1. If the radius of the base and height of a cylinder and cone are each equal to r, and the radius of a hemisphere is also equal to r, then the volumes of the cone, cylinder and hemisphere are in the ratio ?
a) 1 : 2 : 3
b) 1 : 3 : 2
c) 2 : 1 : 3
d) 3 : 2 : 1
Discussion
Explanation: Required ratio :
= Volume of cone : Volume of cylinder : Volume of hemisphere
$$\eqalign{ & = \frac{1}{3}\pi {r^2}:\pi {r^2}r:\frac{2}{3}\pi {r^3} \cr & = \frac{1}{3}:1:\frac{2}{3} \cr & = 1:3:2 \cr} $$
2. The radius of a cylinder is 5 m more than its height. If the curved surface area of the cylinder is 792 m2, what is the volume of the cylinder ?
a) 5712 m3
b) 5244 m3
c) 5544 m3
d) 5306 m3
Discussion
Explanation: Let the height of the cylinder be x cm
Then, radius = (x + 5) m
Curved surface area of the cylinder = $$2\pi rh$$
Now,
$$\eqalign{ & 2\pi \left( {x + 5} \right) \times x = 792 \cr & 2 \times \frac{{22}}{7} \times \left( {{x^2} + 5x} \right) = 792 \cr & {x^2} + 5x = \frac{{792 \times 7}}{{44}} = 126 \cr & {x^2} + 5x - 126 = 0 \cr & {x^2} + 14x - 9x - 126 = 0 \cr & x\left( {x + 14} \right) - 9\left( {x + 14} \right) = 0 \cr & \left( {x - 9} \right)\left( {x + 14} \right) = 0 \cr & x = 9, - 14{\text{(neglect negative value)}} \cr} $$
Height of cylinder = 9 m
Radius of cylinder = 9 + 5 = 14 m
Volume of cylinder :
$$\eqalign{ & = \pi {r^2}h \cr & = \frac{{22}}{7} \times 14 \times 14 \times 9 \cr & = 5544\,{m^3} \cr} $$
3. The radius of a sphere is equal to the radius of the base of a right circular cone, and the volume of the sphere is double the volume of the cone. The ratio of the height of the cone to the radius of its base is :
a) 2 : 1
b) 1 : 2
c) 2 : 3
d) 3 : 2
Discussion
Explanation: Let the radius of cone and the sphere be R and the height of the cone be H
Volume of sphere $$ = \frac{4}{3}\pi {r^3}$$
Volume of cone $$ = \frac{1}{3}\pi {r^2}h$$
$$\eqalign{ & \Rightarrow \frac{4}{3}\pi {R^3} = 2 \times \frac{1}{3}\pi {R^2}H \cr & \Rightarrow 4R = 2H \cr & \Rightarrow \frac{H}{R} = \frac{4}{2}\,Or\,2:1 \cr} $$
4. The dimensions of an open box are 52 cm × 40 cm × 29 cm. It thickness is 2 cm. If 1 cu.cm of metal used in the box weight 0.5 gm, then the weight of the box is :
a) 6.832 kg
b) 37.576 kg
c) 7.76 kg
d) 8.56 kg
Discussion
Explanation: Since the box is an open one, we have :
Internal length = (52 - 4) cm = 48 cm
Internal breadth = (40 - 4) cm = 36 cm
Internal depth = (29 - 2) cm = 27 cm
Volume of the metal used in the box :
= External volume - Internal volume
= [(52 × 40 × 29) - (48 × 36 × 27)] cm3
= (60320 - 46656) cm3
= 13664 cm3
Weight of the box :
$$\eqalign{ & = \left( {\frac{{13664 \times 0.5}}{{1000}}} \right){\text{kg}} \cr & = 6.832\,{\text{kg}} \cr} $$
5. Except for one face of a given cube, identical cubes are glued through their faces to all the other faces of the given cube. If each side of the given cube measures 3 cm, then what is the total surface area of the solid body thus formed ?
a) 225 cm2
b) 234 cm2
c) 270 cm2
d) 279 cm2
Discussion
Explanation: Clearly, each of the 5 faces of the given cube are glued to a face of another cube
Total surface area of the solid :
= 5 × 5a2 + a2 = 26a2
= (26 × 32) cm2
= 234 cm2
6. The length of the longest rod that can be placed in a room of dimensions 10 m × 10 m × 5 m is :
a) 15$$\sqrt 3 $$ m
b) 15 m
c) 10$$\sqrt 2 $$ m
d) 5$$\sqrt 3 $$ m
Discussion
Explanation: Required length :
$$\eqalign{ & = \sqrt {{{\left( {10} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 5 \right)}^2}} \,m \cr & = \sqrt {225} \,m \cr & = 15\,m \cr} $$
7. The sum of the radius and the height of a cylinder is 19 m. The total surface area of the cylinder is 1672 m2, what is the volume of the cylinder ?
a) 3080 m3
b) 2940 m3
c) 3220 m3
d) 2660 m3
Discussion
Explanation: Let the radius of the cylinder be r and height be h
Then, r + h = 19
Again, total surface area of cylinder = $$\left( {2\pi rh + 2\pi {r^2}} \right)$$
$$\eqalign{ & 2\pi r\left( {h + r} \right) = 1672 \cr & 2\pi r \times 19 = 1672 \cr & 38\pi r = 1672 \cr & \pi r = \frac{{1672}}{{38}} = 44\,m \cr & r = \frac{{44 \times 7}}{{22}} = 14\,m \cr} $$
Height = 19 - 14 = 5 m
Volume of cylinder :
$$\eqalign{ & = \pi {r^2}h \cr & = \frac{{22}}{7} \times 14 \times 14 \times 5 \cr & = 22 \times 2 \times 14 \times 5 \cr & = 3080\,{m^3} \cr} $$
8. Given that 1 cu. cm of marble weights 25 gms, the weight of a marble block 28 cm in width and 5 cm thick is 112 kg. The length of the block is :
a) 26.5 cm
b) 32 cm
c) 36 cm
d) 37.5 cm
Discussion
Explanation: Let length = x cm
$$\eqalign{ & x \times 28 \times 5 \times \frac{{25}}{{1000}} = 112 \cr & x = \left( {112 \times \frac{{1000}}{{25}} \times \frac{1}{{28}} \times \frac{1}{5}} \right) \cr & x = 32\,cm \cr} $$
9. An open box is made by cutting the congruent squares from the corners of a rectangular sheet of cardboard of dimension 20 cm × 15 cm. If the side of each square is 2 cm, the total outer surface area of the box is :
a) 148 cm2
b) 284 cm2
c) 316 cm2
d) 460 cm2
Discussion
Explanation: Clearly,
$$l$$ = (20 - 4) cm = 16 cm
b = (15 - 4) cm = 11 cm and
h = 2 cm
Outer surface area of the box :
$$\eqalign{ & = \left[ {2\left( {l + b} \right) \times h} \right] + lb \cr & = \left[ {\left\{ {2\left( {16 + 11} \right) \times 2} \right\} + 16 \times 11} \right] \cr & = \left( {108 + 176} \right) \cr & = 284{\text{ c}}{{\text{m}}^2} \cr} $$
10.V1, V2, V3 and V4 are the volumes of four cubes of side lengths x cm, 2x cm, 3x cm and 4 cm respectively. Some statements regarding these volumes are given below :
a) 1 and 2
b) 2 and 3
c) 1 and 3
d) 1, 2 and 3
Discussion
Explanation: Clearly, we have :
V1 = x3,
V2 = (2x)3 = 8x3
V3 = (3x)3 = 27x3
V4 = (4x)3 = 64x3
(i) V1 + V2 + 2V3
= x3 + 8x3 + 2 × 27x3
= 63x3 < V4
(ii) V1 + 4V2 + V3
= x3 + 4 × 8x3 + 27 x3
= 60x3 < V4
(iii) 2(V1 + V3) + V2
= 2 (x3 + 27x3) + 8x3
= 64x3 = V4
11. 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
Discussion
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 :
$$\eqalign{ & = 154 \times 250 \cr & = 38500{\text{ c}}{{\text{m}}^3} \cr & = 38.5{\text{L}} \cr} $$
12. 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
Discussion
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) m3
= (8x3) m3
8x3 = 512
x3 = 64
x = 4
Length of the room is :
= 4x
= (4 × 4)
= 16 m
13. 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 cm2
b) 376 cm2
c) 456 cm2
d) 696 cm2
Discussion
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] cm2
= (2 × 188) cm2
= 376 cm2
14.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 m3
b) 480 m3
c) 508 m3
d) 520 m3
Discussion
Explanation: Volume of the remaining solid :
= Volume of the cube - Volume of the cuboid cut out from it
= [(8 × 8 × 8) - (3 × 3 × 8)] m3
= (512 - 72) m3
= 440 m3
15. 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
Discussion
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(6x2 + 12x2 + 8x2)]
= 52x2
$$\eqalign{ & 52{x^2} = \frac{{1248}}{{1.50}} \cr & 52{x^2} = 832 \cr & {x^2} = \frac{{832}}{{52}} \cr & {x^2} = 16 \cr & x = 4 \cr} $$
Hence, the diameter of the box are 8 m, 12 m and 16 m
16.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
Discussion
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
17. 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
Discussion
Explanation: Length of rectangle paper = Circumference of the base of cylinder
If r is the radius of the cylinder :
$$\eqalign{ & 44 = 2\pi r \cr & r = \frac{{44 \times 7}}{{2 \times 22}} \cr & r = 7\,cm \cr} $$
18. A school room is be built to accommodate 70 children so as to allow 2.2 m2 of floor and 11 m3 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
Discussion
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 :
$$\eqalign{ & = \left( {14 \times 11 \times h} \right){m^3} \cr & = \left( {154h} \right){m^3} \cr} $$
$$\eqalign{ & \therefore 154h = 11 \times 70 \cr & h = \frac{{11 \times 70}}{{154}} \cr & h = 5 \cr} $$
19. A rectangular water tank is open at the top. Its capacity is 24 m3. 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 m2 is :
a) Rs. 400
b) Rs. 500
c) Rs. 600
d) Rs. 800
Discussion
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 :
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {80 \times 10} \right) \cr & = {\text{Rs}}{\text{. 800}} \cr} $$
20. 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
Discussion
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 m3
Number of men :
$$\eqalign{ & = \left( {\frac{{\frac{{18}}{5}}}{{0.1}}} \right) \cr & = \left( {\frac{{18}}{5} \times 10} \right) \cr & = 36 \cr} $$
