(i) Since EF " GR and AC is a transversal.
∴ \(\angle \)CAF + \(\angle \)ACR = 180°
=> \(\angle \)CAF + 105° = 180°
=> \(\angle \)CAF = (180° - 105°) = 75°
(ii)∵ FE || GR and AC is a transversal.
:. \(\angle \)EAC = \(\angle \)ACR [Alternate angles]
=> \(\angle \)EAC = 105°
=> \(\angle \)BAC + LEAB = 105°
=> \(\angle \)BAC + 60° = 105°
=> \(\angle \)BAC = 105° - 60° = 45°
Thus, \(\angle \)CAF = 75° and \(\angle \)BAC = 45°

Age of Prashant = 40 years.
Let the age of son be x years.
 \(\therefore\)According to the condition,
5 x (Age of son) + 5 = Prashant's age
\(\Rightarrow \)5[x] + 5 = 40
\(\Rightarrow \)5x + 5 = 40
Transposing 5 to R.H.S., we have
5x = 40 - 5
\(\Rightarrow \)5x = 35
\(\Rightarrow \) x=\(\frac{35}{5}\)=7
Hence, the required age of son is 7 years.

We have
\(5\times \frac { 3 }{ 20 } \times \frac { 2 }{ 15 } \)=\(\frac { 5 }{ 1 } \times \frac { 3 }{ 20 } \times \frac { 2 }{ 15 } =\frac { 5\times 3\times 2 }{ 1\times 20\times 15 } =\frac { 1\times 1\times 2 }{ 1\times 4\times 5 } \)
= \(\frac { 1 }{ 2\times 5 } =\frac { 1 }{ 10 } \)

(a) Since, she deposits Rs 2000 per month
∴ Amount deposit by her in six months
= Rs 2000 \(\times\) (6)
= Rs 12,000
She withdraw Rs 1600 and Rs 5000 in need.
So, withdrawal = - (1600 + 5000)
So, present position of account
= 12000- (1600 + 5000)
= 12000- 6600
= 5400
So, she has balance Rs 5400 in her account.
(b) Multiplication and subtraction of integers.
(c) If you save the money you can use it in need.

We have,
n = Number of observations = 40, Mean = 160
\(\therefore Mean=\frac{Sum\ of\ the\ observations}{Number\ of\ observations}\)
\(\Rightarrow 160=\frac{Sum\ of\ the\ observations}{40}\)
\(\Rightarrow 160\times40\)  = Sum of the observations.
Thus, incorrect sum of observations = 160 x 40
Now,
Correct sum of the observations = Incorrect sum of the observations - Incorrect observation + correct observation
\(\Rightarrow\) Correct sum of the observations \(=160\times40-125+165\)
\(\Rightarrow\) Correct sum of the observations
= 6400 + 40 = 6440
\(\therefore\) correct mean
\(=\frac{Correct\ sum\ of\ the\ observations}{Number\ of\ observations}\)
\(=\frac{6440}{40}=161\)

263.52kg

Let the given triangle be \(\triangle ABC\).
Then, we have AB = AC.
So, \(\triangle ABC\) is an isosceles triangle.
\(\therefore \angle A= \angle C=45^{0}\)

[since, the angle opposite to equal sides of an isosceles triangle are equal]
Now, in \(\triangle ABC\), by angle sum property of a triangle,
\(\angle A+\angle B+\angle C=180^0\Rightarrow45^0+x+45^0=180^0\)
\([\angle A=\angle B=45,because\ of\ equal\ side]\)
\(\Rightarrow\) x+90⁰=180°  \(\Rightarrow\) x=180⁰-90° =90°
Hence, the value of x is 90°.

We have,

Sum of the digits along
Ist row = 5 + (-1) + (-4) = 5 - 5 = 0
Ilnd row = (-5) + (-2) + 7= -7 + 7=0
IIIrd row = 0 + 3 + (- 3) = 3 + (-3) = 0
Similarly, sum of the digits along
Ist column = 5 + (-5) +0=5 + (-5)=0
IInd column = (-1) + (-2) + 3= (-3) +3 =0
IIIrd column = (-4) + 7 + (- 3) = 7 + (-7) = 0
Sum of the digits along
Ist diagonal =5 + (-2) + (-3)= 5 + (-5)=0
IInd diagonal = (-4) + (-2) + 0 = - 6 \(\neq\) 0
Since, the sum of digits along the Ilnd diagonal \(\neq\) 0, so it is not a magic square.
(ii)Magic Square:
Sum of the digits along:
1st row = 1+ (-10) + 0 = 1 - 10
= -9
2nd row = (-4) + (-3) + (-2)
= (-7) + (-2) = -9
3rd row = (-6) + 4 + (-7) = (-13) + 4
= -9
1st column = 1 + (-4) + (-6) = 1 - 10
= -9
2nd column = (-10) + (-3) + 4
= (-13) + 4 = -9
3rd column = 0 + (-2) + (-7) = 0 + (-9)
=-9
One diagonal = 1 + (-3) + (-7) = 1 + (-10)
= -9
Second diagonal = (-6) + (-3) + 0
= (-9) + 0 = -9
∵ Each row, column and diagonal have the same sum.
∴ The square (ii) is the magic square.

