The sum of first n odd natural numbers is n².
e.g. 1 + 3 + 5 + 7 = 16 = 4²
1 + 3 + 5 = 9 = 3²
\(\Rightarrow\) 1 + 3 = 4 = 2²

Given, number is 150.
Now, prime factorisation of 150
150 = 2 x 3 x 5 x 5
The prime factors of 150 do not appear in group of triples.
Hence, 150 is not a perfect cube.

We know that, the numbers end with 2, 3, 7 or 8, can never be a perfect square.
In a number 408, end digit is 8, which is one of 2, 3, 7 or 8. So, it never be a perfect square.

We have, 77
One's digit of 77 = 7
Now, cube of one's digit of 77 = (7)³= 7 x 7 x 7=343
Hence,one's digit in the cube of 77 is 3.

The quadrilaterals whose diagonals are perpendicular bisectors of each other can be a rhombus or a square.

= \(\frac { 2 }{ 5 } \times \left( \frac { -3 }{ 7 } \right) -\frac { 1 }{ 4 } +\frac { 1 }{ 14 } \times \frac { 2 }{ 5 } \)
= \(\frac { 2 }{ 5 } \times \left( \frac { -3 }{ 7 } \right) +\frac { 1 }{ 14 } \times \frac { 2 }{ 5 } -\frac { 1 }{ 4 } \) (Using commutativity)
= \(\frac { 2 }{ 5 } \left[ \frac { -3 }{ 7 } +\frac { 1 }{ 14 } \right] -\frac { 1 }{ 4 } \) (using distributivity)
= \(\frac { 2 }{ 5 } \left[ \frac { -6+1 }{ 14 } \right] -\frac { 1 }{ 4 } =\frac { 2 }{ 5 } \times \frac { -5 }{ 14 } -\frac { 1 }{ 4 } \)
= \(\frac { -1 }{ 7 } -\frac { 1 }{ 4 } =\frac { -4-7 }{ 28 } =\frac { -11 }{ 28 } \)

Firstly,find the measure of exterior angle by subtracting measureof interior angle from 180°.

x=36⁰, 36⁰, 72⁰,108⁰,144⁰

We have 6 = z + 2 
\(\Rightarrow\) 6 - 2 = z  [transposing 2 from RHS to LHS]
\(\Rightarrow\) 4 = z \(\Rightarrow\) z = 4, Which is the Required solution

7 x 7 = 49

2 x 12 = 24

Perfect square numbers end with 0, 1, 4, 5, 6 or 9 at unit's place

Central angle for sector A
= \(\frac{30}{100}\)\(\times\) 360° =  360°.

-\(\frac{2}{3}\)=2x          \(\Rightarrow\)x =-\(\frac{1}{3}\)

10000 = 2 \(\times\) 2\(\times\) 2 \(\times\) 2 \(\times\) 5 \(\times\) 5 \(\times\) 5 \(\times\) 5
= 23 \(\times\) 2 \(\times\) 5³ \(\times\) 5.

300 + 400 + 200 + 200 + 100 = 1200

Hindi ⟶ 400

4\(\times\)4\(\times\)4=64

10

Perimeter of a square = 4 \(\times\)Side
\(\Rightarrow \ 4\times Side=96\ m\Rightarrow \ Side=\frac { 96 }{ 4 } =24\ m\)
\(\therefore \ Area\ od\ a\ square={ (Side) }^{ 2 }=Side\times Side\)
= 24 \(\times\)24 = 576 m²

We have, a parallelogram WISH.
Here, WH Il lS and IW II HS
∴ ㄥWIH = ㄥIHS = 35°  [alternate angle]
Similarly,  ㄥHIS = ㄥIHW = 25° [alternate angle]
Now, by using angle sum property of a triangle in ΔWOH, we get
ㄥWOH + ㄥOWH + ㄥOHW = 180°
⇒ 110° + ㄥOWH + 25° =180°
[∵ ㄥOHW = ㄥIHW]
⇒  ㄥOWH =180° - (110° + 25°)
= 180° -135° = 45° = ㄥSWH
In  ΔOHS, we have
ㄥHOS = 180° -110° = 70°
ㄥOHS = 35°     [proved]
∴ 70° + 35° + ㄥOSH =180°
⇒ ㄥOSH =180° -105° =75°.

