Question

One container is filled with mixture that is 25% acid. A second container is filled with a mixture that is 55% acid. The second container is 55% larger than the first, and the two containers are emptied into a third container. What percent of acid is the third container?

Answer 1

Answer:

Therefore the third container contains   $\frac{735}{17}\%$ = 43.23 %  acid.

Step-by-step explanation:

Given that, One container is filled with the mixture that 25% acid. 55% acid is contain by second container.

Let the volume of first container be x cubic unit.

Since the volume of second container is 55% larger than the first.

Then the volume of the second container is

$= (V\times \frac{100+55}{100})$   cubic unit.

$=\frac{155}{100}V$  cubic unit.

The amount of acid in first container is

$=V\times \frac{25}{100}$  cubic unit.

$=\frac{25}{100}V$  cubic unit.

The amount of acid in second container is

$=(\frac{155}{100}V \times \frac{55}{100})$ cubic unit.

$=\frac{8525}{10000}V$ cubic unit.

Total amount of acid $=(\frac{25}{100}V+\frac{8525}{10000}V)$ cubic unit.

                                   $=(\frac{2500+8525}{10000}V)$ cubic unit.

                                   $=(\frac{11025}{10000}V)$cubic unit.

Total volume of mixture $=(V+\frac{155}{100}V)$ cubic unit.

                                       $=\frac{100+155}{100}V$ cubic unit.

                                       $=\frac{255}{100}V$  cubic unit.

The amount of acid in the mixture is

$=\frac{\textrm{The volume of acid}}{\textrm{The amount of mixture}}\times 100 \%$

$=\frac {\frac{11025}{10000}V}{\frac{255}{100}V}\times 100\%$

$=\frac{735}{17}\%$

Therefore the third container contains  $\frac{735}{17}\%$  acid.