We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways | = (^{7}C_{3} x ^{6}C_{2}) + (^{7}C_{4} x ^{6}C_{1}) + (^{7}C_{5}) | |||||||||||

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= (525 + 210 + 21) | ||||||||||||

= 756. |

Let the number be *x*.

Then, | 1 | of | 1 | of x = 15 x = 15 x 12 = 180. |

3 | 4 |

So, required number = | 3 | x 180 | = 54. | ||

10 |

Required decimal = | 1 | = | 1 | = .00027 |

60 x 60 | 3600 |

(.000216)^{1/3} | = | 216 | 1/3 | ||

10^{6} |

= | 6 x 6 x 6 | 1/3 | ||

10^{2} x 10^{2} x 10^{2} |

= | 6 |

10^{2} |

= | 6 |

100 |

= 0.06

Let the third number be *x*.

Then, first number = 120% of x = | 120x | = | 6x |

100 | 5 |

Second number = 150% of x = | 150x | = | 3x |

100 | 2 |

Ratio of first two numbers = | 6x | : | 3x | = 12x : 15x = 4 : 5. | ||

5 | 2 |

Part filled by (A + B + C) in 3 minutes = 3 | 1 | + | 1 | + | 1 | = | 3 x | 11 | = | 11 | . | ||||

30 | 20 | 10 | 60 | 20 |

Part filled by C in 3 minutes = | 3 | . |

10 |

Required ratio = | 3 | x | 20 | = | 6 | . | ||

10 | 11 | 11 |

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 |

5*x* + 24 = 40 - 5*x*

10*x* = 16

x = | 8 | . |

5 |

So, part of the mixture replaced = | 8 | x | 1 | = | 1 | . | ||

5 | 8 | 5 |

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.

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = | n(E) | = | 9 | . |

n(S) | 20 |

T.D. |
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= Rs. 400. |