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 $640 L$ 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{2 x}{100}+\frac{8}{100} \times 640\right]>\frac{4}{100} \times(640+x) \times 100$
$ \Rightarrow 2 x+8 \times 640>4 \times 640+4 x$
Transferring the term $4 x$ to $LHS$ and the term $(8 \times 640)$ to $RHS$,
$2 x-4 x >4 \times 640-8 \times 640$
$\Rightarrow -2 x >640(4-8)$
$\Rightarrow -2 x >-4 \times 640$
Dividing both sides by $-2$,
$\frac{-2 x}{-2}<\frac{-4 \times 640}{-2}$
$\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{2 x}{100}+\frac{8}{100} \times 640\right]<[6 \times 640+6 x] \times \frac{100}{100} $
$\Rightarrow 2 x+8 \times 640<6 \times 640+6 x $
Transferring the term $6 x$ to LHS and the term $(8 \times 640)$ to RHS,
$ 2 x-6 x<6 \times 640-8 \times 640$
$ \Rightarrow -4 x<640(6-8)$
$ \Rightarrow -4 x<-2 \times 640 $
Dividing both sides by $-4$,
$\frac{-4 x}{-4} >\frac{-2 \times 640}{-4}$
$x >320$ ...(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 $320 L$ and less than $1280 L$.