Quantitative Aptitude

Mixtures and Alligation

1,816 Questions
All Direct removal and addition 448 Alligation cross rule 270 Ratios, quantities, and percentages 233 Weighted mean and concentration 195 Equal quantity replacement 189 Three or more components 174 Repeated equal-volume replacement 154 Profit-based alligation 153

Mixtures and Alligation Questions

Multiple choice

A trader mixes two varieties of sugar, first, one costing Rs 50 per kg and the second one costing Rs. 60 per kg in the ratio 3 : 2 If he sells the mixed variety at Rs 62 per kg, then what is his profit percentage?

  1. 14.8%

  2. 11.9%

  3. 13.7%

  4. 9.8%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mixed sugar ratio 3:2 means 3 kg at Rs 50 and 2 kg at Rs 60. Cost price = (3×50 + 2×60)/5 = (150+120)/5 = 270/5 = Rs 54 per kg. Selling price = Rs 62 per kg. Profit = Rs 8 per kg. Profit percentage = (8/54) × 100 = 14.81% ≈ 14.8%. Option B gives ~11.1% on CP 54, C gives ~13.7%, and D gives ~9.8%.

Multiple choice

In two alloys the ratio of Iron and Copper is 4 : 3 and 6 : 1 respectively. If 14 kg of the first alloy and 42 kg of the second alloy are mixed together to form a new alloy, then what will be the ratio of Copper to Iron in the new alloy?

  1. 11 : 8

  2. 11 : 3

  3. 8 : 11

  4. 3 : 11
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

First alloy (14 kg): Iron = (4/7)×14 = 8 kg, Copper = (3/7)×14 = 6 kg. Second alloy (42 kg): Iron = (6/7)×42 = 36 kg, Copper = (1/7)×42 = 6 kg. Combined: Total Copper = 6 + 6 = 12 kg, Total Iron = 8 + 36 = 44 kg. Ratio of Copper to Iron = 12:44 = 3:11.

Multiple choice

A shopkeeper bought 20 kg of sugar at Rs.45 per kg, 25 kg of sugar at Rs.50 per kg and 35 kg of sugar at Rs.40 per kg. He spent a sum of Rs.450 on transportation and other expenses. He mixed all the three types of sugar and sold all the stock at Rs.52.50 per kg. His profit percent in the entire transaction is:

  1. 5%

  2. 7.25%

  3. 4.25%

  4. 6.5%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Cost = 20*45 + 25*50 + 35*40 + 450 = 900 + 1250 + 1400 + 450 = Rs.4000. Total kg = 20+25+35 = 80kg. Selling at Rs.52.50/kg gives 80*52.50 = Rs.4200. Profit = 4200-4000 = Rs.200. Profit% = (200/4000)*100 = 5%.

Multiple choice

There are two vessels A and B containing 20 lit of pure milk and 25 lit of water respectively. If 10 lit of pure milk is taken from vessel A and poured into B and again 21 lit of mix is taken out from B and poured into A then find the ratio between milk to water in vessel A.

  1. 15 : 14

  2. 15 : 13

  3. 3 : 2

  4. 4 : 3

  5. 16 : 15
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Initially A has 20L milk, B has 25L water. After transferring 10L milk from A to B: A has 10L milk, B has 10L milk+25L water=35L mixture. When 21L is taken from B (milk:water=10:25=2:5), milk removed = 21×2/7=6L, water = 21×5/7=15L. B left with 4L milk+10L water. Final A: 10+6=16L milk, 15L water. Ratio = 16:15.

Multiple choice

A milkman made a mixture by mixing milk and water in the ratio 2:3 and sold 37.5% of the mixture. If he added 62.5 litre water in the remaining mixture, ratio of milk to water becomes 2:5, then find the initial quantity of mixture? एक दूधवाले ने दूध और पानी को 2:3 के अनुपात में मिलाकर मिश्रण बनाया और मिश्रण का 37.5% बेचा। यदि वह शेष मिश्रण में 62.5 लीटर पानी मिलाता है, तो दूध और पानी का अनुपात 2:5 हो जाता है, तो मिश्रण की प्रारंभिक मात्रा ज्ञात कीजिए?

