Modified Table 
[Minumum class-]

(ii) The class interval in which the maximum number of surname lie is 6-8.

(i)   2r = 3.5 m
\(\Rightarrow\)  \(r=\frac { 3.5 }{ 2 } m\)
\(\Rightarrow\)   r = 1.75 m
h = 10 m
\(\therefore \) Inner curved surface area of the circular well = \(2\pi rh\)
\(=2\times \frac { 22 }{ 7 } \times 1.75\times 10=110{ m }^{ 2 }.\)
(ii) Cost of plastering the curved surface at the rate of Rs 40 per m² = Rs 110 \(\times\) 40 = Rs 4400.   

a = 18 cm, b = 10 cm
Perimeter = 42 cm
\(\Rightarrow a+b+c=42\)
\(\\ \Rightarrow 18+10+c=42\)
\(\\ \Rightarrow 28+c=42\)
\(\\ \Rightarrow c=42-28\)
\(\Rightarrow \ c=14\quad \)cm
\(s=\frac { 42 }{ 2 } =21\)cm
\(\therefore \)   Area of the triangle \(=\sqrt { s(s-a)(s-b)(s-c) } \quad \quad \)
\(=\sqrt { 21(21-18)(21-10)(21-14) } \)
\(\\ =\sqrt { 21(3)(11)(7) }\)
\( \\ =\sqrt { (7)(3)(3)(11)(7) } \)
\(=(7)(3)\sqrt { 11 } =21\sqrt { 11 } \)cm²

Let BD = x m

Then in right triangle ODB,
OB² = OD² + BD²
By Pythagoras Theorem
⇒ (20)² = OD² + x²
⇒ OD²=  400 - x²⇒ OD =\(\sqrt { 400-x^{ 2 } } \)
Again, area of equilateral triangle ABC
= Area of \(\Delta OBC\) + Area of \(\Delta OCA\) + Area of \(\Delta OAB\)
= 3 Area of \(\Delta OBC\)=  3\(\frac { \left( BC \right) \left( OD \right) }{ 2 } \)
\(=3x\sqrt { 400-x^{ 2 } } \)                       ....(2)
⇒  \(\sqrt { 3 } \sqrt { 400-x^{ 2 } } =x\)
Squaring both sides,
3(400 - x²) = x²
⇒ 1200 - 3x² = x²
⇒  4x² = 1200   ⇒ x² = 300
⇒ \(x=10\sqrt { 3 } \)  ⇒ BD=\(10\sqrt { 3 } \)
⇒  \(2BD=20\sqrt { 3 } \)  ⇒ \(BC=20\sqrt { 3 } \)
Hence, the length of string of each phone is \(20\sqrt { 3 } \) m.

The given equation is x-2y = 4
(2,0)
Put x = 2 and y = 0 in (1), we get
x - 2y = 2 - 2(0) = 2 - 0 = 2, which is not 4.
(2,0) is not a solution of (1)

\(35y^2+13y-12\)
\(=35y^2+28y-15y-12\)
\(=7y(5y+4)-3(5y+4)\)
\(=(5y+4)(7y-3)\)
\(\because\)  The possible expressions for the length and breadth of the rectangle are 7y-3 and 5y+4

0.01001 0001 00001...
0.20 2002 20002 200002...
0.003000300003...

Let x = 1.272727... Since two digits are repeating, we multiply x by 100 to get 
100 x = 127.2727
So, 100 x = 126 + 1.272727 = 126 + x
Therefore, 100 x – x = 126, i.e., 99 x = 126
\(x=\frac{126}{99}=\frac{14}{11}\)
You can check the reverse that \(\frac{14}{11}=1 . \overline{27}\)

Arranging the points scored by the team in ascending order, we get
2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48.
There are 16 terms. So there are two middle terms, i.e. the \(\frac{16}{2} \text { th and }\left(\frac{16}{2}+1\right) \text { th, i.e., }\) the 8th and 9th terms.
So, the median is the mean of the values of the 8th and 9th terms.
i.e, the median = \(\frac{10+14}{2}=12\)
So, the medial point scored by the Kabaddi team is 12.

