- 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 cm2 | |
693 cm2 | |
750 cm2 | |
770 cm2 |
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 +2.2Πr2
= (20p2 x 22)/7 = 770 sq cm
Correct option is (d)
Thus the radius = p.
Thus the height 8p+2p = 35,
10p=35.
=> 2p=7.
Thus the total surface area = 2Πrh +2.2Πr2
= (20p2 x 22)/7 = 770 sq cm
Correct 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 =23/33 = 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)
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 cm2, and BX: XC = 3: 2, CY: YD= 2: 1 and AZ: ZD = 3: 1. Area (in cm2) of pentagon AXCYZ is
47 | |
63 | |
73 | |
79 |
Question 4 Explanation:
Given Area of Parallelogram = 120 cm2
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)= 79cm2
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)= 79cm2
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 |
