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If after 200 grams of water were added to the 24% solution
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14 Aug 2010, 07:42
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If after 200 grams of water were added to the 24%solution of alcohol, the strength of the solution decreased by onethird, how much of the 24%solution was used? A. 180 grams B. 220 grams C. 250 grams D. 350 grams E. 400 grams
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Re: MIXTUREEE
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30 Nov 2010, 11:20
bibha wrote: I don't know why i can't solve even the easiestttt mixture probs ((( If after 200 grams of water were added to the 24%solution of alcohol, the strength of the solution decreased by onethird, how much of the 24%solution was used? 180 grams 220 grams 250 grams 350 grams 400 grams The key to solving mixture problems is to properly identify the solutions you want to mix. Make circles. They let you see what is it that you are mixing with what. e.g. Attachment:
Ques1.jpg [ 23.49 KiB  Viewed 9340 times ]
Now simply use weighted averages formula: \(C_{avg} = \frac{C_1*V_1 + C_2 * V_2}{V_1 + V_2}\) \(16 = \frac{0*200 + 24 * x}{200 + x}\) (You can ignore % because it will just get canceled since it is there on both sides of the equation) x = 400 gm I do prefer the scale method (alligation shown by muralimba above) using the number line since it is faster and much more intuitive but if someone is facing problems in mixtures, I strongly recommend to stick with the formula for as long as it takes to get comfortable.
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Re: MIXTUREEE
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30 Nov 2010, 00:35
jessetoronto wrote: Some times I wonder if I should have Hermione Obliviate all Mixture memory and learning I have since beginning to study and start over!!! The easy ones are getting harder...ahhh For any mixture problem in GMAT, please use the alligation technique: trigger points: From any mixture problem, note down the below 3 pointers first. 1) What is solution 1 2) what is solution2 3) what is the resulting solution For example, i have a solution with 40% milk (==> rest i.e. 60% is water) and i have one more solution which is with 70% milk ==> 30% water qtn: if we mix these two solutions in a particulr ratio to get a resultant solution with 50% milk, how much % solution1 exists in the resultant solution? 4070 50 2010 ==> use the solution 1 and solution2 in the ratio 2:1 to get a resultant solution with 50% milk means if i use 200liters of sultion 1, then i have to use 100liters of solution 2 resulting 300liters of 50%milk..hope it is clear. Back to the Original qtn: solution1: water ==> 100% water solution2: 24% alcohol ==> 76% water resulting solution  alcohol concentarion is decreased by 1/3 ==> 24  24*1/3 = 16% alcohol==> 84% water 10076 84 816 ==> water:24%alcohol solution = 1:2 OR solution1: water ==> 0% alcohol solution2: 24% alcohol resulting solution  alcohol concentarion is decreased by 1/3 ==> 24  24*1/3 = 16% alcohol 024 16 816 ==> water:24%alcohol solution = 1:2 (SAME AS ABOVE) given that 200liters of water ==> 2*200 = 400 liters of 24%alcohol solution ANSWER "E". Regards, Murali. Kudos?




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Re: MIXTUREEE
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14 Aug 2010, 11:31
bibha wrote: If after 200 grams of water were added to the 24%solution of alcohol, the strength of the solution decreased by onethird, how much of the 24%solution was used?
A. 180 grams B. 220 grams C. 250 grams D. 350 grams E. 400 grams Let the weight of 24% solution used be \(x\) grams, weight of alcohol in it would be \(0.24x\). As in final solution strength decreased be 1/3 thus it became 24*2/3=16%. Set the equation: \(0.24x=0.16(x+200)\), the weight of 16% alcohol in \(x+200\) grams of new solution comes only from (equals to) 24% alcohol in \(x\) grams of strong (initial) solution, as there is 0 grams of alcohol in water (0% alcohol solution) > \(0.08x=32\) > \(x=400\). Answer: E.
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Re: MIXTUREEE
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29 Nov 2010, 22:09
Some times I wonder if I should have Hermione Obliviate all Mixture memory and learning I have since beginning to study and start over!!! The easy ones are getting harder...ahhh



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Re: MIXTUREEE
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29 Nov 2010, 22:22
The 24% alcohol solution decreases by 1/3 once the 200 grams of water is added so 200/.08 =2500*.16% (2/3 of 24%) is 400 grams, thus answer is E.



