111 | 1 | m |
9 |
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.
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.
∴ 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.
B runs | 45 | m in 6 sec. |
2 |
∴ B covers 300 m in | ❨ | 6 x | 2 | x 300 | ❩sec | = 80 sec. |
45 |
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.
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 |
∴ |
|
= | 7 | |||||
|
9 |
⟹ | 28x - 21 | = | 7 |
20x + 21 | 9 |
⟹ 252x - 189 = 140x + 147
⟹ 112x = 336
⟹ x = 3.
So, the can contained 21 litres of A.
7 | 4 | m |
7 |
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.
B : 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.
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.
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 |
A : C = 60 : 40.
∴ | B | = | ❨ | B | x | A | ❩ | = | ❨ | 45 | x | 60 | ❩ | = | 45 | = | 90 | = 90 : 80. |
C | A | C | 60 | 40 | 40 | 80 |
∴ B can give C 10 points in a game of 90.
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