- This is an assessment test.
- These tests focus on geometry and mensuration and are meant to indicate your preparation level for the subject.
- Kindly take the tests in this series with a pre-defined schedule.

## Geometry and Mensuration: Test 1

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Question 1 |

What would be the cost of building a 7 metres wide garden around a circular field with diameter equal to 280 metres if the cost per sq. metre for building the garden is Rs.21?

Rs.1, 56, 242 | |

Rs.2, 48, 521 | |

Rs.1, 11, 624 | |

None |

Question 1 Explanation:

The area of the garden

$ \displaystyle \begin{array}{l}\pi {{(140+7)}^{2}}-\pi {{(140)}^{2}}\\=\pi ({{147}^{2}}-{{140}^{2}})\\=\pi (21609-19600)\\=\frac{22}{7}\times 2009=6314\\Therefore\,\,required\,\,amount\\=6314\times 21=132594\end{array}$

$ \displaystyle \begin{array}{l}\pi {{(140+7)}^{2}}-\pi {{(140)}^{2}}\\=\pi ({{147}^{2}}-{{140}^{2}})\\=\pi (21609-19600)\\=\frac{22}{7}\times 2009=6314\\Therefore\,\,required\,\,amount\\=6314\times 21=132594\end{array}$

Question 2 |

The ratio of length and breadth of a rectangular plot Is 8: 5 respectively. If the breadth is 60 metre less than the length, what is the perimeter of the rectangular plot?

260 metres | |

1600 metres | |

500 metres | |

None of these |

Question 2 Explanation:

Let the length and breadth = 8x and 5x.

breadth = length -60,

5x = 8x-60,

=>x= 60/3 = 20

Perimeter = 2 (8x+5x) = 26x = 26X20 = 520 m.

breadth = length -60,

5x = 8x-60,

=>x= 60/3 = 20

Perimeter = 2 (8x+5x) = 26x = 26X20 = 520 m.

Question 3 |

A cistern of dimensions 2.4 m x 2 m x 1.5 m takes 2 h 30 min to get filled with water. The rate at which water flows into the cistern is

0.48000 cu m/h | |

800 cu m/min | |

2.88 cu m/h | |

80 cu m/min |

Question 3 Explanation:

Volume = 2.4X2X1.5 = 7.2 m

Time taken = 5/2 hours.

Rate of flow = 7.2X2/5 = 2.88 cu m /h.

^{3}Time taken = 5/2 hours.

Rate of flow = 7.2X2/5 = 2.88 cu m /h.

Question 4 |

It is required to design a circular pipe such that water flowing through it at a speed of 7 m/min fills a tank of capacity 440 cu m in 10 min. The inner radius of the pipe should be

$ \displaystyle 2\,m$ | |

$ \displaystyle \sqrt{2}$ | |

$ \displaystyle \frac{1}{2}$ | |

$ \displaystyle \frac{1}{\sqrt{2}}$ |

Question 4 Explanation:

To fill a tank of capacity 440 cu m in 10 mins rate should be 44 cu m/min.
Therefore,

$ \begin{array}{l}\pi {{r}^{2}}\times 7=44\\\frac{22}{7}{{r}^{2}}=44\\=>r=\sqrt{2}\end{array}$

$ \begin{array}{l}\pi {{r}^{2}}\times 7=44\\\frac{22}{7}{{r}^{2}}=44\\=>r=\sqrt{2}\end{array}$

Question 5 |

Twenty nine times the area of a square is one square metre less than six times the area of the second square and nine times the side of it exceeds the perimeter of other square by one metre. The difference in sides of these squares is

5 m | |

54/11 m | |

11 m | |

6 m |

Question 5 Explanation:

Let the two sides = a cm and b cm.

29a

Solving the equations or by hit and trial one can observe that ,

a= 5 cmÂ and b= 11 cm is a possible solution. Difference = 11-5=6 cm. Correct option is (d)

29a

^{2}=6b^{2}-1 and 9a â€“ 4b = 1Solving the equations or by hit and trial one can observe that ,

a= 5 cmÂ and b= 11 cm is a possible solution. Difference = 11-5=6 cm. Correct option is (d)

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