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505-555 (Easy)|   Algebra|   Mixture Problems|                              
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Jackie has two solutions that are 2 percent sulfuric acid 27 Dec 2021, 08:59
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Jackie has two solutions that are 2 percent sulfuric acid 30 Dec 2021, 04:04
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KarishmaB
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Alright I'm so frustrated but after reading the article she posted. I try my best to answer using my way of thinking:

7 cups + 3 cups = 10 cups..so you need 10 cups of 5% solution to produce 60 liters. There is 10 cups and there is 60 liters. We have 60, we have 10 so the other number is missing is 6 therefore 6*7 cups = 42 liters of solution


Yes, you are right. Look, for every 10 cups/liters/ml/gallons/units etc of 5% solution, you need 7 cups/liters/ml/gallons/units etc of 2% solution and 3 cups/liters/ml/gallons/units etc of 12% solution.

So for each 10 litres of 5% solution, you need 7 litres of 2% and 3 litres of 12%.
Then how much will you need for 60 litres of 5% solution?

You will need 6 times the original quantity so you will need 7*6 = 42 litres of 2% solution and 3*6 = 18 litres of 12% solution.
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Can you please explain how did you get this - " for every 10 cups/liters/ml/gallons/units etc of 5% solution, you need 7 cups/liters/ml/gallons/units etc of 2% solution and 3 cups/liters/ml/gallons/units etc of 12% solution."
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Jackie has two solutions that are 2 percent sulfuric acid 06 Jan 2022, 08:57
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0,02∗x+0,12∗(60−x)=0,05∗600,02∗x+0,12∗(60−x)=0,05∗60
0,02∗x+7,2−0,12∗x=30,02∗x+7,2−0,12∗x=3
4,2=0,1∗x4,2=0,1∗x
x=0,42
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Jackie has two solutions that are 2 percent sulfuric acid 06 Jan 2022, 10:52
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cant we do this problem in allegation way if so how?
Thanks
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Jackie has two solutions that are 2 percent sulfuric acid 18 Jan 2022, 22:55
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KarishmaB
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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

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/
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KarishmaB the attached links aren't working. Please could you look into it :)
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Jackie has two solutions that are 2 percent sulfuric acid Updated on: 19 Sep 2023, 04:26
Originally posted by KarishmaB on 18 Jan 2022, 23:07.
Last edited by KarishmaB on 19 Sep 2023, 04:26, edited 1 time in total.
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Hoozan
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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

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 :)
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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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Jackie has two solutions that are 2 percent sulfuric acid 23 Feb 2022, 18:11
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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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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Jackie has two solutions that are 2 percent sulfuric acid 23 Feb 2022, 20:04
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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?
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woohoo921 Correct!
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Jackie has two solutions that are 2 percent sulfuric acid 18 Dec 2022, 19:38
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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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Jackie has two solutions that are 2 percent sulfuric acid 19 May 2023, 09:42
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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 25 Jun 2023, 06:26
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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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Jackie has two solutions that are 2 percent sulfuric acid 19 Jul 2023, 22:02
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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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Jackie has two solutions that are 2 percent sulfuric acid 29 Dec 2023, 22:29
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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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Jackie has two solutions that are 2 percent sulfuric acid 10 Nov 2024, 05:17
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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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Jackie has two solutions that are 2 percent sulfuric acid 16 Jul 2025, 19:26
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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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