Olympiad Mathematics - Surface Areas and Volumes
Exam Duration: 45 Mins Total Questions : 30
The total surface area of a cylinder of height 6.5 cm is 220 sq cm. Find its volume.
- (a)
25.025 cm3
- (b)
2.5025 cm3
- (c)
2502.5 cm3
- (d)
250.25 cm3
What is the length of the longest pole that can be put in a room of dimensions 10 m x 10 m x 5 m?
- (a)
15 m
- (b)
16 m
- (c)
10 m
- (d)
12 m
Dimensions of a room = 10 m x 10 m x 5 m
Length of the diagonal \(=\sqrt {(10)^2+(10)^2+(5)^2}=15\ m\)
\(\therefore\) The length of the longest pole that can be put III a room = 15 m
How many litres of water flows out through a pipe having an area of cross section of 5 cm2 in one minute, if the speed of water in pipe is 30 cm/sec?
- (a)
9 litres
- (b)
15 litres
- (c)
30 litres
- (d)
3 litres
The volumes of two spheres are in the ratio 64: 27. Find the difference of their surface areas,if the sum of their radii is 7 units
- (a)
28\(\pi\) sq. units
- (b)
64\(\pi\) sq. units
- (c)
88\(\pi\) sq. units
- (d)
4\(\pi\) sq. units
Into a conical tent of radius 7 m and vertical height 4.5 rn, how many full bags of rice can be emptied, if volume of each bag is 1.5 m3
- (a)
120 bags
- (b)
144 bags
- (c)
154 bags
- (d)
172 bags
Find the curved surface of a right circular cone, whose slant height and the base radius are 25 cm and 7 cm respectively.
- (a)
420 cm2
- (b)
550 cm2
- (c)
460 cm2
- (d)
580 cm2
Slant height l = 25 cm
Radius of the base of the cone, r = 7 cm
C.S.A cone = \(\pi\)rl
=\(\frac { 22 }{ 7 } \)\(\times\)7\(\times\)25cm2 = 550cm2
A right circular cylinder has a height of 21 cm and base radius of 5 cm. Find the curved surface area of the cylinder.
- (a)
230 cm2
- (b)
660 cm2
- (c)
550 cm2
- (d)
450 cm2
C.S.A = 2\(\pi\)rh = 660 cm2
Curved surface area = 660 cm2
The area of the curved surface of a right circular cylinder is 4400 cm2 and the circumference of its base is 110 cm. Find the height of the cylinder
- (a)
30 cm
- (b)
10 cm
- (c)
20 cm
- (d)
40 cm
If A1 A2 and A3 denote the areas of three adjacent faces of a cuboid, find its volume
- (a)
A1 A2 A3
- (b)
2A1 A2 A3
- (c)
\(\sqrt { { A }_{ 1 }{ A }_{ 2 }{ A }_{ 3 } } \)
- (d)
\(\sqrt [ 3 ]{ { A }_{ 1 }{ A }_{ 2 }{ A }_{ 3 } } \)
If I is the length of a diagonal of a cube of volume V, what is the relation between I and V?
- (a)
3V = l3
- (b)
\(\sqrt { 3V } \)=l3
- (c)
3\(\sqrt { 3 } \)V=2l3
- (d)
3\(\sqrt { 3 } \)V=l3
If each edge of a cube, of volume, V, is doubled, find the volume of the new cube.
- (a)
2V
- (b)
4V
- (c)
6V
- (d)
8V
Volume of a cube = V
Edge of a cube is doubled, a = 2a
∴ Volume of new cube = a3 = 8V
Find the cost of constructing a wall 8 m long, 4 m high and 20 cm thick at the rate of Rs.25 per m3.
- (a)
Rs.16
- (b)
Rs.80
- (c)
Rs.160
- (d)
Rs.320
The dimensions of wall = 8 m\(\times\)4
m\(\times\)20 cm = 8 m\(\times\)4 m\(\times\)0.2 m
Volume of the wall = (840.2)m3 = =6.4m3
∴ Cost of constructing a wall at a rate
of Rs.25 per m3 = 6.4\(\times\)Rs.25=\(\times\) 60
The largest sphere is carved out of a cube of side 7 cm. Find the volume of the sphere. (Take \(\pi \) = 3.14.)
- (a)
152.74 cm3
- (b)
243.41 cm3
- (c)
179.67 cm3
- (d)
195.01 cm3
The radius of a wire is decreased to its one-third. If its volume remains the same, by how many times will its length increase
- (a)
3 times
- (b)
6 times
- (c)
9 times
- (d)
27 times
If the lateral surface area of a cube is 1600 cm2 what is its edge?
- (a)
15 cm
- (b)
18 cm
- (c)
20 cm
- (d)
25 cm
If a square card board piece of side 4 cm is rotated 360o about one of its sides, what is the volume of the solid so formed?
