A) 6 | B) 8 |

C) 9 | D) 10 |

Explanation:

Average speed = total distance / total time

Total distance covered = 6 miles; total time = 45 minutes = 0.75 hours

Average speed = 6/ 0.75 = 8 miles/hour

A) 1080 cm | B) 180 m |

C) 108 m | D) 118 cm |

Explanation:

We know the rule that,

At particular time for all object , ratio of height and shadow are same.

Let the height of the pole be 'H'

Then

$\frac{18}{8}=\frac{H}{48}$

=> H = 108 m.

A) 14 cm | B) 13.5 cm |

C) 12.5 cm | D) 11.4 cm |

Explanation:

Let the required height of the building be x meter

More shadow length, More height(direct proportion)

Hence we can write as

(shadow length) 40.25 : 28.75 :: 17.5 : x

⇒ 40.25 × x = 28.75 × 17.5

⇒ x = 28.75 × 17.5/40.25

= 2875 × 175/40250

= 2875 × 7/1610

= 2875/230

= 575/46

= 12.5 cm

A) 42.2 mts | B) 33.45 mts |

C) 66.6 mts | D) 58.78 mts |

Explanation:

From above diagram

AC represents the hill and DE represents the tower

Given that AC = 100 m

angleXAD = angleADB = 30° (∵ AX || BD )

angleXAE = angleAEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1)

tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)

∵ BD = CE and Substitute the value of CE from equation 1

100/√3 = 100−h(√3) => h = 66.66 mts

The height of the tower = 66.66 mts.

A) 21.6 m | B) 23.2 m |

C) 24.72 m | D) None of these |

Explanation:

Draw BE // CD

Then, CE = AB = 1.6 m,

BE = AC =

$\frac{DE}{\mathrm{tan}30}\phantom{\rule{0ex}{0ex}}=DE=\mathrm{tan}30x20\sqrt{3}\phantom{\rule{0ex}{0ex}}=DE=\frac{1}{\sqrt{3}}x20\sqrt{3}=20$

Therefore, CD = CE + DE = (1.6 + 20) m = 21.6 m.

A) Data inadequate | B) 8 units |

C) 12 units | D) None of these |

Explanation:

One of AB, AD and CD must have given.

So, the data is inadequate.

A) 173 m | B) 200 m |

C) 273 m | D) 300 m |

Explanation:

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100m, $\angle ACB=30\xb0and\angle ADB=45\xb0$

$\frac{AB}{AC}=\mathrm{tan}30\xb0=\frac{1}{\sqrt{3}}=AC=AB*\sqrt{3}=100\sqrt{3m}$

$\frac{AB}{AD}=\mathrm{tan}45\xb0=1=AD=AB=100m$

CD=(AC+AD)=$\left(100\sqrt{3}+100\right)m=100\left(\sqrt{3}+1\right)=100*2.73=273m$

A) 30 | B) 3000 |

C) 18 | D) 1800 |

Explanation:

If an object travels at 5 feet per second it covers 5x60 feet in one minute, and 5x60x60 feet in one hour.

Answer = 1800