Here the maximum class frequency is 8, and the class corresponding to this frequency is 3 – 5. So, the modal class is 3 – 5.
Now
modal class = 3 – 5, lower limit (l ) of modal class = 3, class size (h) = 2
frequency ( f₁ ) of the modal class = 8,
frequency ( f₀ ) of class preceding the modal class = 7,
frequency ( f₂) of class succeeding the modal class = 2.
Now, let us substitute these values in the formula :
\(\text { Mode } =l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \)
\(=3+\left(\frac{8-7}{2 \times 8-7-2}\right) \times 2=3+\frac{2}{7}=3.286\)
Therefore, the mode of the data above is 3.286.

Given, pair of linear equations is
x + y = 5 \(\Rightarrow\) x + y - 5 = 0       ....(i)
and 2x + 2y = 10 \(\Rightarrow\) 2x + 2y - 10 = 0 ....(ii)
On comparing with standard form of pair of linear equations, we get
a₁ = 1, b₁ = 1, c₁ = -5
and a₂ = 2, b₂ = 2, c₂ = -10
Here, \(\frac{a_{1}}{a_{2}}=\frac{1}{2},\frac{b_{1}}{b_{2}}=\frac{1}{2}\) and \(\frac{c_{1}}{c_{2}}=\frac{-5}{-10}=\frac{1}{2}\)
Thus, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{2}\)
So, the pair of linear equations is consistent.
Now, table for x + y = 5  or y = 5 - x is  ....(iii)

Here, Eq. (ii) is same as Eq. (i) because on dividing Eq (ii) by 2 on both sides, it becomes same as Eq (i). Therefore., table for both equations is same.
On plotting and joining these points on graph paper, we observe that the lines are coincident., so pair of linear equations have infinitely many solutions and hence consistent. All the points lying on the line ACB satisfy the given pair of linear equations.

There are 40 students, and only one name card has to be chosen.
(i) The number of all possible outcomes is 40
The number of outcomes favourable for a card with the name of a girl = 25  
Therefore, P (card with name of a girl) = P(Girl) = \(\frac{25}{40}=\frac{5}{8}\)
(ii) The number of outcomes favourable for a card with the name of a boy = 15 
Therefore, P(card with name of a boy) = P(Boy) \(=\frac{15}{40}=\frac{3}{8}\)
Note : We can also determine P(Boy), by taking
P(Boy) = 1 – P(not Boy) = 1 – P(Girl) \(=1-\frac{5}{8}=\frac{3}{8}\)

Let us apply the distance formula to find the distances PQ, QR and PR, where P(3, 2), Q(–2, –3) and R(2, 3) are the given points. We have
\(\mathrm{PQ}=\sqrt{(3+2)^{2}+(2+3)^{2}}=\sqrt{5^{2}+5^{2}}=\sqrt{50}=7.07(\text { approx. })\) 
\(\mathrm{QR}=\sqrt{(-2-2)^{2}+(-3-3)^{2}}=\sqrt{(-4)^{2}+(-6)^{2}}=\sqrt{52}=7.21(\text { approx. })\) 
\(P R=\sqrt{(3-2)^{2}+(2-3)^{2}}=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}=1.41 \text { (approx.) }\) 
Since the sum of any two of these distances is greater than the third distance, therefore, the points P, Q and R form a triangle.
Also, PQ² + PR² = QR², by the converse of Pythagoras theorem, we have \(\angle \mathrm{P}=90^{\circ}\)
Therefore, PQR is a right triangle.

Let the number of marbles John had be x
Then the number of marbles Jivanti had be = 45 – x (Why?).
The number of marbles left with john, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marble = 45 – x – 5 = 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – x² – 200 + 5x
= – x² + 45x – 200
So, – x² + 45x – 200 = 124 (Given that product = 124)
i.e., – x² + 45x – 324 = 0
i.e., x² – 45x + 324 = 0
Therefore, the number of marbles John had, satisfies the quadratic equation
x² – 45x + 324 = 0
which is the required representation of the problem mathematically

Let AD = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
\( \frac{A D}{D B}=\frac{A E}{E C} \)
\(\frac{x}{7.2}=\frac{1.8}{5.4} \)
\(x=\frac{1.8 \times 7.2}{5.4}\)
x = 2.4
∴ AD = 2.4 cm

336 and 54
336 = 2 x 168
= 2 x 168
= 2 x 2 x 84
= 2 x 2 x 2 x 42
= 2 x 2 x 2 x 2 x 21
= 2 x 2 x 2 x 2 x 7 x 3 x 1
Therefore 336 = 2 x 2 x 2 x 2 x 7 x 3 .......(A)
54 = 2 x 27
= 2 x 3 x 9
= 2 x 3 x 3 x 3
Therefore 54 = 2 x 3 x 3 x 3 ......(B)
From (A) and (B) HCF of 336 and 54 = 2 x 3 = 6
LCM of 336 and 54 = 2 x 3 x 2 x 2 x 2 x 7 x 3 x 3 = 2⁴ x 3³ x 7 = 3024
Product of 336 and 54 = 18144
Product of LCM and HCF = 6 x 3024 = 18144
Therefore it is proved that LCM X HCF = Product of the two numbers.

