A beaker contains a mixture of acid and water in the ratio 1 : x (acid : water). When 300 ml of this mixture is taken and mixed with an additional 50 ml of water, the new ratio of acid to water becomes 2 : 5. What is the value of x?

Introduction:
This question combines ratio composition with a change caused by adding pure water. The mixture starts with acid and water in the ratio 1:x. From this mixture, 300 ml is taken (so acid and water in that sample remain in the same ratio 1:x). Then 50 ml of pure water is added, changing the acid-to-water ratio to 2:5. The goal is to find x by expressing acid and water amounts using fractions and setting up the final ratio equation.

Given Data / Assumptions:

• Initial ratio (acid:water) = 1 : x• Quantity taken from mixture = 300 ml• Additional water added = 50 ml• Final ratio (acid:water) = 2 : 5

Concept / Approach:
If acid:water = 1:x, then in any amount of this mixture, acid fraction = 1/(1+x) and water fraction = x/(1+x). Multiply these fractions by 300 ml to get the acid and water in the taken sample. Then add 50 ml water and enforce final ratio 2:5 by equating acid/water = 2/5.

Step-by-Step Solution:
Step 1: Compute acid and water in 300 ml of the mixture.Acid in 300 ml = 300 * (1/(1+x)) = 300/(1+x)Water in 300 ml = 300 * (x/(1+x)) = 300x/(1+x)Step 2: Add 50 ml water.New water amount = 300x/(1+x) + 50Acid amount remains = 300/(1+x)Step 3: Apply final ratio 2:5.Acid / Water = 2/5[300/(1+x)] / [300x/(1+x) + 50] = 2/5Step 4: Multiply numerator and denominator to simplify.300 / (300x + 50(1+x)) = 2/5300 / (300x + 50 + 50x) = 2/5300 / (350x + 50) = 2/5Step 5: Cross-multiply and solve.300*5 = 2*(350x + 50)1500 = 700x + 1001400 = 700xx = 2

Verification / Alternative check:
If x=2, initial ratio is 1:2. In 300 ml, acid = 100 ml and water = 200 ml. Add 50 ml water to get water = 250 ml. Then acid:water = 100:250 = 2:5, matching exactly. This confirms x = 2 is correct.

Why Other Options Are Wrong:
x = 1 gives 1:1; then 300 ml contains 150 ml water, adding 50 gives 200, ratio becomes 150:200 = 3:4, not 2:5.x = 3 or 4 makes the mixture too water-heavy initially, and after adding more water the ratio becomes even smaller than 2:5.x = 5 is far too water-heavy, producing acid:water less than 2:5.

Common Pitfalls:
• Treating 300 ml as 300 parts instead of using fractions of the ratio.• Forgetting to add 50 ml only to water.• Setting up the ratio backwards (water/acid instead of acid/water).

Final Answer:
The value of x is 2.