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Re: MIXTUREEE
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05 Dec 2010, 01:59
Quote: I do prefer the scale method (alligation shown by muralimba above) using the number line since it is faster and much more intuitive but if someone is facing problems in mixtures, I strongly recommend to stick with the formula for as long as it takes to get comfortable. Thanks Karishma. Looking at the solution, I think this 'alligation method' is quite handy and cool but I'm not sure I understand completely. Could you please explain? Thanks in advance.



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Re: MIXTUREEE
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05 Dec 2010, 06:38
Werewolf wrote: Quote: I do prefer the scale method (alligation shown by muralimba above) using the number line since it is faster and much more intuitive but if someone is facing problems in mixtures, I strongly recommend to stick with the formula for as long as it takes to get comfortable. Thanks Karishma. Looking at the solution, I think this 'alligation method' is quite handy and cool but I'm not sure I understand completely. Could you please explain? Thanks in advance. The alligation method, or the scale method as we call it, is based on the weighted averages formula itself: \(C_{avg} = \frac{C_1*V_1 + C_2 * V_2}{V_1 + V_2}\) If I rearrange the formula, I get \(\frac{V_1}{V_2} = \frac{C_2  C_{avg}}{C_{avg}  C_1}\) So I get that volume will be in the same ratio as difference between higher concentration and average concentration and difference between average concentration and lower concentration. How does this help? Knowing this, we can directly make a diagram and get the answer. e.g. 9 liters of 40% solution of sugar is mixed with a 90% solution of sugar to obtain 60% solution of sugar. How much 90% solution was used? Draw: Attachment:
Ques1.jpg [ 6.74 KiB  Viewed 9218 times ]
On a scale, mark 40% as concentration1, 60% as average and 90% as concentration2. Now, distance between 40 and 60 is 20 and distance between 60 and 90 is 30, The ratio of Vol1/Vol2 will be 3:2 (Note, the numbers 20 and 30, ratio of 2:3 give ratio 3:2 for V1:V2 as seen by the formula) So if 9 liters of V1 was used, 6 liters of V2 was used. This method is especially useful what you have the average and need to find the volume. Now try the original question using this and get back if there is a doubt.
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Re: MIXTUREEE
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06 Feb 2012, 04:41
using VeritasPrepKarishma's last method  0%16%(0.24*2/3)24% v1/v2=(2416)/(160)=1/2 1/3x=200 x=600 (total) 2/3*600=y y=400 (24%solution) hope I got it right
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Re: MIXTUREEE
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06 Feb 2012, 09:57
LalaB wrote: using VeritasPrepKarishma's last method  0%16%(0.24*2/3)24% v1/v2=(2416)/(160)=1/2 1/3x=200 x=600 (total) 2/3*600=y y=400 (24%solution) hope I got it right Yes you did. Good job! I have discussed this concept in detail here: http://www.veritasprep.com/blog/2011/04 ... mixtures/Check it out if you have any doubts.
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Re: If after 200 grams of water were added to the 24% solution
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29 Apr 2012, 00:38
we can do it this way too 0%16%(0.24*2/3)24% v1/v2=(2416)/(160)=1/2 1/2=200/x x=400
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Re: If after 200 grams of water were added to the 24% solution
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15 Jan 2014, 04:57
Weight of original solution: x Alcohol of Original solution: 24/100x or 0.24 x
Alc. concentration in new solution decreases by 1/3 so its 2/3 of 24 % which is 16 %. Weight of new solution is x+200 (since 200 grams of water are added)
Lets say that Y is the amount of alc. in first solution: 24/100 * x = y Since the amount of alcohol isn't changed, we can calculate the amount of alcohol in the second solution like this: 16/100 * (x+200) = y
Now we can simplify, equalize and calculate: simplify: 24/100x = 6/25x and 16/100*(x+200) = 4/25 * (x+200) equalize: 6/25*x = 4/25 * (x+200) calculate: 6/25x = 4x+800/25 => 6x = 4x + 800 => 2x = 800 => x=400 As we have defined X as the weight of the original soluition, we have our answer here, which is E!
Hope it helps!