- (a)
64\(\pi \)cm3
- (b)
16\(\pi \)cm3
- (c)
32\(\pi \)cm3
- (d)
128\(\pi \)cm3
The radius of a cylinder is doubled and its height is halved. What is the change in its curved surface area?
- (a)
Halved
- (b)
Doubled
- (c)
Remains the same
- (d)
Becomes four times
A cube of edge 'k' is divided into 'n' equal cubes. Determine the edge of the new cube.
- (a)
\(\sqrt { n } k\)
- (b)
\(\frac { k }{ \sqrt [ 3 ]{ n } } \)
- (c)
\(\sqrt [ 3 ]{ nk } \)
- (d)
\(\frac { \sqrt [ 3 ]{ n } }{ k } \)
Edge of big cube = k units
Let the edge of small cube be 'a' units
⇒ Volume of each small cube
= a3 cu.units
⇒ Volume of big cube = k3
Given there are 'n' small cubes
⇒ k3 =n .a3 ⇒ a3 = \(\frac { { k }^{ 3 } }{ n } \)⇒ a= \(\frac { k }{ \sqrt [ 3 ]{ n } } \)
∴ Length of the edge of the new cube is \(\frac { k }{ \sqrt [ 3 ]{ n } } \)
If the height of a cylinder is doubled, by what number must the radius of the base be multiplied so that the resulting cylinder has the same volume as the original cylinder?
- (a)
4
- (b)
\(\frac{1}{\sqrt{2}}\)
- (c)
2
- (d)
\(1 \over 2\)
Let radius and height of original cylinder be r1 and h1, respectively
∴ Volume of original cylinder = ㅠr21h1
Also, let radius of new cylinder be r2 and height of new cylinder
= 2x(height of original cylinder)
= 2xh1=2h1
∴ Volume of new cylinder = r22.2h1
According to question,
Volume of original cylinder = Volume of new cylinder
⇒ \(\pi r^{2}_{1}h_{1}=\pi r^{2}_{1}.2h_{1}\Rightarrow r^{2}{1}=2r^{2}_{2} \Rightarrow r_{2}= \frac{1}{\sqrt{2}}r{1}\)
Hence, radius of base of original cylinder must be multiplied by \(\frac{1}{\sqrt{2}}\) so that the new cylinder has same volume as original.
If the length of diagonal of a cube is \(\sqrt{12}\)cm, then the volume of the cube is
- (a)
8\(\sqrt{12}\) cm3
- (b)
8 cm3
- (c)
16\(\sqrt{2} cm^{3}\)
- (d)
16 cm3
Let 'a' be the side of cube.
Since length of diagonal = \(\sqrt{12}\)cm
∴ \(\sqrt{a^{2}+a^{2}+a^{2}}=\sqrt{12}\)
Squaring both sides, we get
3a2=12 ⇒ a2=4 ⇒ a=2
∴ Volume of the cube = \(2\times 2\times 2=8 cm^{3}\).
The volume of a cylinder of radius r is 1/4 of the volume of a rectangular box with a square base of side length x. If the cylinder and the box have equal heights, what is the value of r in terms of x?
- (a)
\(\frac{x^{2}}{2 \pi}\)
- (b)
\(\frac{x}{2 \sqrt{\pi}}\)
- (c)
\(\sqrt{2x} \over \pi\)
- (d)
\(x \over \sqrt{\pi}\)
Let the height of cylinder and rectangular box be h.
Volume of cylinder = πr2h
∴ Volume of rectangular box = \(x \times x \times \times h=x^{2}h\)
According to question,
Volume of cylinder = \(1 \over 4\)\( \times\) Volume of rectangular box
⇒ πr2h = \(1 \over 4\)\( \times\)x2h ⇒ \(r^{2} = \frac{x^{2}}{4\pi}\) or \(r=\frac{x}{2\sqrt{\pi}}\)
The edge of a cube is 20 cm. How many small cubes of edge 5 cm can be formed from this cube?
- (a)
4
- (b)
32
- (c)
64
- (d)
100
Let 'n' be the number of cubes which can be formed from the given cube.
Volume of big cube = Volume of n smaller cubes
⇒ \(20 \times 20 \times 20 = n \times 5 \times 5 \times 5\)
⇒ n = \(\frac{20\times 20 \times 20}{5 \times 5 \times 5}=4\times 4 \times 4=64\)
The volume of two spheres are in the ratio 64: 27. The difference of their surface areas, if the sum of their radii is 7 units, is __________.
