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- Question
A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is:
Options- A.
1 3 - B.
2 3 - C.
2 5 - D.
3 5 - Correct Answer
2 3
ExplanationBy the rule of alligation, we have:Strength of first jar Strength of 2nd jar 40% Mean
Strength
26%19% 7 14 So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2
∴ Required quantity replaced = 2 3
Alligation or Mixture problems
Search Results
- 1. In what ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg so that by selling the mixture at Rs. 68.20 a kg he may gain 10%?
Options- A. 3 : 2
- B. 3 : 4
- C. 3 : 5
- D. 4 : 5 Discuss
Correct Answer: 3 : 2
Explanation:
S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10%.C.P. of 1 kg of the mixture = Rs. ❨ 100 x 68.20 ❩ = Rs. 62. 110 By the rule of alligation, we have:
Cost of 1 kg tea of 1st kind. Cost of 1 kg tea of 2nd kind. Rs. 60 Mean Price
Rs. 62Rs. 65 3 2 ∴ Required ratio = 3 : 2.
- 2. How many kilogram of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per kg so that there may be a gain of 10% by selling the mixture at Rs. 9.24 per kg?
Options- A. 36 kg
- B. 42 kg
- C. 54 kg
- D. 63 kg Discuss
Correct Answer: 63 kg
Explanation:
S.P. of 1 kg of mixture = Rs. 9.24, Gain 10%.∴ C.P. of 1 kg of mixture = Rs. ❨ 100 x 9.24 ❩ = Rs. 8.40 110 By the rule of allilation, we have:
C.P. of 1 kg sugar of 1st kind Cost of 1 kg sugar of 2nd kind Rs. 9 Mean Price
Rs. 8.40Rs. 7 1.40 0.60 ∴ Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3.
Let x kg of sugar of 1st be mixed with 27 kg of 2nd kind.
Then, 7 : 3 = x : 27
⟹ x = ❨ 7 x 27 ❩ = 63 kg. 3 - 3. Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg.
Options- A. 1 : 3
- B. 2 : 3
- C. 3 : 4
- D. 4 : 5 Discuss
Correct Answer: 2 : 3
Explanation:
By the rule of alligation:Cost of 1 kg of 1st kind Cost of 1 kg of 2nd kind 720 p Mean Price
630 p570 p 60 90 ∴ Required ratio = 60 : 90 = 2 : 3.
- 4. A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is:
Options- A. 4%
- B. 6¼%
- C. 20%
- D. 25% Discuss
Correct Answer: 20%
Explanation:
Let C.P. of 1 litre milk be Re. 1Then, S.P. of 1 litre of mixture = Re. 1, Gain = 25%.
C.P. of 1 litre mixture = Re. ❨ 100 x 1 ❩ = 4 125 5 By the rule of alligation, we have:
C.P. of 1 litre of milk C.P. of 1 litre of water Re. 1 Mean Price
Re. 4 5 0 4 5 1 5 ∴ Ratio of milk to water = 4 : 1 = 4 : 1. 5 5 Hence, percentage of water in the mixture = ❨ 1 x 100 ❩% = 20%. 5 - 5. A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
Options- A. 26.34 litres
- B. 27.36 litres
- C. 28 litres
- D. 29.16 litres Discuss
Correct Answer: 29.16 litres
Explanation:
Amount of milk left after 3 operations = [ 40 ❨ 1 - 4 ❩ 3 ] litres 40 = ❨ 40 x 9 x 9 x 9 ❩ = 29.16 litres. 10 10 10 - 6. A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
Options- A.
1 3 - B.
1 4 - C.
1 5 - D.
1 7
Discuss
Correct Answer:1 5
Explanation:
Suppose the vessel initially contains 8 litres of liquid.Let x litres of this liquid be replaced with water.
Quantity of water in new mixture = ❨ 3 - 3x + x ❩ litres 8 Quantity of syrup in new mixture = ❨ 5 - 5x ❩ litres 8 ∴ ❨ 3 - 3x + x ❩ = ❨ 5 - 5x ❩ 8 8 ⟹ 5x + 24 = 40 - 5x
⟹ 10x = 16
⟹ x = 8 . 5 So, part of the mixture replaced = ❨ 8 x 1 ❩ = 1 . 5 8 5 - 7. In what ratio must water be mixed with milk to gain 16⅔% on selling the mixture at cost price?
Options- A. 1 : 6
- B. 6 : 1
- C. 2 : 3
- D. 4 : 3 Discuss
Correct Answer: 1 : 6
Explanation:
Let C.P. of 1 litre milk be Re. 1.S.P. of 1 litre of mixture = Re.1, Gain = 50 %. 3 ∴ C.P. of 1 litre of mixture = ❨ 100 x 3 x 1 ❩ = 6 350 7 By the rule of alligation, we have:
C.P. of 1 litre of water C.P. of 1 litre of milk 0 Mean Price
Re. 6 7 Re. 1 1 7 6 7 ∴ Ratio of water and milk = 1 : 6 = 1 : 6. 7 7 - 8. A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?
Options- A. 10
- B. 20
- C. 21
- D. 25 Discuss
Correct Answer: 21
Explanation:
Suppose the can initially contains 7x and 5x of mixtures A and B respectively.Quantity of A in mixture left = ❨ 7x - 7 x 9 ❩ litres = ❨ 7x - 21 ❩ litres. 12 4 Quantity of B in mixture left = ❨ 5x - 5 x 9 ❩ litres = ❨ 5x - 15 ❩ litres. 12 4 ∴ ❨ 7x - 21 ❩ 4 = 7 ❨ 5x - 15 ❩ + 9 4 9 ⟹ 28x - 21 = 7 20x + 21 9 ⟹ 252x - 189 = 140x + 147
⟹ 112x = 336
⟹ x = 3.
So, the can contained 21 litres of A.
- 9. 8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16 : 65. How much wine did the cask hold originally?
Options- A. 18 litres
- B. 24 litres
- C. 32 litres
- D. 42 litres Discuss
Correct Answer: 24 litres
Explanation:
Let the quantity of the wine in the cask originally be x litres.Then, quantity of wine left in cask after 4 operations = [ x ❨ 1 - 8 ❩ 4 ] litres. x ∴ ❨ x(1 - (8/x))4 ❩ = 16 x 81 ⟹ ❨ 1 - 8 ❩ 4 = ❨ 2 ❩ 4 x 3 ⟹ ❨ x - 8 ❩ = 2 x 3 ⟹ 3x - 24 = 2x
⟹ x = 24.
- 10. In a 300 m race A beats B by 22.5 m or 6 seconds. B's time over the course is:
Options- A. 86 sec
- B. 80 sec
- C. 76 sec
- D. None of these Discuss
Correct Answer: 80 sec
Explanation:
B runs 45 m in 6 sec. 2 ∴ B covers 300 m in ❨ 6 x 2 x 300 ❩sec = 80 sec. 45
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