21. A cube of length 1 cm is taken out from a cube of length 8 cm. What is the weight of the remaining portion ?
a) $$\frac{7}{8}$$ of the weight of the original cube
b) $$\frac{8}{9}$$ of the weight of the original cube
c) $$\frac{63}{64}$$ of the weight of the original cube
d) $$\frac{511}{512}$$ of the weight of the original cube
Discussion
Explanation: Volume of the bigger cube = (83) cm3 = 512 cm3
Volume of the cut-out cube = (13) cm3 = 1 cm3
Volume of the remaining portion = (512 - 1) cm3 = 511 cm3
$$\frac{{{\text{Weight of the remaining portion}}}}{{{\text{Weight of the original cube}}}}$$ $$ = \frac{{511}}{{512}}$$
22.Capacity of a cylindrical vessel is 25.872 litres. If the height of the cylinder is three times the radius of its base, what is the area of the base ?
a) 336 cm2
b) 616 cm2
c) 1232 cm2
d) None of these
Discussion
Explanation: Volume of cylinder :
$$\eqalign{ & = 25.872\,{\text{litres}} \cr & = \left( {25.872 \times 1000} \right){\text{c}}{{\text{m}}^{\text{3}}} \cr & = 25872\,{\text{ c}}{{\text{m}}^3} \cr} $$
Let the radius of the base of the cylinder be r cm
Then, height = (3r) cm
$$\eqalign{ & \frac{{22}}{7} \times {r^2} \times \left( {3r} \right) = 25872 \cr & {r^3} = \frac{{25872 \times 7}}{{66}} \cr & {r^3} = 2744 \cr & r = \root 3 \of {2744} \cr & r = 14 \cr} $$
Hence, area of the base :
$$\eqalign{ & = \pi {r^2} \cr & = \left( {\frac{{22}}{7} \times 14 \times 14} \right){\text{ c}}{{\text{m}}^2} \cr & = 616\,{\text{ c}}{{\text{m}}^2} \cr} $$
23. What part of a ditch, 48 metres long, 16.5 metres board and 4 metres deep can be filled by the earth got by digging a cylindrical tunnel of diameter 4 metres and length 56 metres ?
a) $$\frac{1}{9}$$
b) $$\frac{2}{9}$$
c) $$\frac{7}{9}$$
d) $$\frac{8}{9}$$
Discussion
Explanation: Volume of earth dug :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times 2 \times 2 \times 56} \right){{\text{m}}^3} \cr & = 704\,{{\text{m}}^3} \cr} $$
Volume of ditch :
$$\eqalign{ & = \left( {48 \times 16.5 \times 4} \right){{\text{m}}^3} \cr & = 3168\,{{\text{m}}^3} \cr} $$
Required fraction :
$$\eqalign{ & = \frac{{704}}{{3168}} \cr & = \frac{2}{9} \cr} $$
24. If the area of the base of a right circular cone is 3850 cm2 and its height is 84 cm, then the curved surface area of the cone is :
a) 10001 cm2
b) 10010 cm2
c) 10100 cm2
d) 11000 cm2
Discussion
Explanation:
$$\eqalign{ & \pi {r^2} = 3850 \cr & \Rightarrow {r^2} = \left( {\frac{{3850 \times 7}}{{22}}} \right) = 1225 \cr & \Rightarrow r = 35 \cr & {\text{Now, }}r = 35{\text{ cm, }}h = 84{\text{ cm}} \cr & {\text{So,}} \cr & l = \sqrt {{{\left( {35} \right)}^2} + {{\left( {84} \right)}^2}} \cr & l = \sqrt {1225 + 7056} \cr & l = \sqrt {8281} \cr & l = 91{\text{ cm}} \cr} $$
Curved surface area :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times 35 \times 91} \right){\text{ c}}{{\text{m}}^2} \cr & = 10010{\text{ c}}{{\text{m}}^2} \cr} $$
25. Ice cream completely filled in a cylinder of diameter 35 cm and height 32 cm is to be served by completely filling identical disposable cones of diameter 4 cm and height 7 cm. The maximum number of persons that can be served this way is :
a) 950
b) 1000
c) 1050
d) 1100
Discussion
Explanation: Volume of cylinder :
$$\eqalign{ & = \left( {\pi \times \frac{{35}}{2} \times \frac{{35}}{2} \times 32} \right){\text{ c}}{{\text{m}}^3} \cr & = 9800\pi {\text{ c}}{{\text{m}}^3} \cr} $$
Volume of 1 cone :
$$\eqalign{ & = \left( {\frac{1}{3} \times \pi \times 2 \times 2 \times 7} \right){\text{ c}}{{\text{m}}^3} \cr & = \frac{{28\pi }}{3}{\text{ c}}{{\text{m}}^3} \cr} $$
Number of persons that can be served :
$$\eqalign{ & = \left( {9800\pi \times \frac{3}{{28\pi }}} \right) \cr & = 1050 \cr} $$
26.A hollow spherical metallic ball has an external diameter 6 cm and is $$\frac{1}{2}$$ cm thick. The volume of metal used in the ball is :
a) $$37\frac{2}{3}{\text{ c}}{{\text{m}}^3}$$
b) $$40\frac{2}{3}{\text{ c}}{{\text{m}}^3}$$
c) $$41\frac{2}{3}{\text{ c}}{{\text{m}}^3}$$
d) $$47\frac{2}{3}{\text{ c}}{{\text{m}}^3}$$
Discussion
Explanation: External radius = 3 cm
Internal radius = (3 - 0.5) cm = 2.5 cm
Volume of the metal :
$$\eqalign{ & = \left[ {\frac{4}{3} \times \frac{{22}}{7} \times \left\{ {{{\left( 3 \right)}^3} - {{\left( {2.5} \right)}^3}} \right\}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times \frac{{91}}{8}} \right){\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{{143}}{3}} \right){\text{ c}}{{\text{m}}^3} \cr & = 47\frac{2}{3}{\text{ c}}{{\text{m}}^3} \cr} $$
27.If a hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cylinder of base diameter 8 cm, then the height of the cylinder is :
a) 4 cm
b) $$\frac{13}{3}$$ cm
c) $$\frac{14}{3}$$ cm
d) 5 cm
Discussion
Explanation: Let the height of the cylinder be h cm
$$\eqalign{ & \frac{4}{3}\pi \left[ {{{\left( 4 \right)}^3} - {{\left( 2 \right)}^3}} \right] = \pi \times {4^2} \times h \cr & \frac{4}{3} \times \pi \times 56 = \pi \times 16h \cr & h = \frac{{4 \times 56}}{{3 \times 16}} \cr & h = \frac{{14}}{3}\,cm \cr} $$
28. The ratio of the volume of a hemisphere and a cylinder circumscribing this hemisphere and having a common base is :
a) 1 : 2
b) 2 : 3
c) 3 : 4
d) 4 : 5
Discussion
Explanation: Let the radius of the hemisphere be be r cm
Then, radius of the cylinder = r cm
Height of the cylinder = r cm
Required ratio :
$$\eqalign{ & = \frac{{{\text{Volume of hemisphere}}}}{{{\text{Volume of cylinder}}}} \cr & = \frac{{\frac{2}{3}\pi {r^3}}}{{\pi {r^2} \times r}} \cr & = \frac{2}{3}\, Or\,2:3 \cr} $$
29.The volume of a right circular cone which is obtained from a wooden cube of edge 4.2 dm wasting minimum amount of wood is :
a) 19404 dm
b) 194.04 dm
c) 19.404 dm
d) 1940.4 dm
Discussion
Explanation: The volume of cone should be maximum
Radius of the base of cone :
$$\eqalign{ & = \frac{{{\text{Edge of cube}}}}{2} \cr & = \frac{{4.2}}{2} \cr & = 2.1\,{\text{dm.}} \cr} $$
Height of cone = Edge of cube = 4.2 dm.
