Geometry and Mensuration: Level 3 Test 1

$ \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}$

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)

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)

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

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