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- Question
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
- Correct Answer
- 21
ExplanationSuppose 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.
Alligation or Mixture problems
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- 1. 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 - 2. 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 - 3. 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
Discuss
Correct Answer:2 3
Explanation:
By 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 - 4. 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.
- 5. 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 - 6. 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.
- 7. 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 - 8. In a 200 metres race A beats B by 35 m or 7 seconds. A's time over the course is:
Options- A. 40 sec
- B. 47 sec
- C. 33 sec
- D. None of these Discuss
Correct Answer: 33 sec
Explanation:
B runs 35 m in 7 sec.∴ B covers 200 m in ❨ 7 x 200 ❩ = 40 sec. 35 B's time over the course = 40 sec.
∴ A's time over the course (40 - 7) sec = 33 sec.
- 9. In a 100 m race, A can give B 10 m and C 28 m. In the same race B can give C:
Options- A. 18 m
- B. 20 m
- C. 27 m
- D. 9 m Discuss
Correct Answer: 20 m
Explanation:
A : B = 100 : 90.A : C = 100 : 72.
B : C = B x A = 90 x 100 = 90 . A C 100 72 72 When B runs 90 m, C runs 72 m.
When B runs 100 m, C runs ❨ 72 x 100 ❩m = 80 m. 90 ∴ B can give C 20 m.
- 10. A can run 22.5 m while B runs 25 m. In a kilometre race B beats A by:
Options- A. 100 m
- B.
111 1 m 9 - C. 25 m
- D. 50 m Discuss
Correct Answer: 100 m
Explanation:
When B runs 25 m, A runs 45 m. 2 When B runs 1000 m, A runs ❨ 45 x 1 x 1000 ❩m = 900 m. 2 25 ∴ B beats A by 100 m.
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