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]

04 Nov 2007, 19:56

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[Reveal] Spoiler: OA

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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04 Nov 2007, 20:14

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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[#permalink]

04 Nov 2007, 20:20

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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: [#permalink]

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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Re: Re: [#permalink]

24 Mar 2009, 22:41

Hi bigfernhead,
pls visit my post in t55740-ps-intensity-red-paint

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

bigfernhead wrote:

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]

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]

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

Consider kudos for the post

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Re: Mixture [#permalink]

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]

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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Re: [#permalink]

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]

24 Mar 2012, 03:01

bmwhype2 wrote:

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.

Answer: C.
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Re: A portion of the 85% solution of chemicals was replaced with [#permalink]

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

Re: A portion of the 85% solution of chemicals was replaced with [#permalink] 25 Mar 2012, 12:04

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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

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