A) 65.25 | B) 56.25 |

C) 65 | D) 56 |

Explanation:

let each side of the square be a , then area = a x a

New side = 125a / 100 = 5a / 4

New area =(5a x 5a) / (4 x 4) = (25a²/16)

increased area== (25a²/16) - a²

Increase %= [(9a²/16 ) x (1/a² ) x 100]% = 56.25%

A) 1 : 4 | B) 4 : 1 |

C) 2 : 1 | D) 1 : 2 |

Explanation:

Let the side of the square be x. Then, its diagonal = .

Radius of incircle = and radius of circumcircle = .

Therefore, required ratio = =

A) 3 cm | B) 4 cm |

C) 6 cm | D) 8 cm |

A) 156 | B) 166 |

C) 176 | D) 186 |

Explanation:

Area of the room=(544 * 374)

size of largest square tile= H.C.F of 544 & 374 = 34 cm

Area of 1 tile = (34 x 34)

Number of tiles required== [(544 x 374) / (34 x 34)] = 176

A) 10cm | B) 20cm |

C) 30cm | D) 40cm |

Explanation:

Let the breadth of the given rectangle be x then length is 2x.

thus area of the given rect is

after dec 5cm from length and inc 5cm breadth , new lenght becomes 2x-5 and breadth is x+5.thus new area =(2x-5)(x+5)=

since new area is 75 units greater than original area thus

5x=75+25

5x=100

therefore x=20

hence length of the rectangle is 40 cm.

A) 56m | B) 65m |

C) 34m | D) 36m |

Explanation:

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres

Two poles will be kept 5 metres apart. Also remember that the poles will be placed

along the perimeter of the rectangular plot, not in a single straight line which is

very important.

Hence number of poles required = 280 / 5 = 56