Question

A solution of $8\%$ boric is to be diluted by adding a $2\%$ boric acid solution to it. The resulting mixture is to be more than $4\%$ but less than $6\%$ boric acid. If we have $640L$ of the $8\%$ solution, of the $2\%$ solution will have to be added is

• A. more than 320 and less than 1000
• B. more than 160 and less than 320
• C. more than 320 and less than 1280
• D. more than 320 and less than 640

Answer 1

Answer: (C)

Let the $2\%$ boric acid solution be $\times L$.
$\therefore \text{ Mixture }=(640+x)L$
Now, according to the question, two conditions arise
I. $2\%$ of $x+8\%$ of $640>4\%$ of $(640+x)$
II. $2\%$ of $x+8\%$ of $640<6\%$ of $(640+x)$
From condition I,
$\frac{2}{100}\times x+\frac{8}{100}\times 640>\frac{4}{100}\times (640+x)$
Multiplying both sides by 100 ,
$100\times \left[\frac{2x}{100}+\frac{8}{100}\times 640\right]>\frac{4}{100}\times (640+x)\times 100$
$\Rightarrow 2x+8\times 640>4\times 640+4x$
Transferring the term $4x$ to $LHS$ and the term $(8\times 640)$ to $RHS$,
$2x-4x>4\times 640-8\times 640$
$\Rightarrow -2x>640(4-8)$
$\Rightarrow -2x>-4\times 640$
Dividing both sides by $-2$,
$\frac{-2x}{-2}<\frac{-4\times 640}{-2}$
$\Rightarrow x<2\times 640$
$\Rightarrow x<1280$...(i)
From condition II,
$\frac{2}{100}\times x+\frac{8}{100}\times 640<\frac{6}{100}\times (640+x)$
$\Rightarrow 100\times \left[\frac{2x}{100}+\frac{8}{100}\times 640\right]<[6\times 640+6x]\times \frac{100}{100}$
$\Rightarrow 2x+8\times 640<6\times 640+6x$
Transferring the term $6x$ to LHS and the term $(8\times 640)$ to RHS,
$2x-6x<6\times 640-8\times 640$
$\Rightarrow -4x<640(6-8)$
$\Rightarrow -4x<-2\times 640$
Dividing both sides by $-4$,
$\frac{-4x}{-4}>\frac{-2\times 640}{-4}$
$x>320$ ...(ii)
Hence, from Eqs. (i) and (ii),
$320<x<1280\text{ i.e., }x\in (320,1280)$
The number of litres to be added should be greater than $320L$ and less than $1280L$.