Home
Class 11
MATHS
A manufacturer has 600 liters of 12% sol...

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

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we need to find how many liters of a 30% acid solution must be added to 600 liters of a 12% acid solution so that the resulting mixture has an acid concentration greater than 15% but less than 18%. ### Step 1: Define the Variables Let \( x \) be the number of liters of the 30% acid solution to be added. ### Step 2: Calculate the Acid Content The total volume of the resulting mixture will be: \[ 600 + x \text{ liters} \] The amount of acid in the 600 liters of 12% solution is: \[ \text{Acid from 12% solution} = 0.12 \times 600 = 72 \text{ liters} \] The amount of acid in \( x \) liters of 30% solution is: \[ \text{Acid from 30% solution} = 0.30 \times x = 0.3x \text{ liters} \] ### Step 3: Set Up the Inequalities The total acid in the mixture will be: \[ \text{Total acid} = 72 + 0.3x \text{ liters} \] We want the concentration of acid in the mixture to be more than 15% and less than 18%. Therefore, we can set up the following inequalities: 1. For more than 15%: \[ \frac{72 + 0.3x}{600 + x} > 0.15 \] 2. For less than 18%: \[ \frac{72 + 0.3x}{600 + x} < 0.18 \] ### Step 4: Solve the First Inequality Multiply both sides by \( 600 + x \) (assuming \( 600 + x > 0 \)): \[ 72 + 0.3x > 0.15(600 + x) \] Expanding the right side: \[ 72 + 0.3x > 90 + 0.15x \] Rearranging gives: \[ 72 - 90 > 0.15x - 0.3x \] \[ -18 > -0.15x \] Dividing by -0.15 (remember to reverse the inequality): \[ x > 120 \] ### Step 5: Solve the Second Inequality Now, we solve the second inequality: \[ 72 + 0.3x < 0.18(600 + x) \] Expanding the right side: \[ 72 + 0.3x < 108 + 0.18x \] Rearranging gives: \[ 72 - 108 < 0.18x - 0.3x \] \[ -36 < -0.12x \] Dividing by -0.12 (again, reverse the inequality): \[ x < 300 \] ### Step 6: Combine the Results From the two inequalities, we have: \[ 120 < x < 300 \] ### Conclusion The manufacturer must add more than 120 liters but less than 300 liters of the 30% acid solution.

To solve the problem step by step, we need to find how many liters of a 30% acid solution must be added to 600 liters of a 12% acid solution so that the resulting mixture has an acid concentration greater than 15% but less than 18%. ### Step 1: Define the Variables Let \( x \) be the number of liters of the 30% acid solution to be added. ### Step 2: Calculate the Acid Content The total volume of the resulting mixture will be: \[ ...
Promotional Banner

Topper's Solved these Questions

  • LINEAR INEQUALITIES

    NAGEEN PRAKASHAN ENGLISH|Exercise EXERCISE|76 Videos
  • LINEAR INEQUALITIES

    NAGEEN PRAKASHAN ENGLISH|Exercise NCERT QUESTION|51 Videos
  • LIMITS AND DERIVATIVES

    NAGEEN PRAKASHAN ENGLISH|Exercise Miscellaneous Exercise|30 Videos
  • MATHEMATICAL REASONING

    NAGEEN PRAKASHAN ENGLISH|Exercise Misellaneous exercise|7 Videos

Similar Questions

Explore conceptually related problems

A manufacturer has 600 litres of a 12% solution of acid. How many litres 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%?

How may litres of water will have to be added to 1125 litres of the 45 % solution of acid so that the resulting mixture will contain more than 25 % but less than 30% acid content?

How many liters of a 25% saline solution must be added to 3 liters of a 10% saline solution to obtain a 15% saline solution?

How many times a fair coin must be tossed, so that the probabillity of getting at least one tail is more than 90 %

Solution of a monobasic acid has a pH=5. If one mL of it is diluted to 1 litre, what will be the pH of the resulting solution ?

How many times must a man toss a fair com so that the probability of having at least one head is more than 90%?

How many times must a man toss a fair coin, so that the probability of having at least one head is more than 80%?

90% acid solution (90% pure acid and 10% water) and 97% acid solution are mixed to obtain 21 litres of 95% acid solution. How many litres of each solution are mixed.

A solution of 8% boric acid 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 there are 640 litres of the 8% solution, how many litres of 2% solution will have to be added?

How much pure alcohol be added to 400 ml of a 15% solution to make its strength 32%?