Linear Inequations-24

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A manufacturer has 600 liters of a 12% solution of acid. How many liters of a 30% acid solution must be added
to it so that acid content in the resulting mixture will be more than 15% but less than 18%?

Solution: Let x liters of 30% acid solution be added to 600 liters of 12% solution of acid. Then,

Total quantity of mixture = (600 + x) liters

Total acid content in the (600 + x) liters of mixture

= 30x/100 + 12/100 X 600

It is given that acid content in the resulting mixture must be more than 15% and less than 18%.

Therefore, 15% of (600 + x) < (30x/100 + 12/100 X 600) < 18% of (600 + x)

=> 15/100 X (600 + x) < 30x/100 + 12/100 X 600 < 18/100 X (600 + x)

=> 15(600 + x) < 30x + 12 X 600 < 18(600 + x) [Multiplying throughout by 100]

=> 9000 + 15x < 30x + 7200 < 10800 + 18x

=> 9000 + 15x < 30x + 7200 and 30x + 7200 < 10800 + 18x

=> 9000 – 7200 < 30x – 15x and 30x – 18x < 10800 – 7200

=> 1800 < 15x and 12x < 3600

=> 15x > 1800 and 12x < 3600

=> x > 120 and x < 300

Hence, the number of liters of the 30% solution of acid must be more than 120 but less than 300.

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