12 - 5)/(5 - 2) = 7/3

You need 7 parts of 2% solution and 3 parts of 12% solution to get 10 parts of 5% solution.
If total 5% solution is actually 60 litres, you need 7*6 = 42 litres of 2% solution and 3*6 = 18 litres of 12% solution.

Answer (E)

The formula and its application in mixtures are discussed in the following two posts:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... -averages/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... -mixtures/

KarishmaB the attached links aren't working. Please could you look into it :)

Not sure why. They are working for me. You can access the weighted average post here: https://anaprep.com/arithmetic-weighted-averages/
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

avigutman

This is so helpful! To confirm, you are using the "seesaw" method starting at around the one-minute mark, correct?
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woohoo921


This is so helpful! To confirm, you are using the "seesaw" method starting at around the one-minute mark, correct?

woohoo921 Correct!
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2% of Sulfuric acid + 12% of Sulfuric acid = 60ltrs of 5% sulfuric acid

0.05 * 60 = 3ltrs of Sulfuric acid

Let the 2% Sulfuric acid solution be represented by x
Let the 12% Sulfuric acid solution be represented by y

x + y = 60 => x = 60 - y [these solutions are mixed in appropriate quantities to produce 60 liters]

(2/100 * x) + (12/100 * y) = 3
=> [2/100 * (60 - y)] + (12/100 * y) = 3
=> 120 + 10y = 300
=> y= 18

Therefore x(or 2% sulfuric acid solution) = 60 - 18 = 42ltrs
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Let's assume that the volume of the 2 percent sulfuric acid solution needed is x liters. Therefore, the volume of the 12 percent sulfuric acid solution needed would be 60−x liters to make a total of 60 liters.

To calculate the amount of sulfuric acid in the final mixture, we can set up the following equation:
(0.02x+0.12(60−x))/60=0.05

Let's solve this equation:

0.02x+0.12(60−x)=0.05∗60

0.02x+7.2−0.12x=3

−0.1x=3−7.2

−0.1x=−4.2

Dividing both sides by -0.1:

x=−0.1/−4.2=42

Therefore, approximately 42 liters of the 2 percent sulfuric acid solution will be required.

The correct answer is (E) 42 liters.
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Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid, approximately how many liters of the 2 percent solution will be required?

A) 18
B) 20
C) 24
D) 36
E) 42

With formula, the resulting mixture % can be calculated as avg= (n1x1+n2x2)/n+n2
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I'll tell you an easy trick to solve these problems quickly.


2%........12%. --Current Mix
.....\...../
........5%. --Final Mix (60L)
...../.....\
7%........3%. -- what ratio of current mix req to form Final mix. (|12-5| =7 : |2-5| = 3)


So from above, we need the solution in 7:3 ratio. TF, 7/10*60 = 42. Ans
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Short video solution here, using the Teeter Totter Timesaver Method (1:42):



Teeter Totter Method Steps:

1) Draw the Endpoints and Weighted Average numbers on the Teeter Totter Diagram (the weighted average represents the final mixture, and goes above the triangle-shaped fulcrum)

2) Draw the weights on each side. The LARGER weight goes on the side with the SHORTER distance from the fulcrum

3) Draw the distances from the Endpoints to the Weighted Average

4) Solve the equation: (W2)/(W1) =(D1)/(D2)
(the weights and distances are in an INVERSE RATIO)­


Teeter Totter Basic Examples Playlist: https://www.youtube.com/playlist?list=PL2exXfCUscn8Hvafet5-IPH1eNNLSjQBP­
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    GMATCoachBen Thank you for the explanation. I just have a quick question: what does "by volume" mean here?
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Allegation formula:

(Final - B)/(A - B) * 1/100%


(5%-12%)/(2%-12%) * 1/100% = 70%

70% of 60 L = 42 L

E.
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