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A portion of the 85% solution of chemicals was replaced with [#permalink]
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04 Nov 2007, 19:56
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A portion of the 85% solution of chemicals was replaced with an equal amount of 20% solution of chemicals. As a result, 40% solution of chemicals resulted. What part of the original solution was replaced? A. 5/13 B. 6/13 C. 9/13 D. 13/9 C. 13/6
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Last edited by Bunuel on 24 Mar 2012, 02:52, edited 1 time in total.
Edited the question, added the answer choices and OA



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Let the original amount of 85 percent solution = x and amount replaced be y
0.85(x  y)+ 0.2y = 0.4 x
0.85x 0.85y+0.2y = 0.4x
0.45x = 0.65y
9x = 13y
y = 9x/13
So, amount replaced = 9/13 of original amount



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See attachment
the new ratio is 20:45 = 4:9 > meaning that if we had 13ml of 85% solution we now have only 4ml so the part that was replaced is 9/13.
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Drawing Number 1.GIF [ 2.76 KiB  Viewed 10460 times ]



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Can someone explain to me killer squirrels method? my head is spinning. Specifically, this part: meaning that if we had 13ml of 85% solution we now have only 4ml so the part that was replaced is 9/13.KillerSquirrel wrote: See attachment the new ratio is 20:45 = 4:9 > meaning that if we had 13ml of 85% solution we now have only 4ml so the part that was replaced is 9/13.



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Hi bigfernhead, pls visit my post in t55740psintensityredpainthope it helps you to do it in a much easier way bigfernhead wrote: Can someone explain to me killer squirrels method? my head is spinning. Specifically, this part: meaning that if we had 13ml of 85% solution we now have only 4ml so the part that was replaced is 9/13.KillerSquirrel wrote: See attachment the new ratio is 20:45 = 4:9 > meaning that if we had 13ml of 85% solution we now have only 4ml so the part that was replaced is 9/13.



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Re: Mixture [#permalink]
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12 Nov 2009, 14:05
KS? Could you please develop your picture?
I haven't seen this kind of picture during my learnings, but it seems to be quite powerful ... I would appreciate having a better view of your method



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Re: Mixture [#permalink]
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17 Nov 2009, 22:31
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A portion of the 85% solution of chemicals was replaced with an equal amount of 20% solution of chemicals. As a result, 40% solution of chemicals resulted. What part of the original solution was replaced?
(1x)0.85 + .20x = .40 .85 .85x + .20x = .40 .45=.65x x=9/13
Alternate way
Let there be x litres of 85% solution initially Suppose y litres of 85% solution is removed
Then the remaining solution is (xy).85 Now, equal amount of solution that is replaced(y) is added, therefore => .20y
(xy).85 + .20y = .40x .45x = .65y y=9/13
Hope this helps
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Re: Mixture [#permalink]
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12 Mar 2011, 23:31
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9/13. The formula for these problems is simple. Whole  part removed + part added = Total 85% (100)  85% (X) + 20% (X) = 40% (100)
solve



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Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
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14 Dec 2011, 10:37



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KillerSquirrel wrote: See attachment the new ratio is 20:45 = 4:9 > meaning that if we had 13ml of 85% solution we now have only 4ml so the part that was replaced is 9/13. KillerSquirrel, i follow the Ratio how you calculated and got o 4/9, and 9+4 = 13 Ml and 85% wold be 4. but how did you determine that 9 the denominator of 4/9 should go over 13? I'm very new to these types of math problems



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Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
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24 Mar 2012, 03:01



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Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
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25 Mar 2012, 12:04
Ans C
Solution Replaced = x
85/100(1x) + 20x/100 = 40/100
85/100  85x/100 + 20x/100 = 40/100
85/100  65x/100 = 40/100
85/100  40/100 = 65x/100
45/100 = 65x/100
45/65 = x 9/13 = x



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Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
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09 Oct 2013, 18:42
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Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
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18 Oct 2013, 21:35
i like the allegation way of resolving problems We all will get to the point  that to achieve 40% of solution, we need to have 4 parts of 85% sol and 9 part of 20% solution. Question asks how many part of 85% was removed? We had to remove 9/13 parts of 85% solution. If its difficult to understand, imagine a container full of 85% solution or we have 13 parts of it. Now we keep replacing 85% with 20% solution till the time it reaches 40%. We know that in final solution both will be in ratio of 4:9. How much did we remove? We need to remove 9 parts of 85% solution.
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Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
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05 Oct 2014, 07:43
I reached the solution this way: Let there be 2 solutions: S1100 units containing 85 u chemical S2100 units containing 20 units of chemical We're given:850.85x+0.20x=40u Solving the equation we get 9/13%. Hence answer C. (Note:Whatever quantity of S1 or S2 we take out, say x in this case, will contain 85% and 20% of chemicals only)



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Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
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06 Oct 2014, 10:25
I kind of got confused with 85% solution....85% of what? Is the question saying that 85% of the solution is made up of chemicals?? it was hard for me to visualize what the question was actually asking



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A portion of the 85% solution of chemicals was replaced with [#permalink]
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08 Oct 2014, 00:58
angelfire213 wrote: I kind of got confused with 85% solution....85% of what? Is the question saying that 85% of the solution is made up of chemicals?? it was hard for me to visualize what the question was actually asking Take it in this way..... Solution I >> 85% solution of chemicals Solution II >> 20% solution of chemicals Resultant Mixture >> 40% solution of chemicals "Replacement" of Solution I means "Addition" of Solution II. They are asking how much Solution II is added? I used method of allegation for this as shown in above posts; works perfect
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Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
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22 Oct 2014, 03:17
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Differential approach is:
854020
45x=20y x/y=20/45=4/9 meaning that final solution (40%) have 4 part of 85% (original) and 9 part of 20%. So, 9/13 was replaced
Weigthed mean:
85x+20y/x+y=40 85x+20y=40x+40y 45x=20y x/y=4/9, so the same as above
C



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Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
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24 Nov 2015, 00:41
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Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
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04 Apr 2016, 16:21
let x=part of original solution replaced .85.85x+.2x=.4 x=45/65=9/13




Re: A portion of the 85% solution of chemicals was replaced with
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