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A 30% solution of barium chloride is mixed with 10 grams of water to f
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Updated on: 27 Oct 2014, 01:12
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73% (01:49) correct 27% (02:10) wrong based on 232 sessions
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A 30% solution of barium chloride is mixed with 10 grams of water to form a 20% solution. How many grams of original solution did we start with ? A) 10 B) 15 C) 20 D) 25 E) 30 I am trained to complete above problem as per following technique. Is it possible to solve in following manner. 30% _________100% ______20% ratio is 10020 : 2030, which is wrong. There should not be 20. Instead, value in the middle should be in between 30% to 100%. Is there any way to solve above problem as per technique.? Thank you
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Originally posted by acc1444 on 26 Oct 2014, 23:06.
Last edited by Bunuel on 27 Oct 2014, 01:12, edited 2 times in total.
Renamed the topic and edited the question.




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Re: A 30% solution of barium chloride is mixed with 10 grams of water to f
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27 Oct 2014, 20:25
Algebraic way: Say x gms of barium chloride solution mixed with water \(\frac{30}{100} x + 0 * 10\) ............. (1) Resultant is 20% of barium chloride \(\frac{20}{100} (x + 10)\) ................. (2) Equating (1) & (2) \(\frac{30}{100} x + 0 * 10 = \frac{20}{100} (x + 10)\) x = 20 Answer = C
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Re: A 30% solution of barium chloride is mixed with 10 grams of water to f
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27 Oct 2014, 20:04
acc1444 wrote: A 30% solution of barium chloride is mixed with 10 grams of water to form a 20% solution. How many grams of original solution did we start with ?
A) 10 B) 15 C) 20 D) 25 E) 30
I am trained to complete above problem as per following technique. Is it possible to solve in following manner.
30%_________100% ______20%
ratio is 10020 : 2030, which is wrong. There should not be 20. Instead, value in the middle should be in between 30% to 100%.
Is there any way to solve above problem as per technique.?
Thank you 10 grams of water means 0% of barium chloride solutionTake it in this way..... 30% _________ 0% ______ 20% ................ 20  0= 20 ............. 3020 = 10 Answer = 20 = C
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Re: A 30% solution of barium chloride is mixed with 10 grams of water to f
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11 Jan 2015, 00:13
Let X gm be the starting solution For 30% solution, the amount of BaCl in gm = 0.3X gm  (I)
Now, 10 g of H2O was added to the solution. Therefore the new quantity of solution =(X + 10) gm This gives 20% solution. Hence the amount of BaCl in gm the new solution = 0.2 (X + 10) gm  (II)
Note that no new BaCl has been added. Hence the amount of BaCl should remain the same. Hence, we can equate I and II
0.3X = 0.2 (X + 10) 0.3X = 0.2X + 2 0.1X = 2 X = 20 = Answer



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Re: A 30% solution of barium chloride is mixed with 10 grams of water to f
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11 Jan 2015, 04:34
just deal with unchanged quantities > here unchanged quantity is Barium Chloride
Say original solution = X New solution will be X + 10
Barium Chloride remains same in both the solution
0.3(X) = 0.2(X + 10)
0.1X = 2
X = 20



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Re: A 30% solution of barium chloride is mixed with 10 grams of water to f
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11 Jan 2015, 16:24
Hi All, These types of "weighted" average questions can be solved in a few different ways. You can use the Weighted Average formula, you can use Allegation (which is a ratiobased approach) and (at least in this question) you can TEST THE ANSWERS.... We're told that 10g of water are mixed with an unknown amount of 30% barium mixture to form a 20% barium final mixture. We're asked to figure out how much barium mixture we have.... Logically, if we had 10g of water mixed with 10g of barium mixture, we'd end up with a 15% overall mixture, so 10 is NOT the answer. We need MORE barium mixture to raise the final mixture up to 20%. Since the numbers in the prompt are all "round" numbers, it's highly likely that the correct answer will ALSO be a round number.... Let's TEST Answer C = 20g (20g of barium mixture)(30% barium) = 6g barium + 10g water..... (6g barium)/(30g total liquid) = .2 = 20% final barium mixture. So, the initial barium mixture MUST be 20g Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: A 30% solution of barium chloride is mixed with 10 grams of water to f
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13 Jan 2015, 15:05
Another quick way for these type of problems:
(old  new)/new = dilution (3020)/(20) = 1/2 = dilution = 10 gm So, original solution = (1/2)*10 gm = 20 gm



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Re: A 30% solution of barium chloride is mixed with 10 grams of water to f
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04 Jun 2016, 03:27
Let's x be the original quantity
we have 20%(10+x) = 30%(x) 200 + 20x = 30x x=20



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Re: A 30% solution of barium chloride is mixed with 10 grams of water to f
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04 Jun 2016, 08:02
acc1444 wrote: A 30% solution of barium chloride is mixed with 10 grams of water to form a 20% solution. How many grams of original solution did we start with ?
A) 10 B) 15 C) 20 D) 25 E) 30
I am trained to complete above problem as per following technique. Is it possible to solve in following manner.
30%_________100% ______20%
ratio is 10020 : 2030, which is wrong. There should not be 20. Instead, value in the middle should be in between 30% to 100%.
Is there any way to solve above problem as per technique.?
Thank you 70% water (x gm) is mixed with 100% water (10gm) to give 80% water solution x/10= 10080/ 8070= 20/10 x= 20 C is the answer
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Re: A 30% solution of barium chloride is mixed with 10 grams of water to f
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07 Jun 2016, 02:19
Simplest way is weighted average:Just remember that water has a 0% concentration.
and use the weighted average formula: C1V1+C2V2/V1+V2=Final concentration.
0.3x+10*0/x+10=0.2
Solving,x=20.



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Re: A 30% solution of barium chloride is mixed with 10 grams of water to f
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27 Oct 2016, 17:40
0.70x+10=0.80(10+x) 0.70x+10=8+0.80x 2=0.10x x=20



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Re: A 30% solution of barium chloride is mixed with 10 grams of water to f
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26 Feb 2018, 22:06
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