Question

A solution of \[9\%\] acid is to be diluted by adding \[3\%\] acid solution to it. The resulting mixture is to be more than \[5\%\] but less than \[7\%\] acid. If there is \[460\] litres of the \[9\%\] solution, how many litres of \[3\%\]solution will have to be added?

Answer 1

Hint: We know that by considering the quantity of the $3$ percent solution added to be x. Given a quantity of $9$ percent solution. Then the resulting mixture will be we litres. Proceeding by forming the appropriate inequalities according to the problem statement and then solving these inequalities, we will get the value of x i.e. quantity of solution to be added.

Complete answer:
Let the required quantity of $3$ percent boric acid solution be litres. Given, quantity of$9$ percent acid solution is $460litres.$
Let us assume that ‘x’ litres of \[3\%\] solution is added to \[460\text{ }L\] of \[9\%\] solution.
Thus, total solution \[=\left( 460+x \right)L\] & total acid content in resulting solution \[=460\left( \dfrac{9}{100} \right)+\text{ }x\left( \dfrac{3}{100} \right)\]
\[=\left( 41.4+0.03x \right)\%\]
Now, according to the question, the resulting mixture we get should be less than \[7\%\] acidic & more than \[5\%\]acidic Thus, we get, \[5\%\text{ }of\text{ }\left( 460\text{ }+\text{ }x \right)\text{ }<\text{ }41.4+0.03x<7\%\text{ }of\text{ }\left( 460+x \right)\]
\[=\text{ }\left( 23+0.05x \right)<\left( 41.4+0.03x \right)<\left( 32.2+0.07x \right)\]
Now we have; \[\left( 23+0.05x \right)<\left( 41.4+0.03x \right)\text{ }\And \text{ }~\left( 41.4+0.03x \right)<\left( 32.2+0.07x \right)=\text{ }0.02x<18.4~~\And ~~0.04x>9.2\]
Thus, \[2x < 1840\text{ }\And \text{ }4x > 920\text{ }=\text{ }230 < x < 920.\]
Therefore, the number of litres of the $3\%$ solution of acid must be more than $230$ and less than $920.$

Note:
Remember that in these types 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.