- This is an assessment test.
- To draw maximum benefit, study the concepts for the topic concerned.
- Kindly take the tests in this series with a pre-defined schedule.

## Geometry and Mensuration: Level 3 Test 1

Congratulations - you have completed

*Geometry and Mensuration: Level 3 Test 1*. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%% Your answers are highlighted below.

Question 1 |

A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the curved surface area of the remainder is 8/9th of the curved surface of the whole cone, the ratio of the line segments into which the cone's altitude is divided by the plane is given by

2:3 | |

1:3 | |

1: 2 | |

1: 4 |

Question 1 Explanation:

$ \begin{array}{l}Let\text{ }the\text{ }radius,\text{ }slant\text{ }height\text{ }and\text{ }height\text{ }of\text{ }the\text{ }bigger\text{ }cone\text{ }be~R,L,H\\\text{ }and\text{ }that\text{ }of\text{ }the\text{ }smaller\text{ }cone\text{ }=\text{ }r,l,h\\Now\,\,we\,\,\,know\,\,that\\\frac{r}{R}=\frac{l}{L}=\frac{h}{H}\\Therefore\,\,,since\,\,\frac{\pi rl}{\pi RL}={{\left( \frac{h}{H} \right)}^{2}}\\=\left( 1-\frac{8}{9} \right)=\frac{1}{9}\\=\frac{h}{H}=\frac{1}{3}\\Thus\text{ }the\text{ }heights\text{ }are\text{ }divided\text{ }in\text{ }the\text{ }ratio\text{ }of\text{ }1:\text{ }\left( 3-1 \right)\text{ }=1:2\end{array}$

Question 2 |

A solid is in the form of a right circular cylinder with hemispherical ends. The total length of the solid is 35 cm. The diameter of the cylinder is 1/4 of its height. The surface area of the solid is.

Take ∏ =22/7

Take ∏ =22/7

462 cm ^{2} | |

693 cm ^{2} | |

750 cm ^{2} | |

770 cm ^{2} |

Question 2 Explanation:

Let the height = 8p and the diameter = 2p.

Thus the radius = p.

Thus the height 8p+2p = 35,

10p=35.

=> 2p=7.

Thus the total surface area = 2Πrh +2.2Πr

= (20p

Correct option is (d)

Thus the radius = p.

Thus the height 8p+2p = 35,

10p=35.

=> 2p=7.

Thus the total surface area = 2Πrh +2.2Πr

^{2}= (20p

^{2}x 22)/7 = 770 sq cmCorrect option is (d)

Question 3 |

A right circular cone of height h is cut by a plane parallel to the base at a distance h/3 from the base, then the volumes of the resulting cone and the frustum are in the ratio :

1: 3 | |

8: 19 | |

1: 4 | |

1: 7 |

Question 3 Explanation:

The resulting cone =2

Thus the total volume of frustum = 1- (8/27) = 19/27

The required ratio = 8:19

The correct option is (b)

^{3}/3^{3}= 8/27 of the bigger cone.Thus the total volume of frustum = 1- (8/27) = 19/27

The required ratio = 8:19

The correct option is (b)

Question 4 |

In the figure, ABCD is a parallelogram with area 120 cm

^{2}, and BX: XC = 3: 2, CY: YD= 2: 1 and AZ: ZD = 3: 1. Area (in cm^{2}) of pentagon AXCYZ is47 | |

63 | |

73 | |

79 |

Question 4 Explanation:

Given Area of Parallelogram = 120 cm

Now Let BC = 3x cm

XC = 2x cm

CY = 2y cm

YZ = y cm

Now 3x+2x = 5x and 2y + y = 3y

Now Area of parallelogram = l x b = (5x)(3y) = 15xy

120 = 15xy

xy = 8

So x and y could be 2 and 4

We cannot divide 10 in ratio of 3:1

So we will take x = 4 and y = 2

BC = 12

XC=8

CY =4

YZ = 2

AZ= 15

ZD =5

Therefore required area = 120 –(area of ABX + area of ZDY)= 79cm

^{2}Now Let BC = 3x cm

XC = 2x cm

CY = 2y cm

YZ = y cm

Now 3x+2x = 5x and 2y + y = 3y

Now Area of parallelogram = l x b = (5x)(3y) = 15xy

120 = 15xy

xy = 8

So x and y could be 2 and 4

We cannot divide 10 in ratio of 3:1

So we will take x = 4 and y = 2

**Therefore**BC = 12

XC=8

CY =4

YZ = 2

AZ= 15

ZD =5

Therefore required area = 120 –(area of ABX + area of ZDY)= 79cm

^{2}Question 5 |

The circumference of a park is 750m. A and B start walking from the same point in the same direction at 6.75 kmph and 4.75 kmph. In what time will they meet each other again?

3 hours | |

2.5 hours | |

3.5 hours | |

4 hours |

Question 5 Explanation:

Here the circumference will act as the total distance

Here the total distance = 750 m = 0.75 hm

Speed of A = 6.75 km/h

Time taken by him t1 = (0.75/6.75)h

Time taken by B t2 = (0.75/4.75)h

Now the L.C.M of both (t1 and t2) will be the required time

So L.C.M of t1 and t2 = (l.c.m. of numerators)/(h.c.f of denominators)

To solve this we will convert it in m/s

Therefore the required time

(750 x 1800)/125 = 1800 x 6 = 10800 sec

This is equal to 3 hours

Here the total distance = 750 m = 0.75 hm

Speed of A = 6.75 km/h

Time taken by him t1 = (0.75/6.75)h

Time taken by B t2 = (0.75/4.75)h

Now the L.C.M of both (t1 and t2) will be the required time

So L.C.M of t1 and t2 = (l.c.m. of numerators)/(h.c.f of denominators)

To solve this we will convert it in m/s

Therefore the required time

(750 x 1800)/125 = 1800 x 6 = 10800 sec

This is equal to 3 hours

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 5 questions to complete.

List |