Volume of cone :
$$\eqalign{ & = \frac{1}{3}\pi {r^2}h \cr & = \left( {\frac{1}{3} \times \frac{{22}}{7} \times 2.1 \times 2.1 \times 4.2} \right){\text{cu}}{\text{.dm}}{\text{.}} \cr & = 19.404\,{\text{cu}}{\text{.dm}}{\text{.}} \cr} $$
30. If the diameter of a sphere is 6 m, its hemisphere will have a volume of :
a) $$18\pi $$ m3
b) $$36\pi $$ m3
c) $$72\pi $$ m3
d) None of these
Discussion
Explanation: Volume of hemisphere :
$$\eqalign{ & = \left( {\frac{2}{3}\pi \times 3 \times 3 \times 3} \right){m^3} \cr & = \left( {18\pi } \right){m^3} \cr} $$
31. A cuboidal water tank contains 216 litres of water. Its depth is $$\frac{1}{3}$$ of its length and breadth is $$\frac{1}{2}$$ of $$\frac{1}{3}$$ of the difference between length and depth. The length of the tank is :
a) 2 dm
b) 6 dm
c) 18 dm
d) 72 dm
Discussion
Explanation:Let the length of the tank be x dm
Then, depth of the tank = $$\frac{x}{3}$$ dm
Breadth of the tank :
$$\eqalign{ & = \left[ {\frac{1}{2}{\text{ of }}\frac{1}{3}{\text{ of }}\left( {x - \frac{x}{3}} \right)} \right]{\text{dm}} \cr & = \left( {\frac{1}{2} \times \frac{1}{3} \times \frac{{2x}}{3}} \right){\text{dm}} \cr & = \frac{x}{9}\,{\text{dm}} \cr} $$
$$\eqalign{ & x \times \frac{x}{9} \times \frac{x}{3} = 216 \cr & {x^3} = 216 \times 27 \cr & x = 6 \times 3 \cr & x = 18\,dm \cr} $$
32. A covered wooden box has the inner measures as 115 cm, 75 cm and 35 cm and the thickness of wood is 2.5 cm. Find the volume of the wood :
a) 81000 cu.cm
b) 81775 cu.cm
c) 82125 cu.cm
d) None of these
Discussion
Explanation: The external measures of the box are (115 + 5) cm, (75 + 5) cm, and (35 + 5) cm i.e., 120 cm, 80 cm and 40 cm
Volume of the wood :
= External volume - Internal volume
= [(120 × 80 × 40) - (115 × 75 × 35)] cm3
= (384000 - 301875) cm3
= 82125 cm3
33. Find the cost of a cylinder of radius 14 m and height 3.5 m when the cost of its metal is Rs. 50 per cubic meter :
a) Rs. 100208
b) Rs. 107800
c) Rs. 10800
d) Rs. 109800
Discussion
Explanation: Volume :
$$\eqalign{ & = \pi {r^2}h \cr & = \left( {\frac{{22}}{7} \times 14 \times 14 \times 3.5} \right){m^3} \cr & = 2156\,{m^3} \cr} $$
Cost of the cylinder :
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {2156 \times 50} \right) \cr & = {\text{Rs}}{\text{. 107800}} \cr} $$
34.The radii of the bases of two cylinders are in the ratio 3 : 4 and their height are in the ratio 4 : 3. The ratio of their volume is :
a) 2 : 3
b) 3 : 2
c) 3 : 4
d) 4 : 3
Discussion
Explanation: Let their radii be 3x, 4x and heights be 4y, 3y
Ratio of their volumes :
$$\eqalign{ & = \frac{{\pi \times {{\left( {3x} \right)}^2} \times 4y}}{{\pi \times {{\left( {4x} \right)}^2} \times 3y}} \cr & = \frac{{36}}{{48}} \cr & = \frac{3}{4}\,Or\,3:4 \cr} $$
35.Find the number of coins 1.5 cm in diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
a) 430
b) 440
c) 450
d) 460
Discussion
Explanation: Volume one coin :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{{75}}{{100}} \times \frac{{75}}{{100}} \times \frac{2}{{10}}} \right){\text{c}}{{\text{m}}^3} \cr & = \frac{{99}}{{280}}{\text{ c}}{{\text{m}}^3} \cr} $$
Volume of larger cylinder :
$$ = \left( {\frac{{22}}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10} \right){\text{ c}}{{\text{m}}^3}$$
Number of coins :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10 \times \frac{{280}}{{99}}} \right) \cr & = 450 \cr} $$
36. For a sphere of radius 10 cm, What percent of the numerical value of its volume would be the numerical value of the surface area ?
a) 24%
b) 26.5%
c) 30%
d) 45%
Discussion
Explanation: Volume of the sphere :
$$ = \left[ {\frac{4}{3}\pi {{\left( {10} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3}$$
Surface area of the sphere :
$$ = \left[ {4\pi {{\left( {10} \right)}^2}} \right]{\text{ c}}{{\text{m}}^2}$$
Required percentage :
$$\eqalign{ & = \left[ {\frac{{4\pi {{\left( {10} \right)}^2}}}{{\frac{4}{3}\pi {{\left( {10} \right)}^3}}}} \times 100 \right]\% \cr & = 30\% \cr} $$
37. How many lead shots each 3 mm in diameter can be made from a cuboid of dimensions 9 cm × 11 cm × 12 cm ?
a) 7200
b) 8400
c) 72000
d) 84000
Discussion
Explanation: Volume of each lead shot :
$$\eqalign{ & = \left[ {\frac{4}{3}\pi \times {{\left( {\frac{{0.3}}{2}} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times \frac{{27}}{{8000}}} \right){\text{ c}}{{\text{m}}^3} \cr & = \frac{{99}}{{7000}}{\text{ c}}{{\text{m}}^3} \cr} $$
Number of lead shots :
$$\eqalign{ & = \left( {9 \times 11 \times 12 \times \frac{{7000}}{{99}}} \right) \cr & = 84000 \cr} $$
38. A metallic sphere of radius 5 cm is melted to make a cone with base of the same radius. What is the height of the cone ?
a) 5 cm
b) 10 cm
c) 15 cm
d) 20 cm
Discussion
Explanation: Let the height of the cone be h cm
$$\eqalign{ & \frac{4}{3}\pi \times {\left( 5 \right)^3} = \frac{1}{3}\pi \times {\left( 5 \right)^2} \times h \cr & \Rightarrow h = 20\,cm \cr} $$
39. A hemisphere of lead of radius 6 cm is cast into a right circular cone of height 75 cm. The radius of the base of the cone is :
a) 1.4 cm
b) 2 cm
c) 2.4 cm
d) 4.2 cm
Discussion
Explanation: Let the radius of the cone be R cm
$$\eqalign{ & \frac{1}{3}\pi \times {R^2} \times 75 = \frac{2}{3}\pi \times 6 \times 6 \times 6 \cr & {R^2} = \left( {\frac{{2 \times 6 \times 6 \times 6}}{{75}}} \right) \cr & {R^2} = \frac{{144}}{{25}} \cr & {R^2} = \frac{{{{\left( {12} \right)}^2}}}{{{{\left( 5 \right)}^2}}} \cr & R = \frac{{12}}{5} \cr & R = 2.4\,cm \cr} $$
40. Base of a right prism is a rectangle, the ratio of whose length and breadth is 3 : 2. If the height of the prism is 12 cm and total surface area is 288 sq.cm the volume of the prism is :
a) 291 cm
b) 288 cm
c) 290 cm
d) 286 cm
Discussion
Explanation: Let the length of base be 3a cm and breadth be 2a cm
Total surface area of prism :
= [Perimeter of base × height] + [2 × Area of base]
= [2 (3a + 2a) × 12 + 2 × 3a × 2a] sq.cm
= (120a + 12a2) sq.cm
According to the question,
120a + 12a2 = 288
a2 + 10a = 24
a2 + 10a - 24 = 0
a2 + 12a - 2a - 24 = 0
a (a + 12) - 2 (a + 12) = 0
(a - 2)(a + 12) = 0
a = 2 because a $$ \ne $$ -12
Volume of prism :
= Area of base × Height
= (3a × 2a × 12)cu.cm
= 72a2 cu.cm
= (72 × 2 × 2)cu.cm
= 288 cu.cm
41.A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is :
a) 90 cm
b) 1 dm
c) 1 m
d) 1.1 m
Discussion
Explanation: Let the thickness of the bottom be x cm
$$\left[ {\left( {330 - 10} \right) \times \left( {260 - 10} \right) \times \left( {110 - x} \right)} \right]$$ $$ = 8000 \times 1000$$
$$ \Rightarrow 320 \times 250 \times \left( {110 - x} \right) = $$ $$8000 \times 1000$$
$$\eqalign{ & \left( {110 - x} \right) = \frac{{8000 \times 1000}}{{320 \times 250}} \cr & \left( {110 - x} \right) = 100 \cr & x = 10\,cm \cr & x = 1\,dm \cr} $$
42. How many bricks, each measuring 25 cm × 11.25 cm × 6 cm, will be needed to build a wall 8 m × 6 m × 22.5 cm ?
a) 5600
b) 6000
c) 6400
d) 7200
Discussion
Explanation:
$$\eqalign{ & {\text{Number of bricks :}} \cr & = \frac{{{\text{Volume of the wall}}}}{{{\text{Volume of 1 brick}}}} \cr & = \left( {\frac{{800 \times 600 \times 22.5}}{{25 \times 11.25 \times 6}}} \right) \cr & = 6400 \cr} $$
43. A solid cube just gets completely immersed in water when a 0.2 kg mass is placed on it. If the mass is removed, the cube is 2 cm above the water level. What is the length of each side of the cube ?