First, we draw a number line and mark a point 0 on it to represent zero. Now, find the points A and B on the number line representing the positive integers 5 and - 5, respectively (as shown below).
Now, divide the segments OA into three equal parts such that AP = PQ = QO.
By construction
 
Q is one third of OA. Therefore, Q represents the rational number  \({5\over3}\)
Further, B represents -5 on the number line. Now, divide OB into three equal parts, such that BS = SR = RO and OR represents one third distance of OB. Therefore, R represents the rational number - \({5\over3}\)

We know that,
Central angle of a component = \((\frac{Component\quad value}{sum\quad of \quad the\quad component\quad values}\times 360)^o\)
Here, total amount = Rs 7200,
 Cntral angle for an item = \((\frac{Amount\quad spent\quad on\quad the\quad item}{Total\quad amount}\times 360)^o\)
The central angles of the sectors representing different items are computed in the following table:

Now, to construct the pie chart representing the above data, we follow the following steps:
Step I Draw a circle of an appropriate radius.
Step II Draw a vertical radius of the circle drawn in Step I.
Step III Choose the largest central angle. Here, largest central angle is of 200°. Draw a sector with central angle 200° in such a way that its one radius coincides with the radius drawn in Step II and another radius is in its clockwise direction.
Step IV Construct other sectors representing other items in clockwise sense in descending order of magnitudes of their central angles except the sector representing miscellaneous expenses. This sector is to be drawn in the last.
Step V Shade the sectors, so obtained by different designs and label them as shown in figure given below to obtain the required pie chart.

4

100

Central angle for oceans = \((\frac{Value\quad of\quad the\quad component}{Sum\quad of\quad the\quad components}\times 360)^o\)
For pacific ocean = 40% = \(\frac{40}{100}\times 360^o\) = 144^o
For atlantic ocean = 30% = \(\frac{30}{100}\times 360^o\) = 108^o
For Indian ocean = 20% = \(\frac{20}{100}\times 360^o\) = 72^o
For others ocean = 10% = \(\frac{10}{100}\times 360^o\) = 36^o
On the basis of above data we can draw the following pie chart:

246016

We have, two parallelograms ABDH and EFGC.
ㄥABD = ㄥAHD = 130°
[opposite angles of a parallelogram]
ㄥGHD = 180° - ㄥAHD = 180° - 130°
⇒ 50⁰= ㄥGHO
Also, ㄥEFG + ㄥFGC = 180°
[adjacent angles of a parallelogram]
⇒ 30° + ㄥFGC = 180°
⇒ ㄥFGC =180° - 30° =150°
∴ ㄥHGC =180° - ㄥFGC =180° -150°
   =30⁰ = ㄥHGO
Now, in ΔHGO, by using angle sum property,
ㄥOHG + ㄥHGO + ㄥHOG =180°
⇒ 50°+ 30°+ x⁰ =180°
⇒ x⁰ =180° - 80° = 100°

Let the required number be x.
Then, according to the question
x³ = x ...(1) and x²≠x ...(2)
From (1), x³ -x = 0
⇒ x(x² - 1) = 0
⇒ x = 0, ± 1
⇒ x = 0,1,-1
If x = 0, then x² = x.
∴ x = is inadmissible
If x = 1 then x² = x.
∴ x = 1 is inadmissible
If x = - 1, then x² = (- 1)2 = 1 ≠ x(= - 1)
Hence, the required number is - 1.

Let the number of plates on the table be y and the number of puris be x.
Then, according to the first condition of the problem
4(y - 1) = x ...(1)
and, according to the second condition of the
problem
3y + 1= x    ... (2)
From (1) and (2), we have
4(y - 1) = 3y + 1
⇒ 4y - 4 = 3y + 1
⇒ 4y - 3y = 1 + 4
⇒ y=5
Put y = 5 in (1), we get
x = 4(5) - 4 = 20 - 4 = 16
Hence, the number of plates and number of puris on the table are 5 and 16 respectively.

The highest observation = 62
The lowest observation = 17
One of the class intervals = 40-48
∴ Class size = Upper class limit - Lower class limit
= 48 - 40 = 8
∴ The appropriate classes can be:
16-24, 24-32, 32-40, 40-48, 48-56, 56-64
Thus, the frequency distribution table for the above data can be shown using the Tally marks

Groups [Class intervals]	Tally marks	Frequency
16-24	||	7
24-32	||||	4
32-40	|	11
40-48	||	2
48-56	||||	9
56-64	||	7
Total		40