Math Questions for SSC CHSL and RRB NTPC EXAM

Q1. An iron pipe 20cm long has exterior diameter 25cm. If the thickness of the pipe is 1 cm, then the whole surface area of the pipe?
(a) 3168 cm^2
(b) 3186 cm^2
(c) 3200 cm^2
(d) 3150 cm^2


Q2. The radius and the height of a cone are in the ratio 4 : 3. The ratio of the curved surface area to the total surface area of the cone is:
(a) 5 : 9
(b) 3 : 7
(c) 5 : 4
(d) 16 : 9


Q3. A solid cylinder has total surface area of 462 sq. cm. Curved surface area is 1/3rd of its total surface area. The volume of the cylinder is:
(a) 530 cm^3
(b) 536 cm^3
(c) 539 cm^3
(d) 545 cm^3




Q5. A right circular cylindrical tunnel of diameter 4 m and length 10 m is to be constructed from a sheet of iron. The area(in metre square) of the iron sheet required.
(a) 280/π
(b) 40𝜋
(c) 80𝜋
(d) None of these


Q7. The base of a right prism is a ⧠ABCD. If the volume of prism is 2070(cubic unit). Then find the lateral surface area(in square unit). AB = 9, BC = 14, CD = 13, DA = 12 DAB = 90°.
(a) 720
(b) 540
(c) 920
(d) 960


Q8. Diagonal of a cube is 6√3 cm. Ratio of its total surface area and volume (numerically) is
(a) 2 : 1
(b) 1 : 6
(c) 1 : 1
(d) 1 : 2


Q9. A sphere and a hemisphere have the same volume. The ratio of their radii is
(a) 1 : 2
(b) 1 : 8
(c) 1 : √2
(d) 1 : ∛2




Q11. The total surface area of a sphere is 8𝜋 square unit. The volume of the sphere is
(a) (8√2)/3 π cubic unit
(b) 8/3𝜋 cubic unit
(c) 8√3𝜋 cubic unit
(d) (8√3)/5 π cubic unit


Q12. A cylinder has ‘r’ as the radius of the base and ‘h’ as the height. The radius of base of another cylinder, having double the volume but the same height as that of the first cylinder must be equal to
(a) r/2
(b) 2r
(c) r√2
(d) √2r


Q13. The ratio of height and the diameter of a right circular cone is 3 : 2 and its volume is 1078 cc, then (taking 𝜋 = 22/7) its height is:
(a) 7 cm
(b) 14 cm
(c) 21 cm
(d) 28 cm


Q14. If S1 and S2 be the surface areas of a sphere and the curved surface area of the circumscribed cylinder respectively, then S1 is equal to
(a) 3/4 S2
(b) 1/2 S2
(c) 2/3 S2
(d) S2


Q15. A metallic hemisphere is melted and recast in the shape of cone with the same base radius (R) as that of the hemisphere. If H is the height of the cone, then:
(a) H = 2R
(b) H = 2/3 R
(c) H = √3R
(d) B = 3R




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