Question

The strength of an indigo solution in percentage is equal to the amount of indigo in grams per 100 cc of water. Two 800 cc bottles are filled with indigo solutions of strengths 33% and 17%, respectively. A part of the solution from the first bottle is thrown away and replaced by an equal volume of the solution from the second bottle. If the strength of the indigo solution in the first bottle has now changed to 21% then the volume, in cc, of the solution left in the second bottle is

Answer 1

Let's break down the problem step-by-step.

1. Suppose \(x\) cc of the solution from the first bottle is thrown away.

The amount of indigo in the solution that is thrown away = \(0.33x\) grams.

After this, there's \(800 - x\) cc of solution left in the first bottle, containing \(0.33(800) - 0.33x = 264 - 0.33x\)  grams of indigo

2. Now, \(x\) cc of the solution from the second bottle is added to the first bottle. 

The amount of indigo added from the second bottle =\(0.17x\) grams

After this addition, the total volume of the solution in the first bottle remains 800 cc. The total amount of indigo in the first bottle = \(264 - 0.33x + 0.17x = 264 - 0.16x\) grams

3. It's given that after these operations, the strength of the solution in the first bottle changes to 21%. So, the amount of indigo in 800 cc of the solution is \(( 0.21 \times 800 )\)

= \(168 \) grams.

Setting up the equation from the above information:

\(264 - 0.16x = 168\)

\(-0.16x = -96\)

\(x = 600\)

So, 600 cc of the solution was taken from the second bottle.

Now, to find the volume of the solution left in the second bottle:

Original volume - Volume taken out =\( 800 cc - 600 cc = 200 cc. \)

Thus, the volume of the solution left in the second bottle is 200 cc

Answer 2

Let Bottle A have an indigo solution of strength 33% while Bottle B have an indigo solution of strength 17%.
The ratio in which we mix these two solutions to obtain a resultant solution of strength 21%:
\(\frac AB=\frac {21-17}{33-21}\)

\(\frac AB=\frac {4}{12}\)

\(\frac AB=\frac {1}{3}\)
Hence, three parts of the solution from Bottle B is mixed with one part of the solution from Bottle A. For this process to happen, we need to displace \(600\ cc\) of solution from Bottle A and replace it with \(600\ cc\) of solution from Bottle B.
Since both bottles have \(800\ cc\), three parts of this volume \(= 600\ cc\)
As a result, 200 cc of the solution remains in Bottle B.

So, the answer is \(200\ cc\).