a) 6 cm
b) 8 cm
c) 10 cm
d) 12 cm
Discussion
Explanation: Let the length of each side of the cube be a cm
Then, Volume of the part of cube outside water = Volume of the mass placed on it
2a2 = 0.2 × 1000
2a2 = 200
a2 = 100
a = 10
44.The volume of a right circular cylinder, 14 cm in height is equal to that of a cube whose edge is 11 cm. The radius of the base of the cylinder is :
a) 5.2 cm
b) 5.5 cm
c) 11 cm
d) 22 cm
Discussion
Explanation: Volume of the cylinder :
= Volume of the cube
= (11)3 cm3
= 1331 cm3
Let the radius of the base be r cm
$$\eqalign{ & \frac{{22}}{7} \times {r^2} \times 14 = 1331 \cr & {r^2} = \frac{{1331}}{{44}} = \frac{{121}}{4} \cr & r = \frac{{11}}{2} \cr & r = 5.5 \cr} $$
45. If two cylinders of equal volumes have their heights in the ratio 2 : 3, then the ratio if their radii is :
a) $$\sqrt 6 :\sqrt 3 $$
b) $$\sqrt 5 :\sqrt 3 $$
c) $$2 : 3 $$
d) $$\sqrt 3 :\sqrt 2 $$
Discussion
Explanation: Let their heights be 2h and 3h and radii be r and R respectively
$$\eqalign{ & \pi {r^2}\left( {2h} \right) = \pi {R^2}\left( {3h} \right) \cr & \frac{{{r^2}}}{{{R^2}}} = \frac{3}{2} \cr & \Rightarrow \frac{r}{R} = \frac{{\sqrt 3 }}{{\sqrt 2 }}\,i.e.,\sqrt 3 :\sqrt 2 \cr} $$
46. The slant height of a right circular cone is 10 m and its height is 8 m. Find the area of its curved surface.
a) 30π m2
b) 40π m2
c) 60π m2
d) 80π m2
Discussion
Explanation:
$$\eqalign{ & l = 10\,m,h = 8\,m \cr & So, \cr & r = \sqrt {{l^2} + {h^2}} \cr & \,\,\,\, = \sqrt {{{\left( {10} \right)}^2} - {{\left( 8 \right)}^2}} \cr & \,\,\,\, = 6\,m \cr} $$
Curved surface area :
$$\eqalign{ & = \pi rl \cr & = \left( {\pi \times 6 \times 10} \right){m^2} \cr & = 60\pi \,{m^2} \cr} $$
47. A solid metallic right circular cylinder of base diameter 16 m and height 2 cm is melted and recast into a right circular cone of height three times that of the cylinder. Find the curved surface area of the cone. [Use $$\pi $$ = 3.14]
a) 196.8 cm2
b) 228.4 cm2
c) 251.2 cm2
d) None of these
Discussion
Explanation: Let the radius of the cone be r cm
$$\eqalign{ & \pi \times {\left( 8 \right)^2} \times 2 = \frac{1}{3} \times \pi \times {r^2} \times 6 \cr & \Rightarrow r = 8 \cr} $$
Slant height,
$$\eqalign{ & l = \sqrt {{r^2} + {h^2}} \cr & \,\,\, = \sqrt {{8^2} + {6^2}} \cr & \,\,\, = \sqrt {100} \cr & \,\,\, = 10\,cm \cr} $$
Curved surface area of cone :
$$\eqalign{ & = \pi rl \cr & = \left( {3.14 \times 8 \times 10} \right){\text{ c}}{{\text{m}}^2} \cr & = 251.2{\text{ c}}{{\text{m}}^2} \cr} $$
48. If the volume of a sphere is divided by its surface area, the result is 27 cm. The radius of the sphere is :
a) 9 cm
b) 36 cm
c) 54 cm
d) 81 cm
Discussion
Explanation:
$$\eqalign{ & \frac{{\frac{4}{3}\pi {r^3}}}{{4\pi {r^2}}} = 27 \cr & \Rightarrow r = 81\,cm \cr} $$
49. A sphere and a cube have equal surface area. The ratio of the volume of the sphere to that of the cube is :
a) $$\sqrt \pi :\sqrt 6 $$
b) $$\sqrt 2 :\sqrt \pi $$
c) $$\sqrt \pi :\sqrt 3 $$
d) $$\sqrt 6 :\sqrt \pi $$
Discussion
Explanation:
$$\eqalign{ & 4\pi {R^2} = 6{a^2} \cr & \frac{{{R^2}}}{{{a^2}}} = \frac{3}{{2\pi }} \cr & \frac{R}{a} = \frac{{\sqrt 3 }}{{\sqrt {2\pi } }} \cr} $$
$$\eqalign{ & \frac{{{\text{Volume of spere}}}}{{{\text{Volume of cube}}}} \cr & = \frac{{\frac{4}{3}\pi {R^3}}}{{{a^3}}} \cr & = \frac{4}{3}\pi {\left( {\frac{R}{a}} \right)^3} \cr & = \frac{4}{3}\pi \frac{{3\sqrt 3 }}{{2\pi \sqrt {2\pi } }} \cr & = \frac{{2\sqrt 3 }}{{\sqrt {2\pi } }} \cr & = \frac{{\sqrt {12} }}{{\sqrt {2\pi } }} \cr & = \frac{{\sqrt 6 }}{{\sqrt \pi }} \cr & \text{or, }\sqrt 6 :\sqrt \pi \cr} $$
50. Some solid metallic right circular cones, each with radius of the base 3 cm and height 4 cm, are melted to form a solid sphere of radius 6 cm. The number of right circular cones is :
a) 6
b) 12
c) 24
d) 48
Discussion
Explanation: Volume of sphere :
$$\eqalign{ & = \left( {\frac{4}{3}\pi \times {6^3}} \right){\text{c}}{{\text{m}}^{\text{3}}} \cr & = \left( {288\pi } \right){\text{c}}{{\text{m}}^{\text{3}}} \cr} $$
Volume of each cone :
$$\eqalign{ & = \left( {\frac{1}{3}\pi \times {3^2} \times 4} \right){\text{c}}{{\text{m}}^{\text{3}}} \cr & = \left( {12\pi } \right){\text{c}}{{\text{m}}^{\text{3}}} \cr} $$
Number of cone :
$$\eqalign{ & = \frac{{288\pi }}{{12\pi }} \cr & = 24 \cr} $$
51. The ratio of the surface area of a sphere and the curved surface area of the cylinder circumscribing the sphere is :
a) 1 : 1
b) 1 : 2
c) 2 : 1
d) 2 : 3
Discussion
Explanation: Let the radius of the sphere be r
Then, radius of the cylinder = r
Height of the cylinder = 2r
Surface area of sphere = $$4\pi {{\text{r}}^2}$$
Surface area of the cylinder = $$2\pi {\text{r}}(2r) = 4\pi {{\text{r}}^2}$$
Required ratio :
= $$4\pi {{\text{r}}^2}$$ : $$4\pi {{\text{r}}^2}$$
= 1 : 1
52. A hemispherical bowl of internal radius 12 cm contains liquid. This liquid is to be filled into cylindrical container of diameter 4 cm and height 3 cm. The number of containers that is necessary to empty the bowl is :
a) 80
b) 96
c) 100
d) 112
Discussion
Explanation: Volume of hemispherical bowl :
$$ = \left( {\frac{2}{3} \times \pi \times 12 \times 12 \times 12} \right)c{m^3}$$
Volume of 1 cylindrical container :
$$ = \left( {\pi \times 2 \times 2 \times 3} \right)c{m^3}$$
Number of containers required :
$$\eqalign{ & = \frac{2}{3} \times \frac{{12 \times 12 \times 12}}{{2 \times 2 \times 3}} \cr & = 96 \cr} $$
53. Length of each edge of a regular tetrahedron is 1 cm. It volume is :
a) $$\frac{{\sqrt 3 }}{{12}}{\text{ }}c{m^3}$$
b) $$\frac{1}{4}\sqrt 3 {\text{ }}c{m^3}$$
c) $$\frac{{\sqrt 2 }}{6}{\text{ }}c{m^3}$$
d) $$\frac{1}{{12}}\sqrt 2 {\text{ }}c{m^3}$$
Discussion
Explanation: Length of each edge of a regular tetrahedron = 1 cm
Volume of regular tetrahedron :
$$\eqalign{ & = \frac{{{a^3}}}{{6\sqrt 2 }}{\text{ c}}{{\text{m}}^3} \cr & = \frac{1}{{6\sqrt 2 }} \cr & = \frac{{\sqrt 2 }}{{6\sqrt 2 \times \sqrt 2 }}{\text{ c}}{{\text{m}}^3} \cr & = \frac{{\sqrt 2 }}{{12}}{\text{ Or }}\frac{1}{{12}}\sqrt 2 {\text{ c}}{{\text{m}}^3} \cr} $$
54. The base of a right prism is a trapezium whose lengths of two parallels sides are 10 cm and 6 cm and distance between them is 5 cm. If the heights of the prism is 8 cm, its volume is :
a) 320 cm3
b) 300 cm3
c) 310 cm3
d) 300.5 cm3
Discussion
Explanation: Length of parallel sides of prism = 10 cm and 6 cm
Height of prism = 8 cm
Volume of prism :
$$\eqalign{ & = \frac{1}{2}\left( {10 + 6} \right) \times 5 \times 8 \cr & = \frac{1}{2} \times 16 \times 5 \times 8 \cr & = 320{\text{ c}}{{\text{m}}^3} \cr} $$
55. A rectangular water reservoir contains 42000 litres of water. If the length of reservoir is 6 m and breadth of the reservoir is 3.5 m, then the depth of the reservoir will be :
a) 2 m
b) 5 m
c) 6 m
d) 8 m
Discussion
Explanation: Volume of the reservoir = 42000 litres = 42 m3
Let the depth of the reservoir be h metres
$$\eqalign{ & 6 \times 3.5 \times h = 42 \cr & h = \frac{{42}}{{6 \times 3.5}} = 2\,m \cr} $$
56. When a ball bounces, it rises to $$\frac{2}{3}$$ of the height from which it fell. If the ball is dropped from a height of 36 m, how high will it rise at the third bounce ?
