Tea A (100) mixed with Tea B (200) to sell at 150. Is mixing ratio 1:1?

Answer: Yes. 150 is the exact arithmetic mean of 100 and 200, which means equal weights (1:1 ratio) were used.

Can you mix two liquids of density d1 and d2 to get a mixture density greater than d2 (where d2 > d1)?

Answer: No. Mixture property must lie strictly between the components. d_1 .

If water cost is Rs 0, and Milk is Rs 50/L. A mixture worth Rs 40/L implies?

Answer: 4 parts Milk, 1 part Water. Alligation: Milk(50) and Water(0) yielding Mean(40). Cross diff: (40-0) and (50-40) -> 40:10 -> 4:1.

Adding water to a milk mixture changes the ratio from 5:1 to 5:2. Did the absolute quantity of milk change?

Answer: No. Only water was added. The milk component remains absolutely constant (though its percentage drops).

In replacement problems, when you remove a cup of mixture, does the ratio of Milk:Water in the cup differ from the container?

Answer: No (Ratio is identical). A homogeneous mixture has the same ratio throughout. The cup has the exact same proportion as the tank.

Can you apply the Alligation 'Cross' directly to the selling prices of two items if their profit percentages are different?

Answer: No, convert to cost price. A fundamental rule of Alligation: All 3 components must be in the same reference frame, usually Cost Price (CP).

A dairy sells milk mixed with water exactly at Cost Price, yet makes a 20% profit. What represents the 20% profit?

Answer: The free water added. Since there is no price markup, the entire 20% profit comes purely from selling the freely added water at the price of milk.

A vessel has 10L Milk. 1L is replaced with water. This is done once more. Will there be exactly 8L Milk left?

Answer: No, more than 8L. First time, 1L milk goes -> 9L left. Second time, you remove 1L mixture (not pure milk), so you remove 8L (actually 8.1L).

Solution X has 50% alcohol. Solution Y has 30%. In what range must the final mixture's strength lie?

Answer: Between 30% and 50%. The concentration of a mixture always falls strictly between the lowest and highest concentrations of its components.

If 10kg of Rice A and 10kg of Rice B are mixed, the mean price is exactly the average of their prices. Why?

Answer: Weights are equal. When the ratio of mixing weights is 1:1, the weighted average reduces to a simple arithmetic mean.

In what ratio mix rice at Rs 60 and Rs 65 to sell at Rs 68.20 earning 10% profit?

Answer: 3:2. Target CP of mix = 68.20 / 1.1 = 62. Alligation: 60 ... 65 ..62.. Diff: 3 : 2.

A mixture of 40L contains Milk:Water in 4:1. How much water to add to make it 2:3?

Answer: 40 L. Initial: Milk=32L, Water=8L. Water is added, Milk stays 32L. Target ratio M:W = 2:3. 2 units = 32L $\to$ 1 unit = 16L. Target Water = 3 units = 3 \times 16 = 48L. Added = 48 - 8 = 40L.

Two vessels contain Milk:Water in 2:3 and 4:1. If mixed in equal quantities, what is the new Ratio?

Answer: 3:2. V1 sum = 5 (2+3). V2 sum = 5 (4+1). Sums are already equal. Mix directly: Milk = 2+4=6, Water = 3+1=4. Ratio = 6:4 = 3:2.

Gold is 19 times heavier than water, Copper 9 times. In what ratio to mix them to get an alloy 15 times heavier?

Answer: 3:2. Gold(19) ... Copper(9) ... Mean(15) ... Diff: (15-9):(19-15) = 6:4 = 3:2

How many kg of sugar at 9/kg must be mixed with 27kg at 7/kg to gain 10% by selling at 9.24/kg?

Answer: 63 kg. 10% Gain on SP 9.24 $\to$ CP = 8.4. Alligation: 9 ... 7 .. 8.4 .. Ratio = 1.4 : 0.6 = 7:3. Since 3 units (cheap) = 27kg, 1 unit = 9. 7 units (dear) = 7 \times 9 = 63kg.

Solution A is 10% Acid, B is 30% Acid. In what ratio to mix to get 25% Acid?

Answer: 1:3. 10 ... 30 ..25.. Diff: 5 : 15 = 1:3.

Pure Milk is Rs 100/L. Water Rs 0. Mixture is sold at Rs 80/L with a 25% Profit. Ratio of Water:Milk?

Answer: 9:16. SP = 80, P = 25%. CP_mix = 80/1.25 = 64. Water(0) ... Milk(100) .. 64 .. Diff: 36 (Milk side) : 64 (Water side). Ratio W:M = 36:64 = 9:16.

729 mL mixture has 7:2 Milk:Water. Adding how much water makes it 7:3?

Answer: 81 mL. Initial Milk:Water = 7:2 (9 units total). Final = 7:3. Milk (7 units) is constant. Water increased by 1 unit. Since 9 units = 729 mL, 1 unit = 729/9 = 81 mL.

Container has 40L pure Milk. 4L removed and replaced by water. Process done 3 times total. Milk left?