See a triangle.

60° + 30° = 90°.

9.82 x 6
\(\therefore\) 982 x 6 = 5892
and there are two decimal places in the decimal part of 9.82.
\(\therefore\) Insert the decimal point in the product such that there are two decimal places in the product.
\(\therefore\) 9.82 x 6 = 58.92

Wehave,
27-[5 + {28-(29-7)}] = 27-[5 + {28-22}]
[Removing the innermost brackets]
= 27- [5 + 6]
[Removing Brackets]
= 27 -11 = 16

Since,
6x + 18 = 8x + 12
\(\therefore\) 6x - 8x + 18 = 12
[On transposing 8x to LHS]
\(\Rightarrow\) 6x - 8x = 12 - 18
[On transposing 18 to RHS]
\(\Rightarrow\) -2x = -6
\(\Rightarrow \quad \frac { -2x }{ -2 } =\frac { -6 }{ -2 } \)
\(\Rightarrow\) x = 3

Rahul score runs per inning are as follows: 34,37,47,49,54,61
\(Mean=\frac{Sum\ of\ the\ data}{Number\ of\ data}\)
\(=\frac{34+37+47+49+54+61}{6}\)
\(=\frac{282}{6}=47\)

In the right angled \(\triangle ABC\)
AC =13 cm, BC =12 cm, AB = x
By using Pythagoras theorem,
(AC)² = (AB)² + (BC)²
\(\Rightarrow\) 13²=x²+12²
\(\Rightarrow\) 169=x²+144
\(\Rightarrow\) x²=169-144 \(\Rightarrow\) x²=25
\(\Rightarrow x=\sqrt 25=5cm\)

(i) x + y = 1200                                  ...(1)
The exterior angle of a triangle is equal to the sum of its two interior opposite angles
x + y + y = 1800
Base angles opposite to the equal sides of an isosceles triangle are equal and the sum of the measures of the three angles of a triangle is 180⁰
\(\Rightarrow\)x + 2y = 1800                                  ... (2)
Subtracting equation (1) from equation (2),
y = 60°
Put y = 60° in equation (1),
x + 60° = 120°
\(\Rightarrow\)x = 120° - 60°
\(\Rightarrow\)x = 60°

Total number of students = 8
Sum of the marks obtained =85 + 62 + 36 + 48 + x + 75 + 39 + 72=417+x
\(\therefore \)Mean marks =\(\frac { (417\quad +x) }{ 8 } \)
\(\Rightarrow \)\(\frac { (417\quad +x) }{ 8 } \)=60
\(\Rightarrow \)417 + x = 60 x 8 = 480
\(\Rightarrow \)x = 480 - 417 = 63
Thus, the required value of x = 63

Given, \(\frac { 2x-3 }{ 5 } +\frac { x+3 }{ 4 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 4(2x-3) }{ 5\times 4 } +\frac { 5(x+3) }{ 5\times 4 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 8x-12 }{ 20 } +\frac { 5x+15 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 8x-12+5x+15 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 13x+3 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow\) 7(13x + 3) = 20(4x + 1)
\(\Rightarrow\) 91x + 21 = 80x + 20
\(\Rightarrow\) 91x - 80x = 20 - 21
\(\Rightarrow\) 11x = -1
Thus, \(x=-\frac { 1 }{ 11 } \)

Let the value of the base angle be x°
Vertex angle = 48°
\(\because\)  Sum of all angles of triangle = 180°
\(\therefore\) x+x+48° = 180°
\(\Rightarrow\)  2x + 48° = 180°
\(\Rightarrow\) 2x = 180° - 48°
\(\Rightarrow\) 2x = 132°
\(\Rightarrow \quad x=\frac { 132° }{ 2 } \)
\(\Rightarrow\) x = 66°
\(\therefore\)   The value of the base angles of the triangle is 66°. 

Let the ladder is represented by AC, wall by BC and AB is the distance of the foot of the ladder from the wall.

\(\because\)\(\angle\)B = 90°
\(\therefore\) \(\triangle\)ABC is a right triangle such that its hypotenuse is AC.
We have AB² + BC² = AC²
[Using Pythagoras theorem]
x² + 8² = 10²\(\Rightarrow\)x² +64= 100
\(\Rightarrow\) x²= 100 - 64 = 36
\(\Rightarrow\) x² ​​​​​​​ = 6² \(\Rightarrow\) X = 6
Thus, the foot of the ladder is 6 m from the wall.

\(\angle BAD=\angle BAE+\angle EAD\)
= 40° + 30°
= 70°
and \(\angle CDA=70^0\)
\(\therefore \angle BAD=\angle CDA\)
But they form a pair of alternate angles.
\(\Rightarrow\) AB II CD .........(i)
Also, \(\angle BAE+\angle AEF=40^0+140^0=180^0\)
But they form a pair of interior opposite angles.
\(\Rightarrow\) AB II EF..........(ii)
From (i) and (ii),we get
AB II CD II EF
\(\Rightarrow\) CD || EF