1. (SSC CGL, Banking) From 100L of pure milk, 10L is removed and replaced with water twice. What is the final quantity of milk left in the mixture?
Correct Answer: a) 81L
Explanation: Using the formula for repeated replacement: Final quantity = Initial × (1 - Replacement/Total)^n = 100 × (1 - 10/100)^2 = 100 × (0.9)^2 = 100 × 0.81 = 81L Short Trick: For 10% replacement twice, multiply by 0.9 twice
2. (IBPS, SSC) An 80L whisky solution contains 40% alcohol. 20L of whisky is replaced with water. What is the new alcohol percentage?
Correct Answer: a) 30%
Explanation: Initial alcohol = 40% of 80L = 32L Alcohol removed = 40% of 20L = 8L Remaining alcohol = 32 - 8 = 24L New percentage = (24/80) × 100 = 30% Short Trick: Replacement ratio = 20/80 = 1/4, so alcohol remains (3/4 × 40%) = 30%
3. (CAT, SSC) A vessel contains 60L of wine. 12L is replaced with water, then 15L of the mixture is replaced again. Find the final ratio of wine to water.
Correct Answer: b) 21:11
Explanation: After first replacement: Wine = 60 × (48/60) = 48L After second replacement: Wine = 48 × (45/60) = 36L Water = 60 - 36 = 24L Ratio = 36:24 = 3:2 Note: There seems to be inconsistency between calculation and options
4. (SSC CHSL, RRB) A container contains 40L of pure milk. 4L is removed and replaced with water three times. What is the final quantity of milk?
Correct Answer: a) 29.16L
Explanation: Using the formula: Final milk = 40 × (1 - 4/40)^3 = 40 × (0.9)^3 = 40 × 0.729 = 29.16L Short Trick: For 10% replacement three times, multiply by 0.9 three times
5. (Bank PO, SSC) A cask contains 125L of wine. 25L is replaced with water twice. How much wine is left?
Correct Answer: a) 80L
Explanation: Using the formula: Final wine = 125 × (1 - 25/125)^2 = 125 × (0.8)^2 = 125 × 0.64 = 80L Short Trick: For 20% replacement twice, multiply by 0.8 twice
6. (SSC CGL, Banking) A whisky jar contains 40% alcohol. A part of it is replaced with 19% alcohol, and the final mixture contains 26% alcohol. What part of the original whisky was replaced?
Correct Answer: b) 2/3
Explanation: Using alligation: (40 - 26):(26 - 19) = 14:7 = 2:1 Replacement ratio = 2/(2+1) = 2/3 Short Trick: The ratio of remaining to replaced is inverse of the differences from mean
7. (RRB NTPC, SSC) A can contains a mixture of two liquids A:B in the ratio 7:5. When 9L is drawn out and replaced with B, the ratio becomes 7:9. Find the initial quantity of A.
Correct Answer: b) 21L
Explanation: Let total mixture = 12x (7x A + 5x B) After replacement: A = 7x - (7/12 × 9) = 7x - 5.25 B = 5x - (5/12 × 9) + 9 = 5x + 5.25 New ratio: (7x - 5.25)/(5x + 5.25) = 7/9 Solving gives x = 3 ⇒ A = 7×3 = 21L Short Trick: Assume total as 12x based on initial ratio
8. (CAT, SSC) A container contains 40L of milk. A process is repeated three times, where 4L is removed and replaced with water. What is the final quantity of milk?
Correct Answer: b) 29.16L
9. (SSC CHSL, Banking) A cask is full of water. 5L is drawn out and replaced with potion. The process is repeated multiple times, resulting in a final water:potion ratio of 144:25. What was the capacity of the cask?
Correct Answer: c) 60L
Explanation: Let capacity = x, replacements = n Final water = x × (1 - 5/x)^n = 144/169 × x (1 - 5/x)^n = 144/169 ⇒ (12/13)^2 (since 144/169 = (12/13)^2) Thus n = 2 and 5/x = 1/13 ⇒ x = 65L Note: There seems to be inconsistency between calculation and options
10. (SSC CGL, CAT) A vessel contains 80L of pure milk. 20L is removed and replaced with water multiple times until only 145.8L of milk is left. How many times (x) was the replacement process repeated?
Correct Answer: b) 3
Explanation: Using the replacement formula: Final milk = Initial × (1 - Replacement/Total)^n 145.8 = 80 × (1 - 20/80)^n ⇒ 145.8/80 = (0.75)^n 1.8225 = (0.75)^n ⇒ Taking log: n ≈ 3 Verification: 80 × (0.75)^3 = 80 × 0.421875 = 33.75L Note: There seems to be inconsistency in the question as 145.8L exceeds initial 80L