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# Maths Quiz for KVS and CTET Exams Q1. An iron pipe 20 cm long has exterior diameter 25 cm. 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

Q4. If ‘h’ be the height of a pyramid standing on a base which is an equilateral triangle of side ‘a’ units, then the slant edge is:
(a) √(h^2)+(a^2/4)
(b) √(h^2)+(a^2/8)
(c) √(h^2)+(a^2/3)
(d) √(h^2+a^2)

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) 40?
(c) 80?
(d) None of these

Q6. The areas of three adjacent faces of a cuboid are x, y, z. If the volume is V, then V^2 will be equal to
(a) xy/z
(b) yz/x^2
(c) (x^2 y^2)/z^2
(d) xyz

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

Q10. The perimeter of the base of a right circular cylinder is ‘a’ unit. If the volume of the cylinder is V cubic unit, then the height of the cylinder is
(a) (4a^2 V)/π unit
(b) (4πa^2)/V unit
(c) (πa^2 V)/4 unit
(d) 4πV/a^2  unit

Solutions

S1. Ans. (a)
Sol. Total surface area of pipe (hollow cylinder)
= 2? (R + r) [h + (R – r)]
here, R – r = thickness = 1 cm, h = 20 cm
R  = 25/2 = 12.5 cm
∴ r = 12.5 – 1 = 11.5
Area = 2 × (22/7) (12.5 + 11.5) (20 + 1)
= 44 ×72 = 3168 cm^2

S2. Ans.(a)
Sol. r/h=4/3⇒r/4=h/3=k
⇒ r = 4k, h = 3k
∴ l = √(r^2+h^2 )=√(16k^2+9k^2 )
=√(25k^2 )=5k
∴(Curved surface area)/(Total surface area)
=πrl/(π r(r+l))=l/(r+l)
=5k/(4k+5k)=5/9

S3. Ans.(c)
Sol. Let the height of cylinder be h cm and radius of base = r cm
∴ 2?r^2 + 2?rh = 462 …(i)
Area of curved surface = 2?rh
= 1/3 × 462 = 154
∴ 2?r^2 + 154 = 462
⇒ 2πr^2 = 462 – 154 = 308
⇒2×(22/7)×r^2=308
⇒r^2=(308 × 7)/(2 × 22) = 49 ⇒ r = 7 cm
∴ 2?rh = 154 ⇒ 2 × (22/7) × 7 × h = 154
⇒h=  154/(2×22)=7/2 cm
∴ Volume of cylinder = πr^2 h
=(22/7)×7 ×7 ×(7/2)=539 cm^3

S5. Ans.(b)
Sol. 2?rh = 2 × ? × 2 × 10 = 40?m^2

S6. Ans.(d)
Sol. x.y.z = lb × bh × lh = (lbh)^2
(V) Volume of a cuboid = lbh
So V^2=(lbh)^2 = xyz

S8. Ans.(c)
Sol. Side of cube = (6√3)/√3
= 6 cm
Required ratio = (6 ×6^2)/6^3  = 1 : 1

S9. Ans.(d)
Sol. Let the radius of sphere and hemisphere be = R and r
⇒4/3 πR^3=2/3 πr^3
2R^3=r^3
R^3/r^3 =1/2
⇒ R : r = 1 : ∛2