Maths Quiz for KVS and NVS Exams 2016_00.1
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Maths Quiz for KVS and NVS Exams 2016

Maths Quiz for KVS and NVS Exams 2016_40.1

Q1. The length and breadth of rectangle are increased by 20% and 25% respectively. The increase in the area of the resulting rectangle will be:
(a) 60%
(b) 50%
(c) 40%
(d) 30%
Sol. 
Length 5  →  6
Breadth 4 5
Area 20 30
% increase = (30 – 20)/20 × 100 = 50%

Q2 The sides of square are increased by 50%. The increase in the area of the resulting square  will be:
(a) 125%
(b) 150%
(c) 100%
(d) 120%
Sol.
Increase in area
= 50 + 50 + (50 + 50)/100
= 100 + 25 = 125%

Q3. The base radius and height of a cone is 5 cm and 25cm respectively. If the cone is cut parallel to its base at a height of h cm above  from the base. If the volume of this frustrum is 110 cm3 find the radius of smaller cone
(a) (104)^(1/3) cm
(b) (104)^(1/2) cm
(c) 5 cm
(d) None of these
Sol.
Volume of cone =1/3 πr^2 h
 =1/3×22/7×5×5×25=644.761
Volume of smaller cone
= 654.76 – 110 = 544.761
∴(Volume of smaller cone)/(Volume of larger cone)
=((radius of smaller cone)/(radius of large cone))
⇒ 544.761/654.761=(r/5)^3
⇒ r=(104)^(1/3) cm

Q4. The adjacent sides of a parallelogram are 36 cm and 27 cm in length. If the distance between the shorter sides is 12 cm, then the distance between the longer sides is:
(a) 10 cm
(b) 48 cm
(c) 64 cm
(d) 81 cm
Sol.
Area of parallelogram=base×height
Required answer=36×27/12=81cm

Q5. A child reshapes a cone made up of clay of height 24 cm and radius 6 cm into a sphere. The radius (in cm) of the sphere is
(a) 6
(b) 12
(c) 24
(d) 48
Sol.
Radius of cone = 6 cm
height of cone = 24 cm
∴ volume  of cone = 1/3 π6^2 × 24 cm^3
cone in converted to sphere
Let radius of sphere = r
∴ Volume of sphere ⇒ 4/3 πr^3
Volume of sphere = volume of cone
∴ 4/3 πr^3=1/3 × ? × 6 × 6 × 24
⇒r^3⇒(1/3)×(6×6×24/4)×3
⇒r^3 = 3 × 3 × 24
= 3 × 3 × 3 × 8
r^3=(3)^3×(2)^3
r = 3 × 2 = 6 cm
∴ radius of sphere = 6 cm

Q6. Base of the cone and the cylinder have the same radius 6 cm. They have also the same height 8 cm. The ratio of the curved surface of the cylinder to that of the cone is
(a) 8:5
(b) 8:3
(c) 4:3
(d) 5:3
Sol.
Slant height of the cone (l)
=√(6^2+8^2 ) = 10 cm
⇒ Required ratio = 2πrh/πrl=2h/l
= (2 × 8)/10 = 8 : 5

Q7. A solid spherical copper ball whose diameter is 14 cm is melted and converted into a wire having diameter equal to 14 cm. The length of the wire is
(a) 27 cm
(b) 16/3 cm
(c) 15 cm
(d) 28/3 cm
Sol.
Volume of the solid sphere = 4/3 πr^3
= 4/3? × 7 × 7 × 7 cm^3
Let the length of wire = h cm
πR^2 h=4/3? × 7 × 7 ×7
7 × 7 × h = 4/3 × 7 × 7 × 7
h = 28/3 cm

Q8. The radius of cross-section of a solid cylindrical rod of iron is 50 cm. the cylinder is melted down and formed into 6 solid spherical balls of the same radius as that of the cylinder. The length of the rod (in metres) is
(a) 0.8
(b) 2
(c) 3
(d) 4
Sol.
Volume of cylinder
= 6 × volume of a sphere
?(50)^2 h = 6 × 4/3 π(50)^3
h = 6 × 4/3 × 50
= 400 cm = 4 m

Q9. If the height of a given cone be doubled and radius of the base remains the same the ratio of the volume of the given cone to that of the second cone will be
(a) 2:1
(b) 1:8
(c) 1:2
(d) 8:1
Sol.
Let height of cone = h radius of cone = r
volume of cone = 1/3 πr^2 h
Now height is doubled, volume of new cone
=1/3 πr^2 (2h)=2/3 πr^2 h
Required ratio = 1 : 2

Q10. If the height of a cylinder is increased by 15 percent and the radius of its base is decreased by 10 percent then by what percent will its curved surface area change?
(a) 3.5 percent decrease
(b) 3.5 percent increase
(c) 5 percent increase
(d) 5 percent decrease
Sol.
use x + y + xy/100
percentage change in area
= 15 – 10 + (15 × (-10))/100
= 5 – 1.5 = 3.5%
(3.5 % increase)

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