**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)