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in Linear Equations by (15.2k points)

How many litres of water will have to be added to 600 litres of the 45% solution of acid so that the resulting mixture will contain more than 25%, but less than 30% acid content?

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1 Answer

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by (15.7k points)

Let x litres of water be added. 

Then total mixture = x + 600 

Amount of acid contained in the resulting mixture is 45% of 600 litres. 

It is given that the resulting mixture contains more than 25% and less than 30% acid content. 

Therefore, 

45% of 600 > 25% of (x + 600) 

And 

30% of (x+600) > 45% of 600 

When, 

45% of 600 > 25% of (x+600) 

Multiplying both the sides by 100 in above equation

\(\frac{45}{100} \times 600 \) > \(\frac{25}{100} \times\) (x + 600)

45 × 600 > 25(x + 600) 

27000 > 25x + 15500 

Subtracting 15500 from both the sides in above equation 

27000 – 15500 > 25x + 15500 – 15500 

11500 > 25x

Dividing both the sides by 25 in above equation

\(\frac{11500}{25} > \frac{25{\text{x}}}{25}\)

460 > x Now when, 

45% of 600 < 30% of (x+600) 

Multiplying both the sides by 100 in the above equation

\(\frac{45}{100} \times 600 < \frac{30}{100} \times ({\text{x}} + 600)\)

45 × 600 < 30(x + 600) 

27000 < 30x + 18000 

Subtracting 18000 from both the sides in above equation 

27000 – 18000 < 30x + 18000 – 18000 

9000 < 30x 

Dividing both the sides by 30 in above equation

\(\frac{900}{30} > \frac{30{\text{x}}}{30}\)

Thus, the amount of water required to be added ranges from 300 litres to 460 litres.

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