GATEWAY TO CGL MAINS 2015 : Advanced Quant Quiz

1.The volume of a spherical shell whose external and internal diameters are 12 cm, and 8 cm respectively: 

(a) 304/3 π cm^3 

(b) 115 π cm^3 

(c) 608 π cm^3 

(d) 608/3 π cm^3

2.The volume of a pyramid of base area 20 cm^2 and height 15 cm is: 

(a) 300 cm^3 

(b) 100 cm^3 

(c) 200 cm^3 

(d) None of these

3.The ratio of the volume of right circular cylinder and a right circular cone of the same base and height will be :

(a) 1 : 3 

(b) 2 : 3 

(c) 3 : 2  

(d) 3 : 1

4.The radius and height of a right circular cone are in the ratio of 5 : 12. If its volume is (3l4) 2/7 m^3, its slant height is : 

(a) 26 m 

(b) 19.5 m 

(c) 13 m  

(d) 20 m

5.The height and radius of a cone is doubled the volume of the cone becomes :

(a) 8 times 

(b) 4 times 

(c) 3 times 

(d) 2 times

6.The base of right pyramid is a square, length of diagonal of the base is 24√2m. If the volume of the pyramid is 1728 cu.m, its height is :

(a) 12 metre 

(b) 8 metre 

(c) 9 metre 

(d) 11 metre

7.A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, the ratio of their radius and height is :

(a) 1 : 2 

(b) 1 : 3 

(c) 2 : 3 

(d) 3 : 4

8.The perimeter of the triangular base of a right prism is 15 cm and radius of the incircle of the triangular base is 3 cm. If the volume of the prism be 270 cm^3, then the height of the prism is :

(a) 6 cm 

(b) 7.5 cm 

(c) 10 cm 

(d) 12 cm

9.The base of a right pyramid is a square of side 40 cm long. If the volume of the pyramid is 8000 cm^3, then the height of the pyramid is : 

(a) 12 cm 

(b) 18 cm 

(c) 15 cm 

(d) 10 cm

10.The areas of curved surface of a right circular cylinder and a sphere are equal. If the radii of the cylinder and the sphere be equal, then the  ratio of their volumes will be : 

(a) 2 : 3 

(b) 3 : 2 

(c) 3 : 4 

(d) 4 : 3

ANSWERS AND SOLUTION
1.(d) V = 4/3π (R^3 – r^3) = 4/3π (6^3 - 4^3)
= 4/3 π(152) = 608/3 π cm^3

2.(b) Volume of pyramid = 1/3 * base area * height
= 1/3 * 20 * 15 = 100 cm^3

3.(d)
Volume of cylinder/volume of cone = (π r^2 h)/((1/3)πr^2h)=3/1

4.(c) 1/3 ×π × (5x)^2 ×(12x)=314 2/7=2200/7
=> x = 1
so r = 5 and h = 12
so l = 13cm

5.(a) V = 1/3πr^2h
New radius = 2r and new height = 2h
so New volume = V1 =  1/3 π(2r)^2(2h)
=8(1/3 πr^2 h)= 8v

6.(c) Area of the base = 1/2 * (diagonal)^2
= 1/2 * 24√2 × 24√2= 576 sq.metre.
so Volume of pyramid = 1/3 * height * area of base
=> 1728 = 1/3 * h * 576
=> h = (1728 * 3)/576 = 9 metre

7.(d) (curved surface of cylinder)/(curved surface of cone)=8/5
=> (2π rh)/( πr√(h^2+ r^2 ) )=8/5
h/(√(h^2+ r^2) )  = 4/5
On squaring both sides,
h^2/( h^2+ r^2 )= 16/25
⇒ (h^2+ r^2)/( h^2 )= 25/16
=> 1 + (r^2)/( h^2 )= 25/16
⇒ r^2/( h^2 )=25/16- 1= 9/16
so r/h = 3/4

8.(d)
9.(c) Area of the base = 40 * 40
= 1600 sq.cm
Volume of pyramid = 1/3 (area of base) *  height
=> 8000 = 1/3 * 1600 * h
=> h = (8000 * 3)/1600 = 15cm

10.(b) Let the radius of cylinder be r cm and its height be h cm.
so 2πrh = 4πr^2 , h = 2r
so Ratio of volumes = πr^2 h ∶ 4/3 πr^3
= 2πr^3 ∶ 4/3 πr^3 = 3∶2