Curved surface area = \(\pi\)rl
\(=\frac{22}{7} \times 7 \times 10 \mathrm{~cm}^{2}\) 
= 220 cm²

Here we have perimeter of the triangle = 32 cm, a = 8 cm and b = 11 cm.
Third side c = 32 cm – (8 + 11) cm = 13 cm
So, 2s = 32, i.e., s = 16 cm,
s – a = (16 – 8) cm = 8 cm,
s – b = (16 – 11) cm = 5 cm,
s – c = (16 – 13) cm = 3 cm.
Therefore, area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{16 \times 8 \times 5 \times 3} \mathrm{~cm}^{2}=8 \sqrt{30} \mathrm{~cm}^{2}\)

(i) We have
(104)³ = (100 + 4)³
= (100)³ + (4)³ + 3(100)(4)(100 + 4)
(Using Identity VI)
= 1000000 + 64 + 124800
= 1124864
(ii) We have
(999)³ = (1000 – 1)³
= (1000)³ – (1)³ – 3(1000)(1)(1000 – 1)
(Using Identity VII)
= 1000000000 – 1 – 2997000
= 997002999

So, we need to make certain modifications in the lengths of the rectangles so that the areas are again proportional to the frequencies.
The steps to be followed are as given below:
1. Select a class interval with the minimum class size. In the example above, the minimum class-size is 10.
2. The lengths of the rectangles are then modified to be proportionate to the class-size 10.
For instance, when the class-size is 20, the length of the rectangle is 7. So when the class-size is 10, the length of the rectangle will be \(\frac{7}{20} \times 10=3.5\)
Similarly, proceeding in this manner, we get the following table:

Since we have calculated these lengths for an interval of 10 marks in each case, we may call these lengths as “proportion of students per 10 marks interval”.
So, the correct histogram with varying width is given in fig

Once again we proceed by successive magnification, and successively decrease the lengths of the portions of the number line in which \(5.3 \overline{7}\) is located. First, we see that \(5.3 \overline{7}\) is located between 5 and 6. In the next step, we locate 5.37 between 5.3 and 5.4. To get a more accurate visualization of the representation, we divide this portion of the number line into 10 equal parts and use a magnifying glass to visualize that \(5.3 \overline{7}\) lies between 5.37 and 5.38. To visualize \(5.3 \overline{7}\) more accurately, we again divide the portion between \(5.3 \overline{7}\) and 5.38 into ten equal parts and use a magnifying glass to visualize that \(5.3 \overline{7}\) lies between 5.377 and 5.378. Now to visualize 5.37 still more accurately, we divide the portion between 5.377 an 5.378 into 10 equal parts, and visualize the representation of 5.37 as in Fig. 1.14 (iv). Notice that \(5.3 \overline{7}\) is located closer to 5.3778 than to 5.3777 [see Fig 1.14 (iv)].

Let p(x) = x³ – 23x² + 142x – 120
We shall now look for all the factors of –120. Some of these are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60.
By trial, we find that p(1) = 0. So x – 1 is a factor of p(x).
Now we see that x³ – 23x² + 142x – 120 = x³ – x² – 22x² + 22x + 120x – 120
= x²(x –1) – 22x(x – 1) + 120(x – 1) 
= (x – 1) (x² – 22x + 120) [Taking (x – 1) common]
We could have also got this by dividing p(x) by x – 1.
Now x² – 22x + 120 can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
x² – 22x + 120 = x² – 12x – 10x + 120
= x(x – 12) – 10(x – 12)
= (x – 12) (x – 10)
So, x³ – 23x² – 142x – 120 = (x – 1)(x – 10)(x – 12)

Let given sides of the wall be a = 15m, b = 11m and c = 6 m.
ஃ  Semi-perimeter, \(s=\frac{a+b+c}{2}=\frac{15+11+6}{2}=\frac{32}{2}=16m\)
Now, area of the wall= \(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{16(16-15)(16-11)(16-6)}\)
\(=\sqrt{16\times1\times5\times10}=\sqrt{4\times4\times5\times5\times2}=5\times4\sqrt2\)
\(=20\sqrt{2}m^2\)
Hence, the area painted in colour is \(=20\sqrt{2}m^2\)