How to Solve: CUET Quant – Mixture and Alligation

How to Solve: CUET Quant – Mixture and Alligation

Introduction

"Imagine you’re running a juice stall—how do you mix two juices to hit the perfect price? That’s exactly what CUET tests in Mixture & Alligation. Master this, and you’ll solve 3-4 questions in under 2 minutes!

What You Need To Know First

1. Ratios – How to split quantities into parts (e.g., 2:3 means 2 parts + 3 parts = 5 total parts).
2. Percentage to Fraction – 20% = 0.2, 50% = 0.5, etc.
3. Basic Algebra – Solving for one variable (e.g., if 2x + 3x = 10, then x = 2).

Key Vocabulary

Formulas To Know

1. Alligation Rule (MEMORISE THIS)

Formula:

(Quantity of Cheaper) / (Quantity of Expensive) = (Price of Expensive - Mean Price) / (Mean Price - Price of Cheaper)

Variables: - Cheaper Price (C) = Lower price per unit. - Expensive Price (E) = Higher price per unit. - Mean Price (M) = Desired average price of the mixture.

Example: If C = ₹10, E = ₹20, M = ₹15, then ratio = (20-15)/(15-10) = 5/5 = 1:1.

2. Replacement Formula (MEMORISE THIS)

Formula:

Final Quantity = Initial Quantity × (1 - Replacement Fraction)^n

Variables: - Replacement Fraction = Amount replaced / Total quantity. - n = Number of replacements.

Example: If you replace 2L of a 10L solution 3 times, final quantity = 10 × (1 - 0.2)³ = 10 × 0.512 = 5.12L.

Step-by-Step Method

Step 1: Identify the Two Ingredients

• Label the cheaper (C) and expensive (E) components.
• Note their prices/percentages.

Step 2: Find the Mean Price (M)

• This is the desired average of the mixture.
• If not given directly, calculate from the problem.

Step 3: Apply Alligation Rule

• Write the formula: (Cheaper Quantity) / (Expensive Quantity) = (E - M) / (M - C)
• Plug in the numbers and simplify to get the ratio.

Step 4: Use the Ratio to Find Quantities

• If total mixture = T, then:
• Cheaper quantity = (Ratio part of C / Total ratio) × T
• Expensive quantity = (Ratio part of E / Total ratio) × T

Step 5: Solve for the Unknown

• If one quantity is missing, set up an equation and solve.

Worked Example (Using Steps Above)

Problem: How many kg of ₹20/kg rice must be mixed with ₹10/kg rice to get a ₹14/kg mixture?

Solution: 1. Cheaper (C) = ₹10, Expensive (E) = ₹20, Mean (M) = ₹14. 2. Apply alligation:
(Cheaper Qty) / (Expensive Qty) = (20 - 14) / (14 - 10) = 6 / 4 = 3:2 3. Ratio = 3:2 (3 parts cheap, 2 parts expensive). 4. If total mixture = 5 kg, then:
- Cheap rice = (3/5) × 5 = 3 kg
- Expensive rice = (2/5) × 5 = 2 kg

Answer: 2 kg of ₹20/kg rice is needed.

Worked Examples

Example 1 – Basic (Alligation)

Problem: A shopkeeper mixes 30% and 50% sugar solutions to get a 40% solution. What is the ratio of the two solutions?

Solution: 1. Cheaper (C) = 30%, Expensive (E) = 50%, Mean (M) = 40%. 2. Apply alligation:
(30% Qty) / (50% Qty) = (50 - 40) / (40 - 30) = 10 / 10 = 1:1 Answer: 1:1 ratio.

What we did and why: - Used alligation to find the ratio of two solutions to reach a desired concentration.

Example 2 – Medium (Replacement)

Problem: A 40L milk solution is 10% water. How much water must be added to make it 20% water?

Solution: 1. Initial water = 10% of 40L = 4L. 2. Final water = 20% of (40 + x)L = 0.2(40 + x). 3. Set up equation:
4 + x = 0.2(40 + x)
4 + x = 8 + 0.2x
0.8x = 4 → x = 5 Answer: 5L water must be added.

What we did and why: - Used percentage change to find how much extra water is needed to dilute the mixture.

Example 3 – Exam Style (Disguised Problem)

Problem: A vessel has 60L of 25% alcohol. If 12L is removed and replaced with water, what is the new alcohol percentage?

Solution: 1. Initial alcohol = 25% of 60L = 15L. 2. After removal = 15L - (25% of 12L) = 15 - 3 = 12L alcohol left. 3. New mixture = 60L (12L alcohol + 48L water). 4. New percentage = (12/60) × 100 = 20%.

Answer: 20% alcohol.

What we did and why: - Used replacement logic to adjust the mixture’s concentration after removal and addition.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, last-minute Mixture & Alligation recap!

1. Alligation Rule: Cheaper to Expensive ratio = (E - M) / (M - C). Write it down now!
2. Replacement: Final quantity = Initial × (1 - fraction)^n. If you replace 2L of 10L, fraction = 0.2.
3. Watch for traps: Hidden mean price, multiple replacements, disguised problems.
4. Practice 3 problems tonight: One basic alligation, one replacement, one exam-style.

You’ve got this—go ace that CUET!

Final Note for Teachers: - Pacing: Spend 5 mins on alligation, 3 mins on replacement, 2 mins on traps. - Visuals: Draw the alligation cross (C-M-E) on screen. - Engagement: Ask students, "What’s the ratio if C=10, E=30, M=20?" (Answer: 1:1).

Student Action Plan: 1. Memorise the alligation formula. 2. Solve 5 problems (3 alligation, 2 replacement). 3. Review mistakes from this guide.

Good luck—you’ll crush it! ?