Answer: 29.16 L. Formula: 40 \times (1 - 4/40)^3 = 40 \times (0.9)^3 = 40 \times 0.729 = 29.16 L.

A jar has 10 red marbles and 30 green. How many red to add to make red 60% of total?

Answer: 35. Green stays constant at 30. Final Mix: Red 60%, Green 40%. 40% = 30 marbles $\to$ Total 100% = 75 marbles. Initial Total = 40. Added = 75 - 40 = 35.

A milkman buys milk at Rs 24/L, adds water, and sells the mix at Rs 30/L making 50% profit. Ratio of Water:Milk?

Answer: 1:5. SP=30, Profit=50% $\to$ CP = 30/1.5 = 20. Alligation: Water(0) ... Milk(24) .. 20 .. Ratio 4:20 = 1:5.

Zinc:Copper in A is 5:3, in B is 5:11. Mix to get 1:1. Ratio of A:B taken?

Answer: 3:2. Fraction of Zinc: A=58, B=516. Target=12. Base 16: A=10/16, B=5/16, T=8/16. Cross rule on numerators: (8-5):(10-8) = 3:2.

Alcohol 50% and 30% are mixed to produce 40%. Mixing Ratio?

Answer: 1:1. 40 is exactly halfway. 50-40=10, 40-30=10. Ratio 1:1.

Sum of Rs 41 divided among 50 students. Boy gets 90p, Girl 65p. Number of Boys?

Answer: 34. Avg = 4100p / 50 = 82p. Alligation: Boy(90) ... Girl(65) .. 82 .. Diff: 17:8. Total units 25 = 50 students. 1u = 2. Boys(17u) = 34.

In 300g Alloy, Silver:Copper is 2:3. How much Silver to add to make it 3:2?

Answer: 150g. Initial Silver=120, Copper=180 (constant). Target ratio S:C = 3:2. Since 2 units = 180, 1 unit = 90. Target Silver(3 units) = 270. Added = 270 - 120 = 150g.

Vessel A (Milk:Water 4:1), B (11:4). Mixed in what ratio to get 2:1?

Answer: 1:2. Milk fractions: A(4/5), B(11/15), Mean(2/3). Base 15: A(12/15), B(11/15), M(10/15). Cross: (11-10):(12-10) = 1:2.

A jar full of whiskey. 40% removed replaced w/ water. Repeated once. Whiskey dropped by %?

Answer: 64%. Remaining = (0.6)^2 = 0.36 = 36\%. It dropped BY 100\% - 36\% = 64\%.

Three identical glasses: 1/2, 2/3, 3/4 full of milk. Filled to brim with water. Mixed. Ratio M:W?

Answer: 23:13. Total Volume = 3 limits. Milk = 1/2 + 2/3 + 3/4 = 23/12. Water = 3 - 23/12 = 13/12. Ratio 23:13.

Cost of Diamond varies as square of weight. 10g diamond breaks into 2:3:5. Loss %?

Answer: 62%. Orig Price \propto 10^2 = 100. New Price \propto 2^2 + 3^2 + 5^2 = 4 + 9 + 25 = 38. Loss = 62 on 100 = 62%.

Two blends of tea 5:3 and 5:7. Mixed 1:1. New Ratio?

Answer: 25:23. Sums are 8 and 12. LCM = 24. Scale A to 24 \to 15:9. Scale B to 24 \to 10:14. Add them: (15+10) : (9+14) = 25:23.

Home/Rajasthan 4th Grade Exam/Quantitative Aptitude/

Mixture & Alligation

The Art of Weighted Averages. Master the 'See-Saw' method for balancing and the 'Cross Method' for instant ratios.

Expert Answer & Key Takeaways

The Art of Weighted Averages. Master the 'See-Saw' method for balancing and the 'Cross Method' for instant ratios.

1. The 'See-Saw' (Weighted Average)

Visualizing Balance.

Concept: If 10kg is at 20Rs and 20kg is at 50Rs.
Balance: Weight Ratio 1:2. The balance point shifts towards 50.

Example:

Q: Avg of Class A (20 students) is 60. Class B (30 students) is 70.

Solution: Ratio 2:3. Gap 10. Break Gap in 3:2 inverse. 35 of 10 is 6.

60+6=66

2. The 'Cross' (Alligation Rule)

To find Ratio when Mean is given.

Important: All units must be same (e.g., all CP or all SP).

Example:

Q: Rice @ 30 mixed with Rice @ 40 to get 33.

Solution: 30 ... 40
..33..
Diff: 7 : 3. Ratio 7:3.

3. Profit on Mixture (Water is Free)

If mixture sold at Cost Price of Milk yields Profit.

Hack: Profit % is exactly the quantity of Water added relative to Milk.

Example:

Q: Sold at CP, gained 25%. Find Water:Milk.

Solution: Gain 25%. Values: Water=25, Milk=100. Ratio 1:4.

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Mixture & Alligation

Expert Answer & Key Takeaways

1. The 'See-Saw' (Weighted Average)

2. The 'Cross' (Alligation Rule)

3. Profit on Mixture (Water is Free)

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