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

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

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.

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

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

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

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

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

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

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

Answer: C.

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

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.

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)

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

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

let x=part of original solution replaced
.85-.85x+.2x=.4
x=45/65=9/13

w1 is the weight 85% chemical solution which has concentration C1 = 85%
w2 is the weight of 20% chemical solution which has concentration C1 = 20%
Concentration of mixture (average) = 40%

using weihted average formula w1/w2 = (c2 - avg) / (Avg-c1)
w1/w2 = (20-40)/(40-85)
w1/w2 = -20/-45= 4/9

w1 is 4 part and w2 is 9 part in a total 13 part of solution.

question is What part of the original solution was replaced?
w2 is what was replaced, hence 9 out of 13 i.e. 9/13 part was replaced.

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

20 sec solution: : if half of solution A was removed the avg of 85% and 20% would be 52,5%. Since the new solution is 40%, which is closer to 20%, it means that more than half of solution B was used. We then have to find the answer that says that more than half of A was replaced.

Answers A and B say that less than half of solution A was removed

Answer D and E are impossible, we can't remove more than what we have.

Answer C says that more than half was removed = correct.