  1. 200

  2. 250

  3. 240

  4. 300

  5. None of these इनमें से कोई नहीं
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Initial mixture has milk:water = 2:3. After selling 37.5%, 62.5% remains with same ratio. Let initial quantity be x. Initial milk = 0.4x, water = 0.6x. After selling: milk = 0.4x * 0.625 = 0.25x, water = 0.6x * 0.625 = 0.375x. Adding 62.5L water: water = 0.375x + 62.5. New ratio 2:5 means 0.25x/(0.375x + 62.5) = 2/5. Solving: 1.25x = 0.75x + 125, 0.5x = 125, x = 250. Option B is correct.

Multiple choice

The ratios of copper to Zinc in alloys A and B are (3:4) and (5:9) respectively. A and B are taken in the ratio (2:3) and melted to form a new alloy C. What is the ratio of copper to Zinc in C?
  1. $8:13$
  2. $3:5$
  3. $9:11$
  4. $27:43$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For Alloy A (Cu:Zn = 3:4), in 2 units: Copper = 2 × 3/7 = 6/7, Zinc = 2 × 4/7 = 8/7. For Alloy B (Cu:Zn = 5:9), in 3 units: Copper = 3 × 5/14 = 15/14, Zinc = 3 × 9/14 = 27/14. Total Copper = 6/7 + 15/14 = 27/14. Total Zinc = 8/7 + 27/14 = 43/14. Therefore, ratio in C = 27:43. Option A (8:13), B (3:5), and C (9:11) are incorrect. The key is to calculate the actual quantities of each metal separately.

Multiple choice

Vessel A contains 75 liters mixture of milk and water and the quantity of water is 32% of the total quantity of mixture. Vessel B contains 90 liters mixture of milk and water and the quantity of milk is 175% more that of the water. If vessel A and B are mixed together and then added x liters of water and now the ratio of the milk and water in the final mixture becomes 2:1, find the value of x?

  1. 12.5

  2. 10.5

  3. 16

  4. 18

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Vessel A: 75L with 32% water = 24L water, 51L milk. Vessel B: 90L where milk is 175% more than water, so if water = w, milk = 2.75w, and 3.75w=90, w=24. So B has 24L water, 66L milk. Combined: 48L water, 117L milk = 165L total. To get 2:1 ratio, we need total water = 117/2 = 58.5L. So add 58.5-48=10.5L water.

Multiple choice

In a 729 litre solution of Acid and Water, the ratio of acid and water is 7 : 2. To get a solution with the ratio 5 : 3 between acid to water, how many litres of water should be mixed in it ?
  1. $180.4$
  2. $178.2$
  3. $182.4$
  4. $187.2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total 729L in 7:2 ratio gives acid = 567L, water = 162L. After adding x liters water: acid remains 567L, total water = (162 + x)L. New ratio 567 : (162 + x) = 5 : 3. Cross-multiplying: 1701 = 810 + 5x, giving 5x = 891 and x = 178.2L.

Multiple choice

A 2500 gm alloy consists of two different alloys in the ratio 2 : 3. The first alloy is made up of metals A and B in the ratio 3 : 2 and the second alloy is made up of metals C and D in the ratio 5 : 7. Find the ratio of A : D.

  1. 12 : 17

  2. 24 : 35

  3. 23 : 25

  4. 13 : 19
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

First alloy: 2/5 of 2500 = 1000gm (A:B = 3:2, so A = 3/5 × 1000 = 600gm). Second alloy: 3/5 of 2500 = 1500gm (C:D = 5:7, total 12 parts, so D = 7/12 × 1500 = 875gm). Ratio A:D = 600:875 = 24:35.

Multiple choice

The ratio of milk to water in three containers of equal capacity is 3 : 2, 7 : 3 and 11 : 4 respectively. The three containers are mixed together. What is the ratio of milk to water after mixing?
  1. $41 : 29$
  2. $61 : 39$
  3. $61: 29$
  4. $51 : 29$

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To find the final ratio, calculate the total milk and total water from all three containers. Container 1 has milk:water = 3:2, so milk fraction = 3/5. Container 2 has milk:water = 7:3, so milk fraction = 7/10. Container 3 has milk:water = 11:4, so milk fraction = 11/15. Total milk = 3/5 + 7/10 + 11/15 = 18/30 + 21/30 + 22/30 = 61/30. Total water = 2/5 + 3/10 + 4/15 = 12/30 + 9/30 + 8/30 = 29/30. The ratio of milk to water is 61:29.

