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- Question
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
- Correct Answer
- 20 m
ExplanationA : 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.
Races and Games problems
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- 1. 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.
- 2. 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 - 3. 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.
- 4. 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.
- 5. 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 - 6. 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.
- 7. 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.
- 8. 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.
- 9. In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then, A wins by:
Options- A. 60 m
- B. 40 m
- C. 20 m
- D. 10 m Discuss
Correct Answer: 20 m
Explanation:
To reach the winning post A will have to cover a distance of (500 - 140)m, i.e., 360 m.While A covers 3 m, B covers 4 m.
While A covers 360 m, B covers ❨ 4 x 360 ❩m = 480 m. 3 Thus, when A reaches the winning post, B covers 480 m and therefore remains 20 m behind.
∴ A wins by 20 m.
- 10. A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him by 8 seconds. The speed of B is:
Options- A. 5.15 kmph
- B. 4.14 kmph
- C. 4.25 kmph
- D. 4.4 kmph Discuss
Correct Answer: 4.14 kmph
Explanation:
A's speed = ❨ 5 x 5 ❩m/sec = 25 m/sec. 18 18 Time taken by A to cover 100 m = ❨ 100 x 18 ❩sec = 72 sec. 25 ∴ Time taken by B to cover 92 m = (72 + 8) = 80 sec.
∴ B's speed = ❨ 92 x 18 ❩kmph = 4.14 kmph. 80 5
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