Wednesday, August 5, 2015

GATEWAY TO CGL MAINS 2015 : Advanced Quant Quiz

1.If the base of right rectangular prism remains constant and the measures of the lateral edges are halved, then its volume will be reduced by: 
(a) 50% 
(b) 33.33% 
(c) 25% 
(d) 66.66% 

2.A conical tent has 60° angle at the vertex. The ratio of its radius and slant height is : 
(a) 1 : 2 
(b) 2 : 1 
(c) 3 : 2 
(d) 1 :3 
3.If the volume of a cube is 512 cm^3, then its total surface area will be : 
(a) 348 cm^2 
(b) 256 cm^2 
(c) 192 cm^2 
(d) 384 cm^2 
4.A curved surface of a cylindrical pillar is 264 m^2 and its volume is 924 m^3. The diameter of the pillar is : 
(a) 12 m 
(b) 10 m 
(c) 14 m 
(d) 16 m
5.The ratio of radii of two right circular cylinders is 2 : 3 and their heights are in the ratio 5:4.The ratio of  their curved surface area is : 
(a) 5 : 6 
(b) 3 : 4
(c) 4 : 5 
(d) 2 : 3
6.A cone, a hemisphere and a cylinder stand on equal base and have the same height. Their volumes are in the ratio: 
(a) 1 : 3 : 2 
(a) 2 : 3 : 1
(c) 1 : 2 : 3 
(d) 3 : 1 : 2
7.The ratio of the areas of the incircle and the circumcircle of a square is : 
(a) 1 : 2 
(b) 2 : 3 
(c) 3 : 4 
(d) 4 : 5
8.The height of a solid right circular cylinder is 6 metres and three times the sum of the areas of its two end faces is twice the area of its curved surface. The radius of its base, in metre is : 
(a) 4 
(b) 2 
(c) 8 
(d) 10 
9.The ratio of the volume of a cube to that of a sphere, which will exactly fit inside the cube is : 
(a) 2 : π   
(b) π ∶ 6    
(c) 6 : π   
(d) 8 : π  
10.A cube of sides 3 cm is melted and smaller cubes of sides 1 cm each are formed. How many such cubes are possible?
(a) 21
(a) 23 
(c) 25
(d) 27

ANSWERS AND SOLUTION
1.(a) Volume of prism = base area * height = A * h
so new volume = A * h/2
{height = h/2}
so % decrease in volume = (Ah – Ah)/2/Ah * 100 = 1/2 * 100 = 50%
2.(a) 
3.(d) Let side of the cube = a
so v = a^3 = 512=> a = 8 cm
so Total surface area = 6a^2
= 6 * 64
= 384 cm^2
4.(c)Let ‘r' be the radius of base and h be the height of the cylinder,
then;
2πrh = 264 =>  rh = 132/π  and πr^2h = 924 => r = 7
2r = 14m

5.(a) First cylinder  Second cylinder
r1 = 2r  r2 = 3r
h1 = 5h  h2 = 4h
so Required ratio = 2πr1h1 : 2πr2 h2
= 2 * 5 : 3 : 4
= 5 : 6

6.(c) Required ratio
= 1/3 π^2h : 2/3 πr^2h : πr^2h
(height of cone= height of hemisphere = r)
= 1/3 : 2/3 : 1 = 1 : 2 : 3

7.(a) If the length of the side of square be a units, then
Radius of incircle = a/2
Radius of circum – circule = √2a/2= a/√2
so Required ratio = π × (a/2)^2 ∶ π ×  (a/√2)^2
= 1/4 : 1/2 = 1:2

8.(a)Let the radius of the base be r metre.
so 3 * 2πr^2 = 2 * 2 πrh => 3r = 2h
=> 3r = 2 * 6
=> r = 4 metre

9.(c) If the edge of cube be x units then diameter of sphere = x units
so Radius of sphere = x/2
so Required ratio = x^3 : 4/3 π x^3/8 = 6∶ π
10.(d) 

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