a) $$10\frac{1}{3}$$ m
b) $$10\frac{2}{3}$$ m
c) $$12\frac{1}{3}$$ m
d) $$12\frac{2}{3}$$ m
Discussion
Explanation: Ball is dropped from the height of 36 m when the ball will rise at the third bounce
Required height :
$$\eqalign{ & = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times 36 \cr & = \frac{{32}}{3} \cr & = 10\frac{2}{3}\,m \cr} $$
57. The sum of perimeters of the six faces of a cuboid is 72 cm and the total surface area of the cuboid is 16 cm2. Find the longest possible length that can be kept inside the cuboid :
a) 5.2 cm
b) 7.8 cm
c) 8.05 cm
d) 8.36 cm
Discussion
Explanation: Sum of perimeters of the six faces :
$$\eqalign{ & = 2\left[ {2\left( {l + b} \right) + 2\left( {b + h} \right) + 2\left( {l + h} \right)} \right] \cr & = 4\left( {2l + 2b + 2h} \right) \cr & = 8\left( {l + b + h} \right) \cr} $$
Total surface area $$ = 2\left( {lb + bh + lh} \right)$$
$$\eqalign{ & 8\left( {l + b + h} \right) = 72 \cr & \Rightarrow l + b + h = 9 \cr & 2\left( {lb + bh + lh} \right) = 16 \cr & \Rightarrow lb + bh + lh = 8 \cr} $$
Now,
$${\left( {l + b + h} \right)^2} = {l^2} + {b^2} + {h^2} + 2$$ $$\left( {lb + bh + lh} \right)$$
$$\eqalign{ & \Rightarrow {\left( 9 \right)^2} = {l^2} + {b^2} + {h^2} + 16 \cr & {l^2} + {b^2} + {h^2} = 81 - 16 \cr & {l^2} + {b^2} + {h^2} = 65 \cr} $$
Required length :
$$\eqalign{ & = \sqrt {{l^2} + {b^2} + {h^2}} \cr & = \sqrt {65} \cr & = 8.05\,cm \cr} $$
58. The surface area of a cube is 150 cm2. Its volume is :
a) 64 $${\text{c}}{{\text{m}}^3}$$
b) 125 $${\text{c}}{{\text{m}}^3}$$
c) 150 $${\text{c}}{{\text{m}}^3}$$
d) 216 $${\text{c}}{{\text{m}}^3}$$
Discussion
Explanation:
$$\eqalign{ & 6{a^2} = 150 \cr & {a^2} = 25 \cr & a = 5 \cr} $$
Volume :
$${a^3} = {5^3} = 125\,{\text{ c}}{{\text{m}}^3}$$
59. If three equal cubes are placed adjacently in a row, then the ratio of the total surface area of the new cuboid to the sum of the surface areas of the three cubes will be ?
a) 1 : 3
b) 2 : 3
c) 5 : 9
d) 7 : 9
Discussion
Explanation: Let the length of each edge of each cube be a
Then, the cuboid formed by placing 3 cubes adjacently has the dimensions 3a , a and a
Surface area of the cuboid :
$$\eqalign{ & = 2\left[ {3a \times a + a \times a + 3a \times a} \right] \cr & = 2\left[ {3{a^2} + {a^2} + 3{a^2}} \right] \cr & = 14{a^2} \cr} $$
Sum of surface area of 3 cubes :
$$\eqalign{ & = \left( {3 \times 6{a^2}} \right) \cr & = 18{a^2} \cr} $$
Required ratio :
$$\eqalign{ & = 14{a^2}:18{a^2} \cr & = 7:9 \cr} $$
60. The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the ratio of its diameter to its height.
a) 3 : 7
b) 7 : 3
c) 6 : 7
d) 7 : 6
Discussion
Explanation:
$$\eqalign{ & \frac{{\pi {r^2}h}}{{2\pi rh}} = \frac{{924}}{{264}} \cr & \Rightarrow r = \left( {\frac{{924}}{{264}} \times 2} \right) \cr & \Rightarrow r = 7\,m \cr} $$
$$\eqalign{ & \therefore 2\pi rh = 264 \cr & h = \left( {264 \times \frac{7}{{22}} \times \frac{1}{2} \times \frac{1}{7}} \right) \cr & h = 6\,m \cr} $$
Required ratio :
$$ = \frac{{2r}}{h} = \frac{{14}}{6} = 7:3$$
61. The height of a closed cylinder of given volume and the minimum surface area is :
a) Equal to its diameter
b) Half of its diameter
c) Double of its diameter
d) None of these
Discussion
Explanation:
$$\eqalign{ & V = \pi {r^2}h{\text{ and }} \cr & S = 2\pi rh + 2\pi {r^2} \cr & \,\,\,\,\,\,\, = 2\pi r\left( {h + r} \right) \cr & {\text{Where, }}h = \frac{V}{{\pi {r^2}}} \cr & S = 2\pi r\left( {\frac{V}{{\pi {r^2}}} + r} \right) \cr & S = \frac{{2V}}{r} + 2\pi {r^2} \cr & \frac{{dS}}{{dr}} = \frac{{ - 2V}}{{{r^2}}} + 4\pi r{\text{ and}} \cr & \frac{{{d^2}S}}{{d{r^2}}} = \left( {\frac{{4V}}{{{r^3}}} + 4\pi } \right){\text{ > 0}} \cr} $$
S is minimum when :
$$\eqalign{ & \frac{{dS}}{{dr}} = 0 \cr & \frac{{ - 2V}}{{{r^2}}} + 4\pi r = 0 \cr & V = 2\pi {r^3} \cr & \pi {r^2}h = 2\pi {r^3} \cr & h = 2r \cr} $$
62. Water is poured into an empty cylindrical tank at a constant rate for 5 minutes. After the water has been poured into the tank. the depth of the water is 7 feet. The radius of the tank is 100 feet. Which of the following is the best approximation for the rate at which the water was poured into the tank ?
