\({7^2-1\over 2}=24, {7^2+1\over 2}=25\)

Fourth angle = 360° - (65° + 75° + 85°)
= 135°.

3x =\(\frac{20}{7}\)-x       \(\Rightarrow\)3x +x =\(\frac{20}{7}\)
\(\Rightarrow\)4x = \(\frac{20}{7}\)x      \(\Rightarrow\) x=\(\frac{20}{7\times4}\)=\(\frac{5}{7}\)

0\(\times\)0\(\times\)0=0

1\(\times\)1\(\times\)1=1.

The required arrangement is
19, 17, 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16, 20

We can write 0 as \(\frac { 0 }{ 10 } \).
We can write (-2) as \(\left( \frac { 20 }{ 10 } \right) \)
Thus the rational numbers between 0 and -2 are 
\(\frac { -1 }{ 10 } ,\frac { -2 }{ 10 } ,\frac { -3 }{ 10 } ,\frac { -4 }{ 10 } ,\frac { -5 }{ 10 } ,\frac { -6 }{ 10 } ,...,\frac { -19 }{ 10 } \)
We may take any four of these

36

The possible one's digits of the square root of the numbers.
In a given number 657666025, one's digit is 5.
We know that, square of5 gives out 5 in one's digit.
Hence, possible one's digit of the square root digit is 5.

We have, 8888
One's digit of 8888 = 8
Now, cube of the one's digit of 8888 = (8)³ = 8 x 8 x 8 =512
Hence,one's digit in the cube of 8888 is 2.

There are ten separate slips having 1 to 10 numbers (one number on one slip) out of which one slip can be chosen in 10ways.
So, total  number of outcomes = 10
Here, the numbers greater than 6 are 7, 8, 9 and 10. So, one slip having number greater than 6 can be chosen in 4 ways.
So, favourable outcomes = 4
\(\therefore\) Probability of getting a number greater than 6 = \(\frac{Favourable\ outcomes}{Total\ number\ of\ outcomes}=\frac{4}{10}=\frac{2}{5}\)

We know that, a convex quadrilateral have angle of measure less than 180° and no portions of their diagonals in their exteriors.
So, a rectangle is a convex quadrilateral because the measure of its each angle is less than 180° and both the diagonals lie in its interior.

The rational numbers 1 and (-1) are equal to their reciprocals, respectively

p = \(\frac { -4 }{ 3 } \)

(i) The proportion of the sector for hours:
in sleeping = \(Number\ of\ sleeping\ hours\over Whole\ day\)
\(={8\ hours\over 24\ hours}={1\over3}\)
in school =\({6\ hours\over 24\ hours}={1\over4}\)
in home work = \({4\ hours\over 24\ hours}={1\over 6}\)
in play = \({3\ hours\over 24\ hours}={1\over 8}\)
in others = \({3\ hours\over 24\ hours}={1\over 8}\)
(ii) ∵ Total angle at the centre of a circle
= 360°
i.e., Central angle for 1 (whole) = 360°
∴ Central angle for play
(\(1\over8\)th part of Whole) = \(1\over8\) x 360° = 45°
Similarly, central angle for home work
(\(1\over6\) th part of Whole) = \(1\over6\) x 360° = 60°

Volume of the cubical tank
=91125 m³
Let the length of each side of cubical tank be x m.
We know that, Volume of a cube
= (Side)³
SO, (x)³ =91125
x = \(\sqrt [ 3 ]{ 91125 } \)
x = \(\sqrt [ 3 ]{ \begin{matrix} \underline { 3\times 3\times 3 } \times \underline { 3\times 3\times 3 } \times \underline { 5\times 5\times 5 } \\ x=3\times 3\times 5 \end{matrix} } \)
x = 3 x 3 x 5
x = 45 m
Hence, the height of the tank will be 45 m.

(b) The value depicted here is that Aditya is conscious environment. He saves water by reserving rain water.

246016

(i) Yes, RE = OM
Given, the above rectangle, opposite sides are equal. .
(ii) Yes, ㄥMYO = ㄥRXE
Here, MY and RX are perpendicular to OE.
Since, ㄥRXO = 90° ⇒  ㄥRXE = 90⁰
and ㄥMYE = 90°  ⇒ ㄥMYO = 90°
(iii) Yes,  ㄥMOY = ㄥREX
Since, these are alternate interior angles.
∵ RE || OM and EO is a transversal.
∴ ㄥMOE = ㄥOER
⇒ ㄥMOY = ㄥREX
(iv) Yes, ΔMYO ≅ RXE
In ΔMYO and  ΔRXE, we see that
MO=RE                   [proved]
ㄥMOY = ㄥREX     [proved]
ㄥMYO = ㄥRXE     [proved]
∴ ΔMYO ≅ ΔRXE    [by AAS]
(v) Yes, MY = RX
Since, these are corresponding part of congruent triangles.

In a parallelogram, the opposite sides have same length.
Therefore, PQ = SR = 12 cm and QR = PS = 7 cm
So, Perimeter = PQ + QR + RS + SP
= 12 cm + 7 cm + 12 cm + 7 cm = 38 cm

\(\frac{-1}{2}+[\frac{3}{7}+(\frac{-4}{3})] = \frac{-1}{2}+(\frac{-19}{21})=\frac{-59}{42}\)
\([\frac{-1}{2}+\frac{3}{7}]+(\frac{-4}{3})=\frac{-1}{14}+(\frac{-4}{3})=\frac{-59}{42}\)
So, Yes ; \(\frac{-1}{2}+[\frac{3}{7}+(\frac{-4}{3})]=[\frac{-1}{2}+\frac{3}{7}]+(\frac{-4}{3})\)
i.e., the two sums are equal. 

In \(\Delta\) ABC and \(\Delta\) ADC
AB = AD
| One pair of consecutive sides BC = DC
| Other pairs of consecutive sides AC = AC | Common
\(\therefore\) \(\Delta\) ABC \(\cong \) \(\Delta\) ADC
| SSS Congruence Axiom

\(\therefore \angle BAC=\angle DAC\) | CPCT
\(\angle BCA=\angle DCA\) | CPCT
i.e., diagonal AC bisects \(\angle BAD\) and \(\angle BCD\) each.

Using the closure property over addition, subtraction, multiplication and division for rational numbers, integers, whole-numbers and natural numbers, we have:

Given, number is 614125.
So, groups of 614125 are

In the first group, the number 125 ends with 5. We know that, 5 comes at the unit's place of a number only when its cube ends in 5.
So, 5 will come at unit's place.
Now, in the second group, the number is 614
∵ (8)³ = 512 and (9)³=729
and 512 < 614 < 729
So, ten's place of required cube root is 8.
Hence, \(\sqrt [ 3 ]{ 614125 } \) = 85

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