153.5-157.5 and 157.5-161.5

800 %

(i)

(ii) Political party A won the maximum number of seats.

(i) Surface area of the sphere = \(4\pi { r }^{ 2 }\)
(ii) For cylinder
Radius of the base = r
Height  = 2r
\(\therefore\) Curved surface area of the cylinder
\(=2\pi \left( r \right) \left( 2r \right) =4\pi { r }^{ 2 }\)
(iii) Ratio of the areas obtained in (i) and (ii)
 \(=\frac { Surface\ area\ of\ the\ sphere }{ Curved\ surface\ area\ of\ the\ cylinder } \)
\(\\ =\frac { 4\pi { r }^{ 2 } }{ 4\pi { r }^{ 2 } } =\frac { 1 }{ 1 } =1:1.\)

Let the diameter of the earth be 2r.
Then diameter of the moon \(=\frac { 1 }{ 4 } \left( 2r \right) =\frac { r }{ 2 } \)
\(\therefore\) Radius of the earth = \(\frac { 2r }{ 2 } =r\)
and,   Radius of the moon \(=\frac { 1 }{ 2 } \left( \frac { r }{ 2 } \right) =\frac { r }{ 4 } \)
\(\therefore\)  Surface area of the earth \(=4\pi { r }^{ 2 }\)
and,  Surface area of the moon
\(=4\pi { \left( \frac { r }{ 4 } \right) }^{ 2 }=\frac { 1 }{ 4 } \pi { r }^{ 2 }\)
\(\therefore\)  Ratio of their surface area
\(=\frac { Surface\ area\ of\ the\ moon }{ Surface\ area\ of\ the\ earth } \)
\(\\ =\frac { \frac { 1 }{ 4 } \pi { r }^{ 2 } }{ 4\pi { r }^{ 2 } } =\frac { 1 }{ 16 } =1:16.\)

(i) Slant height (l) = 14 cm 
Curved surface area = 308 cm²
\(\Rightarrow \pi rl=308\)
\(\\ \Rightarrow \frac { 22 }{ 7 } \times r\times 14=308\)
\(\\ \Rightarrow r=\frac { 308\times 7 }{ 22\times 14 } \)
\(\\ \Rightarrow r=7cm\)
Hence, the radius of the base is 7 cm.
(ii) Total surface area of the cone = \(\pi r\left( l+r \right) \)
\(=\frac { 22 }{ 7 } \times 7\times \left( 14+7 \right)\)
\( \\ =\frac { 22 }{ 7 } \times 7\times 21=462{ cm }^{ 2 }\)
Hence, the total surface area of the cone is 462 cm².

Yes! These postulates contain two undefined terms: Point and Line. Yes! These postulates are consistent because they deal with two different situations
(i) says that given two points A and B, there is a point C lying on the line in between them,
(ii) says that given A and B, we can take C not lying on the line through A and B. These 'postulates' do not follow from Euclid's postulates, however, they follow from Axiom 'Given two distinct lines, there is a unique line that passes through them.

(i) 251.2 cm²
(ii) 452.16 cm²

Here, \(OX=\frac{1}{2}XY, PX=\frac{1}{2}XZ\)
XY = 2(OX), XZ = 2(PX)
Also. OX = PX(Given)
XY = XZ
(because things which are double of the same things are equal to one another)

In the given figure, AC coincides with AB + BC.
Also, Euclid's axiom 4 says that things which coincide with one another, are equal to one another. So, it can be deduced that
AB + BC = AC

Let us first draw a histogram for this data and mark the mid-points of the tops of the rectangles as B, C, D, E, F, G, H, I, J, K, respectively. Here, the first class is 0-10. So, to find the class preceeding 0-10, we extend the horizontal axis in the negative direction and find the mid-point of the imaginary class-interval (–10) - 0. The first end point, i.e., B is joined to this mid-point with zero frequency on the negative direction of the horizontal axis. The point where this line segment meets the vertical axis is marked as A. Let L be the mid-point of the class succeeding the last class of the given data. Then OABCDEFGHIJKL is the frequency polygon, which is shown in Fig.

