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- Question
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
- Correct Answer
- 80 sec
ExplanationB runs 45 m in 6 sec. 2 ∴ B covers 300 m in ❨ 6 x 2 x 300 ❩sec = 80 sec. 45
Races and Games problems
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- 1. 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.
- 2. 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.
- 3. 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 - 4. 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 - 5. 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 - 6. 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.
- 7. 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.
- 8. 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.
- 9. In a race of 200 m, A can beat B by 31 m and C by 18 m. In a race of 350 m, C will beat B by:
Options- A. 22.75 m
- B. 25 m
- C. 19.5 m
- D.
7 4 m 7
Discuss
Correct Answer: 25 m
Explanation:
A : B = 200 : 169.A : C = 200 : 182.
C = ❨ C x A ❩ = ❨ 182 x 200 ❩ = 182 : 169. B A B 200 169 When C covers 182 m, B covers 169 m.
When C covers 350 m, B covers ❨ 169 x 350 ❩m = 325 m. 182 Therefore, C beats B by (350 - 325) m = 25 m.
- 10. In a 100 m race, A can beat B by 25 m and B can beat C by 4 m. In the same race, A can beat C by:
Options- A. 21 m
- B. 26 m
- C. 28 m
- D. 29 m Discuss
Correct Answer: 28 m
Explanation:
A : B = 100 : 75B : C = 100 : 96.
∴ A : C = ❨ A x B ❩ = ❨ 100 x 100 ❩ = 100 = 100 : 72. B C 75 96 72 ∴ A beats C by (100 - 72) m = 28 m.
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