a) 140 cubic feet/sec
b) 440 cubic feet/sec
c) 700 cubic feet/sec
d) 2200 cubic feet/sec
Discussion
Explanation: Volume of water flown into the tank in 5 min :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times 100 \times 100 \times 7} \right){\text{cu}}{\text{.feet}} \cr & = 220000\,{\text{cu}}{\text{.feet}} \cr} $$
Rate of flow of water :
$$\eqalign{ & = \left( {\frac{{220000}}{{5 \times 60}}} \right){\text{cu}}{\text{.feet/sec}} \cr & = 733.3 \approx 700\,{\text{cu}}{\text{.feet/sec}} \cr} $$
63.The curved surface of a right circular cone of height 15 cm and base diameter 16 cm is :
a) 60π cm2
b) 68π cm2
c) 120π cm2
d) 136π cm2
Discussion
Explanation: h = 15 cm, r = 8 cm
So,
$$\eqalign{ & l = \sqrt {{r^2} + {h^2}} \cr & \,\,\,\,\,\, = \sqrt {{8^2} + {{\left( {15} \right)}^2}} \cr & \,\,\,\,\,\, = 17\,cm \cr} $$
Curved surface area :
$$\eqalign{ & = \pi rl \cr & = \left( {\pi \times 8 \times 17} \right){\text{ c}}{{\text{m}}^2} \cr & = 136\pi {\text{ c}}{{\text{m}}^2} \cr} $$
64. A right circular cone and a right circular cylinder have equal base and equal height. If the radius of the base and the height are in the ratio 5 : 12, then the ratio of the total surface area of the cylinder to that of the cone is :
a) 3 : 1
b) 13 : 9
c) 17 : 9
d) 34 : 9
Discussion
Explanation: Let their radius and height be 5x and 12x respectively
Slant height of the cone,
$$l = \sqrt {{{\left( {5x} \right)}^2} + {{\left( {12x} \right)}^2}} = 13x$$
$$\eqalign{ & \frac{{{\text{Total surface area of cylinder}}}}{{{\text{Total surface area of cone}}}} \cr & = \frac{{2\pi r\left( {h + r} \right)}}{{\pi r\left( {l + r} \right)}} \cr & = \frac{{2\left( {h + r} \right)}}{{\left( {l + r} \right)}} \cr & = \frac{{2 \times \left( {12x + 5x} \right)}}{{\left( {13x + 5x} \right)}} \cr & = \frac{{34x}}{{18x}} \cr & = \frac{{17}}{9}\,Or\,17:9 \cr} $$
65.The curved surface area of a sphere is 5544 sq.cm. Its volume is :
a) 22176 cm3
b) 33951 cm3
c) 38808 cm3
d) 42304 cm3
Discussion
Explanation:
$$\eqalign{ & 4\pi {r^2} = 5544 \cr & {r^2} = \left( {5544 \times \frac{1}{4} \times \frac{7}{{22}}} \right) \cr & {r^2} = 441 \cr & r = 21 \cr} $$
Volume :
$$\eqalign{ & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times 21 \times 21 \times 21} \right){\text{ c}}{{\text{m}}^3} \cr & = 38808{\text{ c}}{{\text{m}}^3} \cr} $$
66. The diameter of a spare is 8 cm. It is melted and drawn into a wire of diameter 3 mm. The length of the wire is :
a) 36.9 m
b) 37.9 m
c) 38.9 m
d) 39.9 m
Discussion
Explanation: Let the length of the wire be h
$$\eqalign{ & \pi \times \frac{3}{{20}} \times \frac{3}{{20}} \times h = \frac{4}{3}\pi \times 4 \times 4 \times 4 \cr & h = \left( {\frac{{4 \times 4 \times 4 \times 4 \times 20 \times 20}}{{3 \times 3 \times 3}}} \right)cm \cr & h = \left( {\frac{{102400}}{{27}}} \right)cm \cr & h = 3792.5\,cm \cr & h = 37.9\,m \cr} $$
67. The external and internal diameters of a hemispherical bowl are 10 cm and 8 cm respectively. What is the total surface area of the bowl ?
a) 257 cm2
b) 286 cm2
c) 292 cm2
d) 302 cm2
Discussion
Explanation: Internal radius, r = 4 cm
External radius, R = 5 cm
Total surface area :
$$\eqalign{ & = 2\pi {R^2} + 2\pi {r^2} + \pi \left( {{R^2} - {r^2}} \right) \cr & = 3\pi {R^2} + \pi {r^2} \cr & = \left[ {\pi \left( {3 \times 25 + 16} \right)} \right]{\text{ c}}{{\text{m}}^2} \cr & = \left( {\frac{{22}}{7} \times 91} \right){\text{c}}{{\text{m}}^2} \cr & = 286{\text{ c}}{{\text{m}}^2} \cr} $$
68. A pyramid has an equilateral triangle as its base of which each side is 1 m. Its slant edge is 3 m. The whole surface are of the pyramid is equal to :
a) $$\frac{{\sqrt 3 + 2\sqrt {13} }}{4}sq.m$$
b) $$\frac{{\sqrt 3 + 3\sqrt {13} }}{4}sq.m$$
c) $$\frac{{\sqrt 3 + 3\sqrt {35} }}{4}sq.m$$
d) $$\frac{{\sqrt 3 + 2\sqrt {35} }}{4}sq.m$$
Discussion
Explanation: Area of base :
$$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times {1^2}} \right){m^2} \cr & = \frac{{\sqrt 3 }}{4}{m^2} \cr} $$
Clearly, the pyramid has 3 triangular faces each with sides 3m, 3m and 1 m
So, area of each lateral face :
$$\eqalign{ & = \sqrt {\frac{7}{2} \times \left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 1} \right)} {m^2} \cr & \left[ { s = \frac{{3 + 3 + 1}}{2} = \frac{7}{2}} \right] \cr & = \sqrt {\frac{7}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{5}{2}} {m^2} \cr & = \frac{{\sqrt {35} }}{4}{m^2} \cr} $$
Whole surface area of the pyramid :
$$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} + 3 \times \frac{{\sqrt {35} }}{4}} \right){m^2} \cr & = \frac{{\sqrt 3 + 3\sqrt {35} }}{4}{m^2} \cr} $$
69. A hemisphere and a cone have equal bases. If their heights are also equal, then the ratio of their curved surface will be :
a) $$\sqrt 2 :1$$
b) $$1:\sqrt 2 $$
c) 2 : 1
d) 1 : 2
Discussion
Explanation:
Let,
$$\eqalign{ & OP = OQ = OR = r \cr & OR = h = r \cr} $$
Curved surface area of the hemisphere = $$2\pi {r^2}$$
Curved surface area of a cone = $$\pi rl$$
Where,
$$\eqalign{ & l = \sqrt {{h^2} + {r^2}} \cr & \,\,\,\,\, = \sqrt {{r^2} + {r^2}} \cr & \,\,\,\,\, = r\sqrt 2 \cr} $$
Required ratio :
$$\eqalign{ & = \frac{{2\pi {r^2}}}{{\pi rl}} \cr & = \frac{{2\pi {r^2}}}{{\pi r \times r\sqrt 2 }} \cr & = \frac{2}{{\sqrt 2 }} \cr & = \frac{{2 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }} \cr & = \frac{{2\sqrt 2 }}{2} \cr & = \frac{{\sqrt 2 }}{1}\,Or\,\sqrt 2 :1 \cr} $$
70. A swimming pool 9 m wide and 12 m long and 1 m deep on the shallow side and 4 m deep on the deeper side. Its volume is :
a) 360 m3
b) 270 m3
c) 420 m3
d) None of these
Discussion
Explanation: Given length of width of swimming pool is 9 m and 12 m respectively
Volume of swimming pool :
$$\eqalign{ & = 9 \times 12 \times \left( {\frac{{1 + 4}}{2}} \right) \cr & = 9 \times 12 \times \frac{5}{2} \cr & = 270\,\text{cu. metre} \cr} $$
71. A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the man is:
a) 12 kg
b) 60 kg
c) 72 kg
d) 96 kg
Discussion
Explanation: Volume of water displaced
= (3 x 2 x 0.01) m3
= 0.06 m3.
Mass of man = Volume of water displaced x Density of water
= (0.06 x 1000) kg
= 60 kg.
72. 50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m3, then the rise in the water level in the tank will be:
a) 20 cm
b) 25 cm
c) 35 cm
d) 50 cm
Discussion
Explanation:
$$\eqalign{ & {\text{Total}}\,{\text{Volume}}\,{\text{of}}\,{\text{water}}\,{\text{displaced}} \cr & = \left( {4 \times 50} \right){m^3} = 200\,{m^3} \cr & {\text{Rise}}\,{\text{in}}\,{\text{water}}\,{\text{level}} \cr & = \left( {\frac{{200}}{{40 \times 20}}} \right)m \cr & = 0.25\,m \cr & = 25\,cm \cr} $$
73. The slant height of a right circular cone is 10 m and its height is 8 m. Find the area of its curved surface.
a) 30π m2
b) 40π m2
c) 60π m2
d) 80π m2
Discussion
Explanation:
$$\eqalign{ & l = 10\,m \cr & h = 8\,m \cr & So,\,r = \sqrt {{l^2} - {h^2}} = \sqrt {{{\left( {10} \right)}^2} - {8^2}} = 6\,m \cr & {\text{Curved}}\,{\text{surface}}\,{\text{area}} \cr & \pi \,rl = \left( {\pi \times 6 \times 10} \right){m^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 60\pi \,{m^2} \cr} $$
74. A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:
a) 49 m2
b) 50 m2
c) 53.5 m2
d) 55 m2
Discussion
Explanation:
$$\eqalign{ & {\text{Area}}\,{\text{of}}\,{\text{the}}\,{\text{wet}}\,{\text{surface}} \cr & = \left[ {2\left( {lb + bh + lh} \right) - lb} \right] \cr & = 2\left( {bh + lh} \right) + lb \cr & = \left[ {2\left( {4 \times 1.25 + 6 \times 1.25} \right) + 6 \times 4} \right]{m^2} \cr & = 49\,{m^2} \cr} $$
75. A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m3) is:
a) 4830
b) 5120
c) 6420
d) 8960
Discussion
Explanation: Clearly, l = (48 - 16)m = 32 m,
b = (36 -16)m = 20 m,
h = 8 m.
