Question

A solution of 8 percent boric acid is to be diluted by adding a 2 percent boric acid solution to it. The resulting mixture is to be more than 4 percent but less than 6 percent boric acid. If we have 640 litres of the 8 percent solution, how many litres of the 2 percent solution will have to be added?

Answer 1

Hint: Here, we will be proceeding by forming the appropriate inequalities according to the problem statement and then solving these inequalities.

Complete step-by-step answer:

Let the required quantity of 2 percent boric acid solution be $x$ litres.

Given, quantity of 8 percent boric acid solution\[ = 640\] litres

Now, when the 8 percent boric acid solution is mixed with 4 percent boric acid, the quantity of the resulting mixture will be \[\left( {640 + x} \right)\] litres.

Also, given that the resulting mixture should be more than 4 percent and less than 6 percent boric acid solution i.e., 6 percent of \[\left( {640 + x} \right)\]\[ > \]8 percent of 640\[ + \]2 percent of $x$\[ > \]4 percent of \[\left( {640 + x} \right)\]

\[\Rightarrow \left( {\dfrac{6}{{100}}} \right) \times \left( {640 + x} \right) > \left( {\dfrac{8}{{100}}} \right) \times 640 + \left( {\dfrac{2}{{100}}} \right) \times x > \left( {\dfrac{4}{{100}}} \right) \times \left( {640 + x} \right)\]

\[\Rightarrow 6\left( {640 + x} \right) > 8 \times 640 + 2x > 4\left( {640 + x} \right) \]

\[\Rightarrow 3840 + 6x > 5120 + 2x > 2560 + 4x \]

Now, solving first and second inequalities separately, we get

\[ \Rightarrow 3840 + 6x > 5120 + 2x \Rightarrow 4x > 1280 \Rightarrow x > 320\] and \[5120 + 2x > 2560 + 4x \Rightarrow 2560 > 2x \Rightarrow x < {\text{1280}}\]

By combining above two inequalities, we will get the range of \[x\] as

\[1280 > x > 320\]

Therefore, the quantity of 2 percent boric acid solution should be between 320 litres and 1280 litres.

Note: In these type of problems, the question statement is very crucial. According to the problem statement, all the inequalities are formed and hence these inequalities are further reduced to the simplest form and evaluation of the variable is aimed. Here, after solving we are getting a range of the values instead of a particular value.