**Question:**

**A solution of 9% acid is to be diluted by adding 3% acid solution to it. The resulting mixture is to be more than 5% but less than 7% acid. If there is 460 litres of the 9% solution, how many litres of 3% solution will have to be added?**

**Solution:**

According to the question,

Let x litres of 3% solution is to be added to 460 liters of the 9% of solution

Then, we get,

Total solution = (460 + x) litres

Total acid content in resulting solution

= (460 × 9/100 + x × 3/100)

= (41.4 + 0.03x)%

According to the question, we have,

Resulting mixture should be more than 5% acidic but less than 7% acidic

So we get,

⇒ 5 % of (460 + x) < 41.4 + 0.03x < 7% of (460 + x)

⇒ 5/100 × (460 + x) < 41.4 + 0.03x < 7/100 × (460 + x)

⇒ 23 + 0.05 x < 41.4 + 0.03x < 32.2 + 0.07x

Now, we have

⇒ 23 + 0.05x < 41.4 + 0.03x and 41.4 + 0.03x < 32.2 + 0.07x

i.e., 0.02x < 18.4 and 0.04x > 9.2

⇒ 2x < 1840 and 4x > 920

⇒ x < 920 and x > 230

⇒ 230 < x < 920

Hence, solution between 230 *l* and 920 *l* should be added**.**

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