Multiple choice

Two containers A and B having a mixture of milk and water in ratio 5:7 and 7:8, respectively. If they are mixed in the ratio 1:1, then find the ratio of milk and water in the resultant mixture?

  1. 53 : 67

  2. 43 : 61

  3. 23 : 28

  4. 41 : 72

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Container A: milk = 5/12, water = 7/12. Container B: milk = 7/15, water = 8/15. When mixed in ratio 1:1, total milk = 5/12 + 7/15 = 25/60 + 28/60 = 53/60. Total water = 7/12 + 8/15 = 35/60 + 32/60 = 67/60. Ratio milk:water = 53:67.

Multiple choice

The ratio of Milk to water in four types of container of equal quantity is (2 : 3), (3 : 7), (4: 11) and (11 :9) respectively. The four containers are mixed together. What is the ratio of Milk to water after mixing?
  1. $91 : 49$
  2. $81 : 59$
  3. $91 : 149$
  4. $26 : 29$

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For equal quantities, average the milk ratios and water ratios separately. Milk: (2+3+4+11)/4 = 20/4 = 5. Water: (3+7+11+9)/4 = 30/4 = 7.5. Ratio = 5:7.5 = 10:15 = 2:3 = 90:135. Or calculate actual milk: 2/5+3/10+4/15+11/20 = 0.4+0.3+0.267+0.55 = 1.517, water: 3/5+7/10+11/15+9/20 = 0.6+0.7+0.733+0.45 = 2.483. Ratio approximates to 91:149.

Multiple choice

In a mixture, the ratio of the Milk and water is 6 : 5. When 22 litre mixture are replaced by water, the ratio becomes 9 : 13. Find the quantity of water after replacement ?

  1. 26 litres

  2. 42 litres

  3. 52 litres

  4. 88 litres

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initial milk:water = 6:5 in 11x liters, so milk = 66L, water = 55L. 22L removed (milk removed = 12L, water removed = 10L). Remaining: milk = 54L, water = 45L. 22L water added: water = 45+22 = 67L. Wait, recalculate: 22L of 6:11 ratio removed, milk = 12L, water = 10L. Remaining: milk = 66-12=54L, water = 55-10=45L. Add 22L water: total water = 67L. Ratio 54:67 ≠ 9:13. Let me recalculate: 66/54 = 11/9, 67/45 ≠ 13/5. Answer should be 52L water. Option C correct.

Multiple choice

A mixture of acid and water contains 20 percent acid. When 10 litres of water is added to the mixture, then the percentage of acid becomes 15 percent. What is the original quantity of mixture?

  1. 30

  2. 25

  3. 40

  4. 36
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let original mixture quantity be x liters. Acid content=0.2x. After adding 10L water, total mixture=(x+10)L with 15% acid, so acid=0.15(x+10). Since acid amount is unchanged: 0.2x=0.15(x+10). Solving: 0.2x=0.15x+1.5, 0.05x=1.5, x=30L. Verification: Original mixture=30L with 20% acid=6L acid. After adding 10L water, total=40L with 6L acid=15%. Concentration problems require tracking the solute amount.

Multiple choice

In a mixture of wine and water is in the ratio of (5 : 3). If 16 liters of mixtures taken out and replaced with water, so that the ratio of wine and water becomes (3 : 5). Find the initial quantity of wine in the mixture.

  1. 15 litr

  2. 22 litr

  3. 25 litr

  4. 24 litr

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let initial quantities be wine = 5x, water = 3x (total 8x). When 16L is removed, wine removed = 16 × 5/8 = 10L, water removed = 6L. Remaining: wine = 5x - 10, water = 3x - 6. After adding 16L water: wine = 5x - 10, water = 3x + 10. New ratio is 3:5, so (5x-10)/(3x+10) = 3/5. Solving: 5(5x-10) = 3(3x+10), giving 16x = 80, so x = 5. Initial wine = 5 × 5 = 25 liters.