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A portion of the 85% solution of chemicals was replaced with [#permalink]
04 Nov 2007, 18:56

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Difficulty:

25% (medium)

Question Stats:

78% (02:26) correct
22% (02:13) wrong based on 210 sessions

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?

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

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?

Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
24 Mar 2012, 02:01

Expert's post

1

This post was BOOKMARKED

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.

Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
09 Oct 2013, 17:42

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Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
18 Oct 2013, 20:35

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. _________________

“Confidence comes not from always being right but from not fearing to be wrong.”

Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
05 Oct 2014, 06:43

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%. 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)

Re: A portion of the 85% solution of chemicals was replaced with [#permalink]
06 Oct 2014, 09:25

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

A portion of the 85% solution of chemicals was replaced with [#permalink]
07 Oct 2014, 23:58

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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