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Re: If after 200 grams of water were added to the 24% solution
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05 Mar 2014, 09:17
This is a differentials problem. Let;s see we have 16(200) + 8x = 0. Therefore we have that x = 400
Answer is clearly E
Hope this helps Cheers J



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Re: If after 200 grams of water were added to the 24% solution
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06 Mar 2014, 01:17
A:W RatioOriginal 6:19 New 4:21 solving for original mixture grams(x) from W perspective [200+(19x/25)]/(x+200)=21/25 25(200)+19x=21x+21(200) 800=2x x=400
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Re: If after 200 grams of water were added to the 24% solution
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09 Jun 2014, 09:46
Dear All,
Why doesn't this work?
24x ………. 16  =  100x +200 100
I've been trying to figure this one out for an hour, just can't seem to understand why this method won't work! Thank you so much! (PS Ignore the ……., that's just for formatting)



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Re: If after 200 grams of water were added to the 24% solution
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09 Jun 2014, 20:33
vak3ee wrote: Dear All,
Why doesn't this work?
24x ………. 16  =  100x +200 100
I've been trying to figure this one out for an hour, just can't seem to understand why this method won't work! Thank you so much! (PS Ignore the ……., that's just for formatting) When a solution is 24% alcohol solution, it means it has 24 parts alcohol and 76 parts water. In the denominator, you are adding 200 (pure water) to 100x (total solution). This doesn't make sense. You need to add 200 to only water. So if you set up the equation like this, it will work: \(\frac{24x}{(76x + 200)} = \frac{16}{84}\) Again, remember that the final solution has 16% alcohol so that means 16 parts alcohol and 84 parts water. You have assumed that the weight of the original solution is 100x. So once you solve this equation, you get x = 4 and hence weight of original solution is 100*4 = 400 gms.
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Re: If after 200 grams of water were added to the 24% solution
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10 Jun 2014, 07:19
Thank you for the reply! I just tried solving for the following: (fractions are reduced from original problem) 6x………………..4  =  25x +200 …….25 and x=16 16*25=400 so it also works! There's so many ways to solve this problem apparently!



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Re: If after 200 grams of water were added to the 24% solution
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26 Jun 2014, 14:24
J can you please explain this in more detail. This was really quick. Thanks, jlgdr wrote: This is a differentials problem. Let;s see we have 16(200) + 8x = 0. Therefore we have that x = 400
Answer is clearly E
Hope this helps Cheers J



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If after 200 grams of water were added to the 24% solution
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16 Oct 2014, 08:30
bibha wrote: If after 200 grams of water were added to the 24%solution of alcohol, the strength of the solution decreased by onethird, how much of the 24%solution was used?
A. 180 grams B. 220 grams C. 250 grams D. 350 grams E. 400 grams sol: 24 0 \ / 16 / \ 16 8 why 16 in middle? since the lost strength is 1/3 so remaining strength is 2/3 (which is resultant strength), there 2*24/3 = 16 why zero? strength is amount of solute present in solvent. here we have 100% solvent and 0% solute. Hence strength of alcohol in pure water is 0. ratio is 16: 8 or 2:1 = x: 200 there x = 400 20sec approach! Is this really a 700+ question?



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Re: If after 200 grams of water were added to the 24% solution
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23 Oct 2015, 19:16
x=weight of original solution .76x+200=.84(x+200) x=400 grams




Re: If after 200 grams of water were added to the 24% solution &nbs
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