SSC CGL Preparation: Concepts and Practice

Outcome

Calculate solute remaining after repeated replacement, determine an unknown replacement volume, and find how many replacements achieve a target.

Concept and assumptions

Concentrations use percentage by volume (% v/v). Assume additive volumes, no reaction or evaporation, and thorough mixing before every withdrawal. From V L, remove x L, then add exactly x L of pure water. Thus total volume remains V.

Uniform mixing makes each withdrawal remove x/V of the solute currently present. The retained fraction is q = 1 − x/V. Every subsequent replacement multiplies what remains by q, so:

Remaining solute = initial solute × qⁿ. Final concentration = initial concentration × qⁿ.

The second formula holds because volume is constant. Here n is a positive integer and 0 ≤ x ≤ V. With 0 < x < V, finite replacements never remove all solute.

For unequal withdrawals, multiply their separate retention factors. These formulas assume pure-water refills; other refill concentrations need fresh solute accounting. Adding water without withdrawal increases volume and is a different operation. Keep fractions exact; round only final answers, marking approximations with ≈.

Worked examples

Example 1 — One replacement. A 40 L vessel contains 30% solution. Remove 10 L and replace it with 10 L water. Find the concentration.

Initial solute = 40 × 0.30 = 12 L. Removed solute = 10 × 0.30 = 3 L. Remaining solute = 12 − 3 = 9 L. Final concentration = 100 × 9/40 = 22.5%.

Example 2 — Three replacements. A 60 L vessel contains 50% solution. Replace 12 L with water three times, mixing between operations.

Retained fraction q = 1 − 12/60 = 4/5. Initial solute = 60 × 0.50 = 30 L. After successive operations, solute is 30 × 4/5 = 24 L, then 24 × 4/5 = 19.2 L, then 19.2 × 4/5 = 15.36 L. Final concentration = 100 × 15.36/60 = 25.6%.

Example 3 — Finding the replacement volume. Two equal-volume water replacements reduce 80 L of 64% solution to 36%. Find each replacement volume.

Let each replacement be x L. 36 = 64(1 − x/80)². Divide by 64: (1 − x/80)² = 36/64 = 9/16. Since the retained fraction is nonnegative, 1 − x/80 = 3/4. Thus x/80 = 1/4, giving x = 20 L. Check: concentrations become 64 × 3/4 = 48%, then 48 × 3/4 = 36%.

Common mistakes

Removing the same solute amount each round; combining rounds into one withdrawal; forgetting to mix; changing the fixed-volume denominator; rounding early.

Practice questions

1. A 50 L vessel contains 24% solution. Replace 10 L with water once. Find the concentration.
2. A 40 L vessel initially contains only a liquid concentrate. Replace 5 L with water twice. How much original concentrate remains?
3. A 54 L vessel contains 30% solution. Replace 18 L with water twice. Find the final concentration.
4. A 20 L vessel contains 80% solution. Replace 5 L with water each round. Find the minimum number of rounds needed to fall below 40%.

Worked answers

1. Initial solute = 50 × 0.24 = 12 L. Retained fraction = 1 − 10/50 = 4/5. Solute remaining = 12 × 4/5 = 9.6 L. Concentration = 100 × 9.6/50 = 19.2%.
2. Initial concentrate = 40 L; retained fraction = 1 − 5/40 = 7/8. Remaining concentrate = 40 × (7/8)² = 40 × 49/64 = 30.625 L, exactly.
3. Retained fraction = 1 − 18/54 = 2/3. Final concentration = 30 × (2/3)² = 30 × 4/9 = 13⅓%, exactly, or ≈ 13.33% rounded to two decimal places.
4. Retained fraction = 1 − 5/20 = 3/4. Successive concentrations are 80 × 3/4 = 60%, 60 × 3/4 = 45%, and 45 × 3/4 = 33.75%. The first two exceed 40%; the third is below it. Therefore the minimum is 3 rounds.