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 $x \,L$.
$\therefore $ 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 $L . H . S$. and the term $(8 \times 640)$ to $R , H , S$
$2 x-4 x > 4 \times 640-8 \times 640$
$\Rightarrow -2 x > 640(4-8)$
$ \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$
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 $L . H . S$. and the term $(8 \times 640)$ to R.H.S.,
$2 x-6 x < 6 \times 640-8 \times 640$
$\Rightarrow -4 x < 640(6-8)$
$ \Rightarrow -4 x < -2 \times 640$
Dividing both sides by $-4$,
$\frac{-4 x}{-4} > \frac{-2 \times 640}{-4} $
$\Rightarrow x > 320$
Hence, from equations (i) and (ii),
$320 < x < 1280$
i.e., $x \in(320,1280)$
The number of litres to be added should be greater than $320\, L$ and less than $1280\, L$.