0
0

Replacement or Repeated Dilution

Introduction

कुछ mixture problems में solution का एक हिस्सा निकालकर उसकी जगह बार-बार pure liquid या कोई दूसरा mixture डाला जाता है। हर replacement के बाद concentration धीरे-धीरे बदलती रहती है।

Replacement or Repeated Dilution pattern आपको बिना हर step अलग-अलग solve किए सीधे final concentration या remaining मात्रा निकालने में मदद करता है।

Pattern: Replacement or Repeated Dilution

Pattern

मुख्य विचार: हर replacement के बाद original liquid का बचा हिस्सा = (1 - x / total), जहाँ x हर बार निकाली जाने वाली मात्रा है।

n replacements के बाद बची हुई original liquid:
Remaining = Initial × (1 - x / Total)n

इसी से आप final concentration या जोड़ा गया दूसरा liquid कितना है-यह निकाल सकते हैं।

Step-by-Step Example

Question

एक vessel में 80 L milk है। 20 L milk निकालकर उसकी जगह पानी डाला जाता है। यह प्रक्रिया 3 बार दोहराई जाती है। 3 operations के बाद vessel में कितना milk बचेगा?

Solution

  1. Step 1: Given data पहचानें

    Total = 80 L; हर बार निकाला गया = 20 L; Replacements = 3.

  2. Step 2: Dilution formula लागू करें

    एक बार में बचा fraction = (1 - 20/80) = 3/4
    3 बार के बाद remaining fraction = (3/4)3 = 27/64

  3. Step 3: Final amount निकालें

    Milk left = 80 × (27/64) = 33.75 L

  4. Final Answer:

    33.75 L milk vessel में बचेगा।

  5. Quick Check:

    हर बार ¼ milk निकलता है → बचा fraction = (¾)³ = ≈0.422 → 80 का 42.2% ≈ 33.76 L ✅

Quick Variations

1. पानी से replace करने पर original substance की concentration घटती है।

2. Same concentration वाले mixture से replace करने पर ratio नहीं बदलता।

3. किसी दूसरे solution से replace करने पर weighted ratio या stepwise composition का उपयोग करें।

4. यही logic solid mixtures या alloy replacement problems में भी लागू होता है।

Trick to Always Use

  • Step 1: (1 - replaced / total) निकालें → एक replacement के बाद बचा fraction।
  • Step 2: n operations के लिए इसी fraction को power n करें।
  • Step 3: Initial quantity से multiply करें → remaining amount।
  • Step 4: यदि दूसरे liquid की quantity चाहिए हो, तो total से remaining subtract करें।

Summary

Summary

Replacement or Repeated Dilution pattern में:

  • मुख्य formula: Remaining = Initial × (1 - x / Total)n
  • हर operation remaining liquid का fixed fraction हटाता है।
  • Concentration हर बार exponentially घटती है।
  • Quick check: जितना कम fraction replace होगा, dilution उतना ही धीमा होगा।

Practice

(1/5)
1. A 60-litre vessel is full of milk. 10 litres of milk are removed and replaced with water. The process is repeated once more. Find the amount of milk left after 2 operations.
easy
A. 41.67 L
B. 45 L
C. 42.25 L
D. 48 L

Solution

  1. Step 1: Note the data

    Total = 60 L; replaced each time = 10 L; repetitions = 2.

  2. Step 2: Use the dilution formula

    Fraction remaining after one replacement = 1 - (10/60) = 5/6. After 2 ops: (5/6)^2.

  3. Step 3: Calculate amount left

    Milk left = 60 × (5/6)^2 = 60 × 25/36 = 1500/36 = 41.666... L ≈ 41.67 L.

