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Re: Mixture Problems - I collected from GMAT Club [#permalink]
11 Jan 2010, 00:44

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

Re: Mixture Problems - I collected from GMAT Club [#permalink]
02 Feb 2010, 17:43

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?

Re: Mixture Problems - I collected from GMAT Club [#permalink]
15 Mar 2010, 16:17

Wow, I am amazed yet again for GMAT clubbers! +1! _________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own. I am the force; I can clear any obstacle before me or I can be lost in the maze. My choice; my responsibility; win or lose, only I hold the key to my destiny - Elaine Maxwell

Re: Mixture Problems - I collected from GMAT Club [#permalink]
31 Aug 2010, 22:54

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!

Re: Mixture Problems - I collected from GMAT Club [#permalink]
11 Mar 2011, 04:35

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

Re: Mixture Problems - I collected from GMAT Club [#permalink]
11 Mar 2011, 08:13

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 %

Re: Mixture Problems - I collected from GMAT Club [#permalink]
02 Jul 2011, 19:32

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.

Re: Seed mixture X is 40 percent ryegrass and 60 + More Problems [#permalink]
28 Jan 2014, 23:27

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