1. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients ?

a) 38L

b) 40L

c) 39.5L

d) 38.5L

Explanation: Diameter of bowl = 7 cm

Radius of bowl = $$\frac{2}{7}$$ cm

Height = 4 cm

Volume of cylindrical bowl :

$$\eqalign{ & = \pi {r^2}h \cr & = \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2} \times 4 \cr & = 154\,cu.cm \cr} $$

Hence, volume of soup for 250 patients :

$$\eqalign{ & = 154 \times 250 \cr & = 38500{\text{ c}}{{\text{m}}^3} \cr & = 38.5{\text{L}} \cr} $$

2. The breadth of a room is twice its height and half its length. The volume of the room is 512 cu.m. The length of the room is :

a) 16 m

b) 18 m

c) 20 m

d) 32 m

Explanation: Let the height of the room be x metres

Then, breadth = 2x metres and length = 4x metres

Volume of the room :

= (4x × 2x × x) m

^{3}

= (8x

^{3}) m

^{3}

8x

^{3}= 512

x

^{3}= 64

x = 4

Length of the room is :

= 4x

= (4 × 4)

= 16 m

3. A closed box made of wood of uniform thickness has length, breadth and height 12 cm, 10 cm and 8 cm respectively. If the thickness of the wood is 1 cm, the inner surface area is :

a) 264 cm^{2}

b) 376 cm^{2}

c) 456 cm^{2}

d) 696 cm^{2}

Explanation: LInternal length = (12 - 2) cm = 10 cm

Internal breadth = (10 - 2) cm = 8 cm

Internal height = (8 - 2) cm = 6 cm

Inner surface area :

= 2 [10 × 8 + 8 × 6 + 10 × 6] cm

^{2}

= (2 × 188) cm

^{2}

= 376 cm

^{2}

4.From a cube of side 8 m, a square hole of 3 m side is hollowed from end to end. What is the volume of the remaining solid ?

a) 440 m^{3}

b) 480 m^{3}

c) 508 m^{3}

d) 520 m^{3}

Explanation: Volume of the remaining solid :

= Volume of the cube - Volume of the cuboid cut out from it

= [(8 × 8 × 8) - (3 × 3 × 8)] m

^{3}

= (512 - 72) m

^{3}

= 440 m

^{3}

5. The dimensions of a rectangular box are in the ratio 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rate of Rs. 8 and Rs. 9.50 per square metre is Rs. 1248. Find the dimensions of the box in metres.

a) 2 m, 12 m, 8 m

b) 4 m, 9 m, 16 m

c) 8 m, 12 m, 16 m

d) None of these

Explanation: Let the length, breadth and height of the box be 2x, 3x and 4x respectively

Then, surface area of the box :

= 2 [2x.3x + 3x.4x + 2x.4x]

= [2(6x

^{2}+ 12x

^{2}+ 8x

^{2})]

= 52x

^{2}

$$\eqalign{ & 52{x^2} = \frac{{1248}}{{1.50}} \cr & 52{x^2} = 832 \cr & {x^2} = \frac{{832}}{{52}} \cr & {x^2} = 16 \cr & x = 4 \cr} $$

Hence, the diameter of the box are 8 m, 12 m and 16 m

6.The radius of base and curved surface area of a right cylinder is 'r' units and 4πrh square units respectively. The height of the cylinder is :

a) $$\frac{{\text{h}}}{2}$$ units

b) 1h units

c) 2h units

d) 4h units

Explanation: Radius of the base = r units

Curved surface area of a right cylinder = $$4\pi {\text{r}}h$$

Curved surface area of cylinder = $$2\pi {\text{RH}}$$

$$2\pi {\text{rH = }}4\pi {\text{rh}}$$

⇒ Height of cylinder = 2h units

7. A rectangular paper of 44 cm long and 6 cm wide is rolled to form a cylinder of height equal to width of the paper. The radius of the base of the cylinder so rolled is :

a) 3.5 cm

b) 5 cm

c) 7 cm

d) 14 cm

Explanation: Length of rectangle paper = Circumference of the base of cylinder

If r is the radius of the cylinder :

$$\eqalign{ & 44 = 2\pi r \cr & r = \frac{{44 \times 7}}{{2 \times 22}} \cr & r = 7\,cm \cr} $$

8. A school room is be built to accommodate 70 children so as to allow 2.2 m^{2} of floor and 11 m^{3} of space for each child. If the room be 14 metres long, what must be its breadth and height ?

a) 11 m, 4 m

b) 11 m, 5 m

c) 12 m, 5.5 m

d) 13 m, 6 m

Explanation: Let the breadth and height of the room be b and h metres respectively.

Area of the floor $$ = \left( {14b} \right)\,{m^2}$$

$$\eqalign{ & \therefore 14b = 2.2 \times 70 \cr & b = \frac{{2.2 \times 70}}{{14}} \cr & b = 11 \cr} $$

Volume of the room :

$$\eqalign{ & = \left( {14 \times 11 \times h} \right){m^3} \cr & = \left( {154h} \right){m^3} \cr} $$

$$\eqalign{ & \therefore 154h = 11 \times 70 \cr & h = \frac{{11 \times 70}}{{154}} \cr & h = 5 \cr} $$

9. A rectangular water tank is open at the top. Its capacity is 24 m^{3}. Its length and breadth are 4 m and 3 m respectively. Ignoring the thickness of the material used for building the tank, the total cost of painting the inner and outer surface of the tank at the rate of Rs. 10 per m^{2} is :

a) Rs. 400

b) Rs. 500

c) Rs. 600

d) Rs. 800

Explanation: Depth of the tank :

$$\eqalign{ & = \left( {\frac{{24}}{{4 \times 3}}} \right)m \cr & = 2\,m \cr} $$

Since the tank is open and thickness of material is to be ignored, we have

Sum of inner and outer surface :

$$\eqalign{ & = 2\left[ {\left\{ {2\left( {l + b} \right) \times h} \right\} + lb} \right] \cr & = 2\left[ {\left\{ {2\left( {4 + 3} \right) \times 2} \right\} + 4 \times 3} \right]{m^2} \cr & = 80\,{m^2} \cr} $$

Cost of painting :

$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {80 \times 10} \right) \cr & = {\text{Rs}}{\text{. 800}} \cr} $$

10. A swimming bath is 24 m long and 15 m broad. When a number of men dive into the bath, the height of the water rises by 1 cm. If the average amount of water displaced by one of the men be 0.1 cu.m, how many men are there in the bath ?

a) 32

b) 36

c) 42

d) 46

Explanation: Volume of water displaced :

$$\eqalign{ & = \left( {24 \times 15 \times \frac{1}{{100}}} \right){m^3} \cr & = \frac{{18}}{5}{m^3} \cr} $$

Volume of water displaced by 1 man = 0.1 m

^{3}

Number of men :

$$\eqalign{ & = \left( {\frac{{\frac{{18}}{5}}}{{0.1}}} \right) \cr & = \left( {\frac{{18}}{5} \times 10} \right) \cr & = 36 \cr} $$