- (a)
28π sq. units
- (b)
88 sq. units
- (c)
88π sq. units
- (d)
4π sq. units
Let r1 and r2 be radii of two spheres. According to question,
\(\frac{\frac{4}{3}\pi r^{3}_{1}}{\frac{4}{3}\pi r^{3}_{2}}=\frac{64}{27} \Rightarrow (\frac{r_{1}}{r_{2}})^{3}=\frac{64}{27}\Rightarrow \frac{r_{1}}{r_{2}}=\frac{4}{3}\) --- (i)
Given, r1 + r2 = 7
From (i) and (ii), we get r1 = 4 units, r2 = 3 units
∴ Required difference = \(4\pi r^{2}_{1}-4\pi r^{2}_{2}\)
=\(4 \pi (4^{2}-3^{2})=4 \pi \times 7=28 \pi\)sq. units.
The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio of 5 : 3. The ratio of their volumes is __________.
- (a)
10 : 17
- (b)
20 : 27
- (c)
17 : 27
- (d)
20 : 37
Let r1, r2 be the radius of two cylinders
ஃ \(\frac{r_{1}}{r_{2}}=\frac{2}{3}\)
Let h1 and h2 be height of two cylinders
∴ \(\frac{h_{1}}{h_{2}}=\frac{5}{3}\)
Now, \(\frac{V_{1}}{V_{2}}=\frac{\pi r^{2}_{1}h_{1}}{\pi r^{2}_{2}h_{2}}=(\frac{r_{1}}{r_{2}})^{2}.\frac{h_{1}}{h_{2}}=(\frac{2}{3})^{2}.\frac{5}{3}=\frac{4}{9}\times \frac{5}{3}=\frac{20}{27}\)
∴ Required ratio = 20 : 27.
A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere.If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the wooden toy.
- (a)
266.11 cm3
- (b)
301.12 cm3
- (c)
242.36 cm3
- (d)
278.34 cm3
Let the radius of the hemisphere and base of cone be r and the height of the conical part of the toy be h. Then,
r = OA = 4.2 cm, h = VO = VO' - OO' = (10.2 - 4.2) cm
⇒ h=6cm
∴ Volume of the wooden toy
= Volume of the conical part
Volume of the hem i-spherical part
= \((\frac{1}{3} \pi r^{2}h+\frac{2\pi}{3}r^{3})cm^{3}\)
=\([\frac{\pi r^{2}}{3}(h+2r)]cm^{3}\)
=\([\frac{1}{3} \times \frac{22}{7} \times 4.2 \times 4.2 \times (6+2\times 4.2)]cm^{3}\)
=\([\frac{1}{3} \times \frac{22}{7} \times 4.2 \times 4.2 \times 14.4]cm^{3} = 266.11 cm^{3}\)
Water flows in a tank 150 m\(\times\) 100 m at the base, through a pipe whose cross-section is 2 dm by 1.5 dm at the speed of 15 km per hour. In what time, will the water be 3 metres deep?
- (a)
50 hours
- (b)
150 hours
- (c)
100 hours
- (d)
200 hours
Suppose in x hours water will be 3 metres deep in the tank.
Volume of water in the tank in x hours
= \((150 \times 100 \times 3)m^{3}=45000 m^{3}\)
Area of the cross-section of the pipe
=\((\frac{2}{10}\times \frac{1.5}{10})m^{2}=\frac{3}{100}m^{2}\)
Volume of water that flows in the tank in \(\times\) hours
= (Area of cross-section of the pipe) \(\times\) (Speed of water) \(\times\) (Time)
= \((\frac{3}{100}\times 15000 \times x)m^{3}\) [∵ Speed = 15 km/hr = 15000 m/hr]
= (450x) m3
Since, the volume of water in the tank is equal to the volume of water that flows in the tank in x hours
∴ 450 \(\times\) = 45000 ⇒ x = 100 hours.
A circus tent is cylindrical to a height of 3 metres and conical above it. if its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.
- (a)
1996 m
- (b)
2096 m
- (c)
1947 m
- (d)
1800 m
For cylinderical part,
Radius (r) = \(105 \over 2\)m
Height (h) = 3 m
For conical part,
Slant height (I) = 53 m
Radius (r) = \(105 \over 2\)m
∴ Total curved surface area of tent = 2πrh + πrl
= \(\pi r(2h+l)= \frac{22}{7}\times \frac{105}{2}\times (6+53)=(11\times 15 \times 59)m^{2}\)
Hence, length of canvas
= \(\frac{Total \ curved \ surface \ area \ of\ tent =}{Width \ of \ cloth}=\frac{11 \times 15 \times 59}{5}=1947 m\)
Match the following.