Volume of the box
= (32 x 20 x 8) m3
= 5120 m3
76. If the volume and surface area of a sphere are numerically the same, then its radius is :
a) 1 unit
b) 2 units
c) 3 units
d) 4 units
Discussion
Explanation:
$$\eqalign{ & \frac{4}{3}\pi {r^3} = 4\pi {r^2} \cr & r = 3 \text{ units} \cr} $$
77. A copper wire of length 36 m and diameter 2 mm is melted to form a sphere. The radius of the sphere (in cm) is :
a) 2.5 cm
b) 3 cm
c) 3.5 cm
d) 4 cm
Discussion
Explanation: Let the radius of the sphere be r cm
$$\eqalign{ & \frac{4}{3}\pi {r^3} = \pi \times {\left( {0.1} \right)^2} \times 3600 \cr & {r^3} = 36 \times \frac{3}{4} \cr & {r^3} = 27 \cr & r = 3\,cm \cr} $$
78.The capacities of two hemispherical vessels are 6.4 litres and 21.6 litres. The areas of inner curved surfaces of the vessels will be in the ratio of :
a) $$\sqrt 2 $$ : $$\sqrt 3 $$
b) 2 : 3
c) 4 : 9
d) 16 : 81
Discussion
Explanation: Let their radii be R and r
$$\eqalign{ & \frac{{\frac{2}{3}\pi {R^3}}}{{\frac{2}{3}\pi {r^3}}} = \frac{{6.4}}{{21.6}} \cr & {\left( {\frac{R}{r}} \right)^3} = \frac{8}{{27}} \cr & {\left( {\frac{R}{r}} \right)^3} = {\left( {\frac{2}{3}} \right)^3} \cr & \frac{R}{r} = \frac{2}{3} \cr} $$
Ratio of curved surface area :
$$ = \frac{{2\pi {R^2}}}{{2\pi {r^2}}} = {\left( {\frac{R}{r}} \right)^2} = \frac{4}{9}\,or\,4:9$$
79. The base of a pyramid is an equilateral triangle of side 1 m. If the height of the pyramid is 4 metres, then the volume is :
a) 0.550 m3
b) 0.577 m3
c) 0.678 m3
d) 0.750 m3
Discussion
Explanation: Area of the base :
$$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times {1^2}} \right){m^2} \cr & = \frac{{\sqrt 3 }}{4}{m^2} \cr} $$
Volume of pyramid :
$$\eqalign{ & = \left( {\frac{1}{3} \times \frac{{\sqrt 3 }}{4} \times 4} \right){m^3} \cr & = \left( {\frac{{\sqrt 3 }}{3}} \right){m^3} \cr & = \left( {\frac{{1.732}}{3}} \right){m^3} \cr & = 0.577\,{m^3} \cr} $$.
80.Each side of a cube is decreased by 25%. Find the ratio of the volume of the original cube and the resulting cube = ?
a) 64 : 1
b) 27 : 64
c) 64 : 27
d) 8 : 1
Discussion
Explanation: Let the side of cube = 10 cm
Original volume = 10 × 10 × 10 = 1000 cm3
Now, side of new cube : = 10 - 25% of 10 = 7.5
New volume = 7.5 × 7.5 × 7.5 = 421.875 cm3
Required ratio :
$$\eqalign{ & = \frac{{1000}}{{421.875}} \cr & = \frac{{1000000}}{{421875}} \cr & = \frac{{64}}{{27}}\,Or\,64:27 \cr} $$
81. A closed aquarium of dimensions 30 m × 25 cm × 20 cm is made up entirely of glass plates held together with tapes. The total length of tape required to hold the plates together (ignore the overlapping tapes) is :
a) 75 cm
b) 120 cm
c) 150 cm
d) 300 cm
Discussion
Explanation: Total length of tape required :
= Sum of lengths of edges
= (30 × 4 + 25 × 4 + 20 × 4) cm
= 300 cm
82. A swimming pool 9 m wide and 12 m long is 1 m deep on the shallow side and 4 m deep on the deeper side. It volume is :
a) 208 m3
b) 270 m3
c) 360 m3
d) 408 m3
Discussion
Explanation: Volume :
$$\eqalign{ & = \left[ {12 \times 9 \times \left( {\frac{{1 + 4}}{2}} \right)} \right]{m^3} \cr & = \left( {12 \times 9 \times 2.5} \right){m^3} \cr & = 270\,{m^3} \cr} $$
83. An aluminium sheet 27 cm long, 8 cm broad and 1 cm thick is melted into a cube. The difference in the surface areas of the two solids would be :
a) Nil
b) 284 cm2
c) 286 cm2
d) 296 cm2
Discussion
Explanation: Volume of cube = Volume of sheet = (27 × 8 × 1) cm3 = 216 cm3
Edge of cube :
$$\root 3 \of {216} \,cm = 6\,cm$$
Surface area of sheet :
$$\eqalign{ & = 2\left( lb + bh + lh \right) \cr & = 2\left( {27 \times 8 + 8 \times 1 + 27 \times 1} \right){\text{ c}}{{\text{m}}^2} \cr & = \left( {216 + 8 + 27} \right){\text{ c}}{{\text{m}}^2} \cr & = 502{\text{ c}}{{\text{m}}^2} \cr} $$
Surface area of cube :
$$\eqalign{ & = 6{a^2} \cr & = \left( {6 \times {6^2}} \right){\text{ c}}{{\text{m}}^2} \cr & = 216{\text{ c}}{{\text{m}}^2} \cr} $$
Required difference :
$$\eqalign{ & = \left( {502 - 216} \right){\text{ c}}{{\text{m}}^2} \cr & = 286{\text{ c}}{{\text{m}}^2} \cr} $$
84. The volumes of two cubes are in the ratio 8 : 27. The ratio of their surface areas is :
a) 2 : 3
b) 4 : 9
c) 12 : 9
d) None of these
Discussion
Explanation: Let their edges be a and b
$$\eqalign{ & \frac{{{a^3}}}{{{b^3}}} = \frac{8}{{27}} \cr & {\left( {\frac{a}{b}} \right)^3} = {\left( {\frac{2}{3}} \right)^3} \cr & \frac{a}{b} = \frac{2}{3} \cr & \frac{{{a^2}}}{{{b^2}}} = \frac{4}{9} \cr & \frac{{6{a^2}}}{{6{b^2}}} = \frac{4}{9}\,Or\,4:9 \cr} $$
85. The height of a right circular cylinder is 6 m. If three times the sum of the areas of its two circular faces is twice the area of the curved surface, then the radius of its base is :
a) 1 m
b) 2 m
c) 3 m
d) 4 m
Discussion
Explanation:
$$\eqalign{ & 3 \times 2\pi {r^2} = 2 \times 2\pi rh \cr & \Rightarrow 6r = 4h \cr & r = \frac{2}{3}h \cr & r = \left( {\frac{2}{3} \times 6} \right)m \cr & r = 4\,m \cr} $$
86.The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the ratio of its diameter to its height.
a) 3 : 7
b) 7 : 3
c) 6 : 7
d) 7 : 6
Discussion
Explanation:
$$\eqalign{ & \frac{{\pi \,{r^2}h}}{{2\pi \,rh}} = \frac{{924}}{{264}}\, \cr & \Rightarrow r = {\frac{{924}}{{264}} \times 2} = 7m \cr & {\text{And}},\,2\pi rh = 264 \cr & \Rightarrow h = {264 \times \frac{7}{{22}} \times \frac{1}{2} \times \frac{1}{7}} = 6m \cr & {\text{Required}}\,{\text{ratio}} \cr & = \frac{{2r}}{h} = \frac{{14}}{6} = 7:3 \cr} $$
87. A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is:
a) 90 cm
b) 1 dm
c) 1 m
d) 1.1 cm
Discussion
Explanation: Let the thickness of the bottom be x cm
$${\mkern 1mu} {\left( {330 - 10} \right) \times \left( {260 - 10} \right) \times \left( {110 - x} \right)} = $$ $$8000 \times $$ $$1000$$
$$ \Rightarrow 320 \times 250 \times \left( {110 - x} \right) = 8000 \times 1000$$
$$\eqalign{ & \left( {110 - x} \right) = \frac{{8000 \times 1000}}{{320 \times 250}} = 100 \cr & x = 10\,{\text{cm}} = 1\,{\text{dm}} \cr} $$
88. What is the total surface area of a right circular cone of height 14 cm and base radius 7 cm?
a) 344.35 cm2
b) 462 cm2
c) 498.35 cm2
d) None of these
Discussion
Explanation:
$$\eqalign{ & h = 14\,cm,\,r = 7\,cm \cr & {\text{So}},\,l = \sqrt {{{\left( 7 \right)}^2} + {{\left( {14} \right)}^2}} = \sqrt {245} = 7\sqrt 5 \,cm \cr & {\text{Total}}\,{\text{surface}}\,{\text{area}} \cr & = \pi \,rl + \pi \,{r^2} \cr & = \left( {\frac{{22}}{7} \times 7 \times 7\sqrt 5 + \frac{{22}}{7} \times 7 \times 7} \right)c{m^2} \cr & = \left[ {154\left( {\sqrt 5 + 1} \right)} \right]c{m^2} \cr & = \left( {154 \times 3.236} \right)c{m^2} \cr & = 498.35\,c{m^2} \cr} $$
89. A large cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. What is the ratio of the total surface areas of the smaller cubes and the large cube?
