# A manufacturer has 640 litres of an 8% solution of boric acid.

Question:

A manufacturer has 640 litres of an 8% solution of boric acid. How many litres of 2% boric and acid solution be added to it so that the boric acid content in the resulting mixture will be more than 4% but less than 6%.

Solution:

Let x litres of 2% boric and acid solution be added to 640 litres of 8% solution of boric acid.

$\%$ Strength $=\frac{\frac{8}{100} \times 640+\frac{2}{100} \times x}{640+x}$

$=\frac{5120+2 x}{100(640+x)}$

It is given that boric acid content in the resulting mixture ranges from 4% to 6% \

Therefore,

$\frac{4}{100}<\frac{5120+2 x}{100(640+x)}<\frac{6}{100}$

Multiplying throughout by 100 in the above equation

$\frac{4}{100}(100)<\frac{5120+2 x}{100(640+x)}(100)<\frac{6}{100}(100)$

$4<\frac{5120+2 x}{640+x}<6$

$\frac{5120+2 x}{640+x}>4$ and $\frac{5120+2 x}{640+x}<6$

When,

$\frac{5120+2 x}{640+x}>4$

Multiplying both the sides by (640 + x) in the above equation

$\frac{5120+2 x}{640+x}(640+x)>4(640+x)$

$5120+2 x>2560+4 x$

Subtracting 2x from both the sides in above equation

$5120+2 x-2 x>2560+4 x-2 x$

$5120>2560+2 x$

Subtracting 2560 from both the sides in above equation

$5120-2560>2560+2 x-2560$

$2560>2 x$

Dividing both the sides by 2 in above equation

$\frac{2560}{2}>\frac{2 x}{2}$

$1280>x$

Now when,

$\frac{5120+2 x}{640+x}<6$

Multiplying both the sides by (640 + x) in the above equation

$\frac{5120+2 x}{640+x}(640+x)<6(640+x)$

$5120+2 x<3840+6 x$

Subtracting 2x from both the sides in above equation

$5120+2 x-2 x<3840+6 x-2 x$

$5120<3840+4 x$

Subtracting 3840 from both the sides in above equation

$5120-3840<3840+4 x-3840$

$1280<4 x$

Dividing both the sides by 4 in above equation

$\frac{1280}{4}<\frac{4 x}{4}$

\$320

Thus, the value of 2% boric acid solution to be added ranges from:

320 to 1280 litres

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