(i) 15
(ii) 30-35
(iii) 47.5

The volume of water the bowl can contain
\(=\frac{2}{3} \pi r^{3}\)
\(=\frac{2}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5 \mathrm{~cm}^{3}=89.8 \mathrm{~cm}^{3}\)

From l² = r² + h², we have
\(r=\sqrt{l^{2}-h^{2}}=\sqrt{28^{2}-21^{2}} \mathrm{~cm}=7 \sqrt{7} \mathrm{~cm}\) 
So, volume of the cone \(=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \times \frac{22}{7} \times 7 \sqrt{7} \times 7 \sqrt{7} \times 21 \mathrm{~cm}^{3}\)
= 7546 cm³

consider the classes 31 - 35 and 36 - 40.
The lower limit of 36 - 40 = 36
The upper limit of 31 - 35 = 35
The difference = 36 – 35 = 1
So, half the difference = 1/2 = 0.5
So the new class interval formed from 31 - 35 is (31 – 0.5) - (35 + 0.5), i.e., 30.5 - 35.5.
Similarly, the new class formed from the class 36 - 40 is (36 – 0.5) - (40 + 0.5), i.e., 35.5 - 40.5.
Continuing in the same manner, the continuous classes formed are:
30.5-35.5, 35.5-40.5, 40.5-45.5, 45.5-50.5, 50.5-55.5, 55.5-60.5, 60.5 - 65.5, 65.5 - 70.5, 70.5 - 75.5.
Now it is possible for us to include the weights of the new students in these classes. But, another problem crops up because 35.5 appears in both the classes 30.5 - 35.5 and 35.5 - 40.5.
If it is considered in both classes, it will be counted twice.
By convention, we consider 35.5 in the class 35.5 - 40.5 and not in 30.5 - 35.5. Similarly, 40.5 is considered in 40.5 - 45.5 and not in 35.5 - 40.5.
So, the new weights 35.5 kg and 40.5 kg would be included in 35.5 - 40.5 and 40.5 - 45.5, respectively. Now, with these assumptions, the new frequency distribution table will be as shown below:

(a) Diameter of a cone = 40 cm
∴ Radius of a cone (r) = \(\frac{40}{2}\) cm = 20 cm = \(\frac{20}{100}\)m
= 0.2m
Height of a cone (h) = 1 m
⇒ Slant height of a cone (I) = \(\sqrt{r^{2}+h^{2}}\)
l = \(\sqrt{(0.2)^{2}+(1)^{2}}\)
\(=\sqrt{0.04+1} \mathrm{~m}\)
\(=\sqrt{1.04} \mathrm{~m}\)
= 1.02 m
[∵ It is given that \(\sqrt{1.04}\) = 1.02]
Now, curved surface area of a cone = πrl
[∵ Base of the cone is hollow]
⇒ Curved surface area of 1 cone =  3.14 x 0.2 x 1.02m²
\(=\frac{314}{100} \times \frac{2}{10} \times \frac{102}{100}\) m²
Cost of Painting :
∵  The rate of painting is 12 per m²
∴ Total cost of 50 cones = 12 x \(\left[\frac{314}{10} \times \frac{102}{100}\right]\)
=  \( \frac{384336}{1000}\) = 384.34 (approx.)
(b) Surface Areas and Volumes [mensuration]
(c) (i) Care for the public convenience.
(ii) Betterment of environment.

(i) False
[If we mark a point O on the surface of a paper and using pencil and ruler, we can draw indefinite number of straight lines passing through O.]
(ii) False
[∵ In the following figure, there are many straight lines passing through 'P'. There are many lines, passing through 'Q'. But there is one and only one line which is passing through 'P' as well as 'Q'.]

(iii) True
[∵ The postulate 2 says that "A terminated line can be produced indefinitely."]
(iv) True
[∵ Superimposing the region of one circle on the other, we find them coinciding. So, their centres and boundaries coincide. Thus, their radii will coincide.]
(v) True
[∵ According to Euclid's axiom, things which are equal to the same thing are equal to one another.]

Let r and R be the radii of the smaller and larger spheres respectively, we have
\(r=\frac { 5 }{ 2 } cm\)
Volume of the smaller sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi { \left( \frac { 5 }{ 2 } \right) }^{ 3 }\)
\(=\frac { 4 }{ 3 } \times \pi \times \frac { 125 }{ 8 } { cm }^{ 3 }\)
Density of metal\(=\frac { mass }{ Valume } \)
\(=\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } g\quad { cm }^{ 3 }\)      ...........(i)
Volume of larger sphere = \(\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
Density of metal=\(\frac { mass }{ Volume } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)    ......(ii)
From (i) and (ii), we have
\(\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)
\(\Rightarrow \quad { R }^{ 3 }=\frac { 5920\times 125 }{ 740\times 8 } \)
= 125
\(\Rightarrow\) R = 5 cm.