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Re: Mixture Problems  I collected from GMAT Club [#permalink]
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11 Jan 2010, 01:44
I think the answer for 13 is B
From the question, we have below equation: x+y = 10(1)
From 1ST, y>4 > insufficent From 2ST, 3x + 5y <40 (3) Assume 3x+5y = 40 (2) From (1) & (2): we have x= 5, y=5 (3) < 40 when y<5, y= 4,3,2,1 x>5, x=6,7,8,9 therefore x>y. We don't need 1ST to answer this question.
Pls correct me, because I am not sure my solution is right. Thanks



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Re: Mixture Problems  I collected from GMAT Club [#permalink]
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02 Feb 2010, 18:43
samrus98 wrote: tejal777 wrote: Wanna confirm my answer for 2..(B)? Yep, the answer for Q2 is B i.e. 2.5 => 3*0.25 = 0.75 gallons of white paint and 5*0.25 = 1.25 gallons of black paint are needed. But since only 0.5 and 1 gallon cans are available, 1 gallon of white paint and 1.5 gallon of black paint would be needed => 1 + 1.5 = 2.5 Thank you so much for your explanation. I am not too sure I am even understanding the question correctly. Why did you multiply the parts of paint by 0.25. Where does that number come from? Thanks!



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Re: Mixture Problems  I collected from GMAT Club [#permalink]
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03 Mar 2010, 17:47
Can anyone explain number 3 and how to solve??



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Re: Mixture Problems  I collected from GMAT Club [#permalink]
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15 Mar 2010, 17:17
Wow, I am amazed yet again for GMAT clubbers! +1!
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Re: Mixture Problems  I collected from GMAT Club [#permalink]
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31 Aug 2010, 23:54
Hi, Thanks a lot for this, I am really weak at mixture and work problems. I was planning on running through the OG and Kaplan looking for mixture problems. This saved me a lot of work!



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Re: Mixture Problems  I collected from GMAT Club [#permalink]
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04 Oct 2010, 17:35
I really enjoyed this set. Thanks a lot.



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Re: Mixture Problems  I collected from GMAT Club [#permalink]
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10 Mar 2011, 03:02
I guess the answer for the 13th question is B.Please correct me if you differ.



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Re: Mixture Problems  I collected from GMAT Club [#permalink]
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10 Mar 2011, 11:56
great stuff, thanks!



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Re: Mixture Problems  I collected from GMAT Club [#permalink]
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11 Mar 2011, 05:35
8.Solution Y is 30 percent liquid X and 70 percent water. If 2 kilograms of water evaporate from 8 kilograms of solution Y and 2 kilograms of solution Y are added to the remaining 6 kilograms of liquid, what percent of this new solution is liquid X? (A) 30% (B) 33 1/3% (C) 37 1/2% (D) 40% (E) 50%
Step 1: Given: Y = 30% L + 70% W
Given: 8 kg solution. Therefore: 30% L = 2.4 kg Liquid X and 70% W = 5.6 kg Water.
Given: 2 kg water evaporate, therefore remaining water = 5.6 kg  2kg = 3.6 kg water.
Now, total solution weight = 3.6 kg (water) + 2.4 kg (liquid X) = 6 kg Liquid (this is already stated in the problem)
Next step, the problem states that 2 kg solution Y are added to the 6 kg of liquid...Therefore in 2 kg of Solution Y, we have 30% Liquid X = 0.6 kg Liquid X AND 70% Water = 1.4 kg Water. (Add these up, they total to 2 kg).
The question asks, what percent of this NEW solution is liquid X?
So: Liquid X = Original solution + added solution = 2.4 kg + 0.6 kg = 3 kg in NEW solution.
New solution total = (Original + added) liquid X + (Original + added) water = (2.4+0.6) + (3.6+1.4) = 8 kg
(Liq X / total soln) x 100 = (3/8) x 100 = 37.5%
Hollaaaaaaaaaaa!



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Re: Mixture Problems  I collected from GMAT Club [#permalink]
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11 Mar 2011, 09:13
15.Two different solutions of alcohol with respective proportions of alcohol to water of 3:1 and 2:3 were combined. What is the concentration of alcohol in the new solution if the first solution was 4 times the amount of the second solution?
Solution: Always start these problems off by assuming a smart number...assume 40 for the first solution here...the second solution is 10 (based on the question) now: 3x + x =40, therefore we have 30 units of alcohol and 10 units of water in the first solution. 2x + 3x = 10, therefore we have 4 units of alcohol to 6 units of water in the 2nd solution. Now adding alcohol up = 30 + 4 = 34. Percent of alcohol in new solution = alcohol / new total = (34/50) X 100 = 68 %



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Re: Mixture Problems  I collected from GMAT Club [#permalink]
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16 May 2011, 22:17
Step 1:
Y = 30% (x) + 70% of Water
So, the 8 Kg of Solution Y contains:
30% of 8 Kg is Liquid x + 70% of 8 Kg is Water
i.e 2.4 Kg of Liquid x + 5.6 Kg of Water
Step 2:
If 2 Kg of water is evaporated, then the remaining solution contains:
2.4 Kg of Liquid x + 3.6 Kg of Water
Step 3:
Now 2 Kg of Solution Y is added to the remaining 6 Kg of Solution
In 2 Kg of Solution Y, it contains:
0.6 Kg of Liquid x + 1.4 Kg of Water
Step 4:
Now, lets add the contents of the remaining 6 Kg of Solution and the fresh 2 Kg of Solution Y. The final contents are:
3 Kg of Liquid x + 5 Kg of Water
The percentage of Liquid x in the new solution is 3Kg/8 Kg i.e 37 ½ %
The answer is (C)



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Re: Mixture Problems  I collected from GMAT Club [#permalink]
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26 Jun 2011, 17:55
What a great thread! Attempted all the 30 problems  got 26 correct.
Messed up the "pig" problem  which I think is a superb problem  (problem number 9 on the list).
I found problem number 5 to be difficult as well (Another DS question).
Problem 25 is a very crafty one as well  tests the concept of weighted averages.



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Re: Mixture Problems  I collected from GMAT Club [#permalink]
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02 Jul 2011, 20:32
How much water (in grams) should be added to a 35%solution of acid to obtain a 10%solution?
1)There are 50 grams of the 35%solution. 2)In the 35%solution the ratio of acid to water is 7:13.
Using statement 1) we can solve the question.. in 50g of 35% solution we have 17.5g acid and 32.5g water. If we add x gram of water to the solution to obtain a 10% solution, the equation becomes 17.5 / (50+x) =10% ...Thus we can solve for x which is 125g.
Statement 2) talks about ratio and not the exact quantity..We will have multiple solutions for the same and hence not sufficient.
Hence, the answer is A.



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Re: Mixture Problems  I collected from GMAT Club [#permalink]
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14 Dec 2011, 12:43



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Re: Seed mixture X is 40 percent ryegrass and 60 + More Problems [#permalink]
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