  4. Final Answer:

    41.67 L → Option A.

  5. Quick Check:

    After 1st op milk = 50 L; after 2nd op remove 10 L of mixture (milk fraction 50/60 = 5/6) → milk removed ≈ 8.333 → left ≈ 41.667 L ✅

Hint: Use Remaining = Total × (1 - replaced/total)^n.
Common Mistakes: Subtracting replaced volume directly each time instead of using fractional removal.
2. A 100-litre tank of acid solution has 10 litres replaced by water each time. The operation is performed 3 times. Find the fraction of the original acid left.
easy
A. 0.70
B. 0.729
C. 0.80
D. 0.81

Solution

  1. Step 1: Fraction retained per operation

    Per op retained fraction = 1 - (10/100) = 0.9.

  2. Step 2: Apply power for 3 operations

    Fraction left = (0.9)^3 = 0.729.

  3. Final Answer:

    0.729 → Option B.

  4. Quick Check:

    0.9×0.9×0.9 = 0.729 ✅

Hint: Raise the retained fraction to the power n: (1 - x/Total)^n.
Common Mistakes: Multiplying linearly instead of using exponential power.
3. A vessel contains 80 L of milk. 20 L of milk is removed and replaced with water. The process is repeated twice. How much milk remains after the two operations?
easy
A. 33.75 L
B. 36 L
C. 45 L
D. 30 L

Solution

  1. Step 1: Find retained fraction per op

    Retained fraction = 1 - (20/80) = 3/4.

  2. Step 2: After 2 operations

    Fraction left = (3/4)^2 = 9/16.

  3. Step 3: Amount of milk left

    Milk left = 80 × 9/16 = 80 × 0.5625 = 45 L.

  4. Final Answer:

    45 L → Option C.

  5. Quick Check:

    After 1st op milk = 60 L; after 2nd op remove 20 L of mixture (milk fraction 60/80 = 3/4) → milk removed = 15 → left = 45 L ✅

Hint: Use (1 - replaced/total)^n then multiply by initial volume.
Common Mistakes: Treating each operation as removing the same absolute amount of the pure substance.
4. A 40-litre vessel contains pure alcohol. 8 litres are taken out and replaced with water. The operation is repeated twice. Find the amount of alcohol left.
medium
A. 25.60 L
B. 23.04 L
C. 24.00 L
D. 20.50 L

Solution

  1. Step 1: Compute retained fraction per op

    Retained fraction = 1 - (8/40) = 0.80.

  2. Step 2: Apply for 2 operations

    Fraction left = (0.80)^2 = 0.64.

  3. Step 3: Amount left

    Alcohol left = 40 × 0.64 = 25.6 L.

  4. Final Answer:

    25.60 L → Option A.

  5. Quick Check:

    After 1st op alcohol = 32 L; after 2nd op remove 8 L of mixture (alcohol fraction 32/40 = 0.8) → removed alcohol = 6.4 → left = 25.6 L ✅

Hint: Multiply the initial amount by (1 - x/Total)^n for n replacements.
Common Mistakes: Rounding too early - keep decimals until final step.
5. A 90-litre mixture contains 80% alcohol. 30 litres of the mixture are removed and replaced with water. Find the new concentration of alcohol (to two decimal places).
medium
A. 50.00%
B. 60.00%
C. 70.00%
D. 53.33%

Solution

  1. Step 1: Compute initial alcohol (litres)

    Initial alcohol = 0.80 × 90 = 72 L.

  2. Step 2: Fraction retained after removal

    Removed fraction = 30/90 = 1/3 → retained fraction = 2/3.

    3. Step 3: Alcohol left in litres

    Alcohol left = 72 × (2/3) = 48 L.

  3. Step 4: New concentration

    New % = (48 ÷ 90) × 100 = 53.333...% → 53.33%.

  4. Final Answer:

    53.33% → Option D.

  5. Quick Check:

    After removal, alcohol = 48 L; total remains 90 L → 48/90 = 0.5333 → 53.33% ✅

Hint: Alcohol left = initial_alcohol × (1 - removed/total); divide by total for %.
Common Mistakes: Forgetting that replacement reduces the pure component proportionally to the removed fraction.

Mock Test

Ready for a challenge?

Take a 10-minute AI-powered test with 10 questions (Easy-Medium-Hard mix) and get instant SWOT analysis of your performance!

10 Questions
5 Minutes