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# A portion of the 85% solution of chemicals was replaced with

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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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Question Stats:

79% (01:32) correct 21% (02:01) wrong based on 609 sessions

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Mar 2012, 02:52, edited 1 time in total.

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04 Nov 2007, 20:14
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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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04 Nov 2007, 20:20
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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 10687 times ]

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24 Mar 2009, 19:47
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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24 Mar 2009, 22:41
pls visit my post in t55740-ps-intensity-red-paint

hope it helps you to do it in a much easier way

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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12 Nov 2009, 14:05

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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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?

(1-x)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 (x-y).85
Now, equal amount of solution that is replaced(y) is added, therefore => .20y

(x-y).85 + .20y = .40x
.45x = .65y
y=9/13

Hope this helps

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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
x part is replaced so 1-x is not replaced
(1-x)0.85+0.20x = 0.40
0.85 - 0.85x + 0.20x = 0.40
0.65x = 0.45
x = 9/13
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23 Mar 2012, 12:38
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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bmwhype2 wrote:
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

This is a weighted average question. Say x% of the solution was replaced --> equate the amount of chemicals: 0.85(1-x)+0.2*x=0.4 --> x=9/13.

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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(1-x) + 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: S1-100 units containing 85 u chemical
S2-100 units containing 20 units of chemical
We're given:85-0.85x+0.20x=40u
Solving the equation we get 9/13%.
(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:

85-------40-----20

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
Hello from the GMAT Club BumpBot!

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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

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Re: A portion of the 85% solution of chemicals was replaced with   [#permalink] 04 Apr 2016, 16:21

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