Column-I | Column-II | |
---|---|---|
(P) | A cylinder of radius 3 cm is inscribed in a sphere of radius 5 cm, then volume of cylinder is _________. |
38.5 cm3 |
(Q) | A conical pit of top diameter 3.5 cm is 12 cm deep, the capacity of pit is _________. |
512 cm3 |
(R) | The length of a diagonal of a cube is 8\(8 \sqrt{3}\)cm, then volume of cube is ________. |
72π cm3 |
(S) | The capacity of a conical vessel with height 12 cm and slant height 13 cm is____. | 100π cm3 |
- (a)
(P)⟶(2); (Q)⟶(3); (R)⟶(4); (S)⟶(1)
- (b)
(P)⟶(1); (Q)⟶(3) ; (R)⟶(2) ; (S)⟶(4)
- (c)
(P)⟶(3); (Q)⟶(1) ; (R)⟶(2) ; (S)⟶(4)
- (d)
(P)⟶(4); (Q)⟶(1) ; (R)⟶(3) ; (S)⟶(2)
(P) Let height and radius of the cylinder be h r respectively.
Let R be the radius of sphere.
In ΔOAB
(OA)2 = (OB)2 + (AB)2
⇒ \(R^{2}=(\frac{h}{2})^{2}+r^{2}\)
⇒ \(R^{2}-r^{2}=\frac{h^{2}}{4} \Rightarrow h=2\sqrt{R^{2}-r^{2}}\)
=2\(\sqrt{(5)^{2}-(3)^{2}}=2\times 4=8cm\)
ஃ Volume of cylinder = \(\pi r^{2}h=\pi\times 3\times 3 \times 8 =72\pi \ cm^{3}\)
(Q) Radius of conical pit (r) = \(\frac{3.5}{2}cm\)
Depth of conical pit (h) = 12 cm
Volume of conical pit = \(\frac{1}{3}\pi r^{2}h\)
= \(\frac{1}{3}\times\frac{22}{7}\times\frac{3.5}{2}\times\frac{3.5}{2}\times12=38.5\ cm^{3}\)
(R) Length of diagonal of cube of side'a' = \(\sqrt{3}\)a
∴ \(\sqrt{3}\)a-8\(\sqrt{3}\) ⇒ a=8 cm
Volume of cube = (side)3 = 512 cm3
(S) Volume of conical vessel
=\(\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times \pi \times (r^{2}-h^{2})\times 12=\frac{1}{3}\times \pi \times ((13)^{2}-(12)^{2})\times 12\)
= \(\frac{1}{3}\times \pi \times 25 \times 12=100 \pi \ cm^{3}\)
Read the statement carefully and write 'T' for true and 'F' for false.
(i) Volume of a cylinder is three times the volume of a cone on the same base and of same height.
(ii) Volume of biggest sphere in cube of edge 6 cm is 36ㅠ cm3.
(iii) Cuboids and cubes are special forms of right prisms.
- (a)
(i) (ii) (iii) T F T - (b)
(i) (ii) (iii) T T T - (c)
(i) (ii) (iii) F T F - (d)
(i) (ii) (iii) F T T
The internal and external radii of a hollow hemispherical bowl are 15 cm and 16 cm respectively, find the cost of painting the bowl at the rate of 35 paise per cm2, if
(i) the area of the edge of the bowl is ignored.
(ii) the area of the edge of the bowl is taken into account.
- (a)
(i) (ii) Rs 1058.20 Rs 1092.30 - (b)
(i) (ii) Rs 1020.50 Rs 1045 - (c)
(i) (ii) Rs 1092.50 Rs 1058.20 - (d)
(i) (ii) Rs 1086.20 Rs 1095.2
Internal radius of the bowl (r) = 15 cm
External radius of the bowl (R) = 16 cm
(i) If area of edge is ignored, then Surface area of bowl = \(2\pi R^{2}+2 \pi r^{2}\)
=\(2\pi( R^{2}+ r^{2})=2\pi(16^{2}+15^{2})\)
=\(2 \times \frac{22}{7}\times 481=\frac{21164}{7} cm^{2}\)
Cost of painting 1 cm2 area of the bowl = Rs 0.35
Cost of painting \(\frac{21164}{7} \ cm^{2} \) area of the bowl
= \((0.35 \times \frac{21164}{7}cm^{2})\)=Rs 1058.20
(ii) If area of edge is counted, then
Surface area of bowl = \(2\pi R^{2}+2\pi r^{2}+\pi(R^{2}-r^{2})\)
=\(2\pi (R^{2}+r^{2})+\pi (R^{2}-r^{2})=\pi[2(16^{2}+15^{2})+(16^{2}-15^{2})]\)
=\(\pi(2\times 481 +31)=\frac{22}{7}\times 993=\frac{21846}{7}cm^{2}\)
Cost of painting area \(\frac{21846}{7} \ cm^{2}=\frac{21846}{7}\times 0.35 = Rs \ 1092.30\)