a) 2 : 1
b) 3 : 2
c) 25 : 18
d) 27 : 20
Discussion
Explanation:
$$\eqalign{ & {\text{Volume}}\,{\text{of}}\,{\text{the}}\,{\text{large}}\,{\text{cube}} \cr & = \left( {{3^3} + {4^3} + {5^3}} \right) = 216\,c{m^3} \cr & {\text{Let}}\,{\text{the}}\,{\text{edge}}\,{\text{of}}\,{\text{the}}\,{\text{large}}\,{\text{cube}}\,{\text{be}}\,a \cr & So,\,{a^3} = 216\,\,\,\,\, \Rightarrow \,\,\,\,\,a = 6\,cm \cr & {\text{Required}}\,{\text{ratio}} = {\frac{{6 \times \left( {{3^2} + {4^2} + {5^2}} \right)}}{{6 \times {6^2}}}} \cr & = \frac{{50}}{{36}} \cr & = 25:18 \cr} $$
90. How many bricks, each measuring 25 cm x 11.25 cm x 6 cm, will be needed to build a wall of 8 m x 6 m x 22.5 cm?
a) 5600
b) 6000
c) 6400
d) 7200
Discussion
Explanation:
$$\eqalign{ & {\text{Number}}\,{\text{of}}\,{\text{bricks}} = \frac{{{\text{Volume}}\,{\text{of}}\,{\text{the}}\,{\text{wall}}}}{{{\text{Volume}}\,{\text{of}}\,{\text{1}}\,{\text{brick}}}} \cr & = {\frac{{800 \times 600 \times 22.5}}{{25 \times 11.25 \times 6}}} \cr & = 6400 \cr} $$
91.A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. The volume of the cone so formed is:
a) 12$$\pi$$ cm3
b) 15$$\pi$$ cm3
c) 16$$\pi$$ cm3
d) 20$$\pi$$ cm3
Discussion
Explanation:
$$\eqalign{ & {\text{We have }}r = 3\,{\text{cm}}\,{\text{and}}\,h = 4\,{\text{cm}} \cr & {\text{Volume}} \cr & = \frac{1}{3}\pi {r^2}h \cr & = \left( {\frac{1}{3} \times \pi \times {3^2} \times 4} \right){\text{c}}{{\text{m}}^3} \cr & = 12\pi \,{\text{c}}{{\text{m}}^3} \cr} $$
92. In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:
a) 75 cu. m
b) 750 cu. m
c) 7500 cu. m
d) 75000 cu. m
Discussion
Explanation:
$$\eqalign{ & {\text{1}}\,{\text{hectare}} = 10000\,{m^2} \cr & {\text{So,}}\,{\text{Area}} = \left( {1.5 \times 10000} \right){m^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 15000\,{m^2} \cr & {\text{Depth}} = \frac{5}{{100}}m = \frac{1}{{20}}m \cr & {\text{Volume}} = \left( {{\text{Area}}\, \times \,{\text{Depth}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {15000 \times \frac{1}{{20}}} \right){m^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 750\,{m^3} \cr} $$
93. A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:
a) 720 m3
b) 900 m3
c) 1200 m3
d) 1800 m3
Discussion
Explanation:
$$\eqalign{ & 2\left( {15 + 12} \right) \times h = 2\left( {15 \times 12} \right) \cr & \Rightarrow h = \frac{{180}}{{27}}m = \frac{{20}}{3}m \cr & {\text{Volume}} = \left( {15 \times 12 \times \frac{{20}}{3}} \right){m^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1200\,{m^3} \cr} $$
94. 66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be:
a) 84 meters
b) 90 meters
c) 168 meters
d) 336 meters
Discussion
Explanation:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{wire}}\,{\text{be h}} \cr & {\text{Radius}} = \frac{1}{2}mm = \frac{1}{{20}}cm.\,{\text{Then}}, \cr & \frac{{22}}{7} \times \frac{1}{{20}} \times \frac{1}{{20}} \times h = 66 \cr & h = {\frac{{66 \times 20 \times 20 \times 7}}{{22}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8400\,cm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 84\,meters \cr} $$
95. A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weights 8 g/cm3, then the weight of the pipe is:
a) 3.6 kg
b) 3.696 kg
c) 36 kg
d) 36.9 kg
Discussion
Explanation:
$$\eqalign{ & {\text{External}}\,{\text{radius}} = 4\,cm \cr & {\text{Internal}}\,{\text{radius}} = 3\,cm \cr & {\text{Volume}}\,{\text{of}}\,{\text{iron}} \cr & = \left( {\frac{{22}}{7} \times \left[ {{{\left( 4 \right)}^2} - {{\left( 3 \right)}^2}} \right] \times 21} \right)c{m^3} \cr & = \left( {\frac{{22}}{7} \times 7 \times 1 \times 21} \right)c{m^3} \cr & = 462\,c{m^3} \cr & {\text{Weight}}\,{\text{of}}\,{\text{iron}} = \left( {462 \times 8} \right)gm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3696\,gm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3.696\,kg \cr} $$
96. The diameter of the base of a cylindrical drum is 35 dm and the height is 24 dm. It is full of kerosene. How many tins each of size 25 cm × 22 cm × 35 cm can be filled with kerosene from the drum ?
a) 120
b) 600
c) 1020
d) 1200
Discussion
Explanation:
$$\eqalign{ & {\text{Number of tins}} = \frac{{{\text{Voulme of the drum}}}}{{{\text{Volume of each tin}}}} \cr & = \frac{{\left( {\frac{{22}}{7} \times \frac{{35}}{2} \times \frac{{35}}{2} \times 24} \right)}}{{\left( {\frac{{25}}{{10}} \times \frac{{22}}{{10}} \times \frac{{35}}{{10}}} \right)}} \cr & = 1200 \cr} $$
97. The radius of a cylindrical cistern is 10 metres and its height is 15 metres. Initially the cistern is empty. We start filling the cistern with water through a pipe whose diameter is 50 cm. Water is coming out of the pipe with a velocity of 5 m/sec. How many minutes will it take in filling the cistern with water ?
a) 20 min
b) 40 min
c) 60 min
d) 80 min
Discussion
Explanation: Volume of cistern :
$$\eqalign{ & = \left( {\pi \times {{10}^2} \times 15} \right){m^3} \cr & = 1500\pi {m^3} \cr} $$
Volume of water flowing through the pipe in 1 sec :
$$\eqalign{ & = \left( {\pi \times 0.25 \times 0.25 \times 5} \right){m^3} \cr & = 0.3125\pi {m^3} \cr} $$
Time taken to fill the cistern :
$$\eqalign{ & = \left( {\frac{{1500\pi }}{{0.3125\pi }}} \right) \cr & = \left( {\frac{{1500 \times 10000}}{{3125}}} \right) \cr & = 4800\,\sec \cr & = \left( {\frac{{4800}}{{60}}} \right)\min \cr & = 80\,\min \cr} $$
98. What length of solid cylinder 2 cm in diameter must be taken to cast into a hollow cylinder of external diameter 12 cm, 0.25 cm thick and 15 cm long ?
a) 42.3215 cm
b) 44.0123 cm
c) 44.0625 cm
d) 44.6023 cm
Discussion
Explanation: External radius = 6 cm
Internal radius = (6 - 0.25) = 5.75 cm
Volume of material in hollow cylinder :
$$\eqalign{ & = \left[ {\frac{{22}}{7} \times \left\{ {{{\left( 6 \right)}^2} - {{\left( {5.75} \right)}^2}} \right\} \times 15} \right]{\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{22}}{7} \times 11.75 \times 0.25 \times 15} \right){\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{22}}{7} \times \frac{{1175}}{{100}} \times \frac{{25}}{{100}} \times 15} \right){\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{11 \times 705}}{{56}}} \right){\text{c}}{{\text{m}}^3} \cr} $$
Let the length of solid cylinder be h
$$\eqalign{ & \frac{{22}}{7} \times 1 \times 1 \times h = \left( {\frac{{11 \times 705}}{{56}}} \right) \cr & h = \left( {\frac{{11 \times 705}}{{56}} \times \frac{7}{{22}}} \right) \cr & h = 44.0625\,{\text{cm}} \cr} $$
99. If the height of a right circular cone is increased by 200% and the radius of the base is reduced by 50%, then the volume of the cone :
a) remains unaltered
b) decreases by 25%
c) increases by 25%
d) increases by 50%
Discussion
Explanation: Let the original radius and height of the cone be r and h respectively
Then, Original volume = $$\frac{1}{3}\pi {r^2}h$$
New radius = $$\frac{r}{2}$$ and new hight = 2h
New volume :
$$\eqalign{ & = \frac{1}{3} \times \pi \times {\left( {\frac{r}{2}} \right)^2} \times 3h \cr & = \frac{3}{4} \times \frac{1}{3}\pi {r^2}h \cr} $$
Decrease % :
$$\eqalign{ & = \left( {\frac{{\frac{1}{4} \times \frac{1}{3}\pi {r^2}h}}{{\frac{1}{3}\pi {r^2}h}} \times 100} \right)\% \cr & = 25\% \cr} $$
100. A bucket is in the from of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket.
a) 15 cm
b) 20 cm
c) 25 cm
d) 30 cm
Discussion
Explanation: Volume of bucket :
= 28.490 litres
= (28.490 ×1000) cm3
= 28490 cm3
Let the height of the bucket be h cm
We have : r = 21 cm. and R = 28 cm
$$ \frac{\pi }{3}h\left[ {{{\left( {28} \right)}^2} + {{\left( {21} \right)}^2} + 28 \times 21} \right]$$ $$ = 28490$$
$$ h\left( {784 + 441 + 588} \right) = $$ $$\frac{{28490 \times 21}}{{22}}$$
$$\eqalign{ & 1813h = 27195 \cr & h = \frac{{27195}}{{1813}} \cr & h = 15\,cm \cr} $$