The given equations can be rewritten as
3x + 2y - 5 = 0 and 2x - 3y - 7 = 0
On comparing with standard form of pair of linear equations, we get a₁ = 3, b₁ = 2, c₁ = -5
and a₂ = 2, b₂ = -3, c₂ = -7
Now, \(\frac{a_{1}}{a_{2}}=\frac{3}{2}, \frac{b_{1}}{b_{2}}=-\frac{2}{3}\) and \(\frac{c_{1}}{c_{2}}=\frac{5}{7}\)
Thus, \(\frac{3}{2}\neq -\frac{2}{3},i.e.\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}\)
Hence, the pair of linear equations is consistent.

Given, P(E) = 0.05
we know that P(E) + P(\(\bar{E}\)) = 1
\(\therefore\) P(\(\bar{E}\)) = 1 - P(E)  \(\Rightarrow\) P(\(\bar{E}\)) = 1 - 0.05 = 0.95

\(3 x^{2}-4 \sqrt{3} x+4=0\)
Comparing it with ax² + bx + c = 0, we get
a = 3, b = \(-4 \sqrt{3}\)  and c = 4
Discriminant = b² - 4ac
\(=(-4 \sqrt{3})^{2}-4(3)(4)\)
= 48 - 48 = 0
As b² - 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be \(\frac{-b}{2 a}\)
Therefore, the roots are \(\frac{2}{\sqrt{3}} \text { and } \frac{2}{\sqrt{3}}\)

Given, ΔABC is an isosceles triangle with AB = AC. Also,
we have AD \(\perp\) BC and EF \(\perp\)AC.
To prove ΔABD \(\sim\) ΔECF
Proof Since, in ΔABC, AB = AC
∴ \(\angle\)B = \(\angle\)C     [\(\therefore\) angle opposite to equal sides are equal]
Now consider ΔABD and ΔECF. In this we have
∠ABD = ∠ECF   [\(\because\) \(\angle\)B = \(\angle\)C proved above]
and ∠ADB = ∠EFC       [Each 90°]
∴  ΔABD ∼ ΔECF    [by AA similarly criterion]
Hence proved.

6 +\(\sqrt 2\)
if possible let a =  6 +\(\sqrt 2\) be a rational number.
Squaring a² = \((6+\sqrt{2})^{2}\)
a² =38 + 12\(\sqrt 2\)
\(\sqrt{2}=\frac{a^{2}-38}{12}-(1)\)
Since a is a rational number the expression \(\frac{a^{2}-38}{12}\) is also rational number.
⇒ \(\sqrt 2\) is a rational number.
This is a contradiction. Hence, 6 +\(\sqrt 2\) is irrational.
Hence proved.

Draw a parallelogram ABCD and produce the line AD to E and join BE.

In parallelogram ABCD, \(\angle A=\angle C\)       ...(i)
[since, opposite angles of a parallelogram are equal]
In \(\triangle ABE\) and \(\triangle CFB\)
\(\angle EAB=\angle BCF\)      [from Eq.(i)]
\(\angle ABE=\angle CFB\)  [alternate interior angles as AB || FC and BE is transversal]
\(\therefore\triangle ABE\sim \triangle CFB\)   [by AA similarity criterion]
Hence proved.

The given pair of linear equations is
5x-4y+8=0   ....(i)
and 7x+6y-9=0    ....(ii)
On comparing with standard form of pair of linear equations, we get
a₁=5, b₁=-4, c₁=8
and a₂=7, b₂=6, c₂=-9
Here, \(\frac { 5 }{ 7 } \neq \frac { -4 }{ 6 } \)    i.e., \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \frac { { b }_{ 1 } }{ { b }_{ 2 } } \)
So, lines (i) and (ii) are intersecting lines.

Here, a = 21, d = 18 – 21 = – 3 and a_n = – 81, and we have to find n.
As a_n = a + ( n – 1) d,
we have – 81 = 21 + (n – 1)(– 3)
– 81 = 24 – 3n
– 105 = – 3n
So, n = 35
Therefore, the 35th term of the given AP is – 81.
Next, we want to know if there is any n for which a_n = 0. If such an n is there, then
21 + (n – 1) (–3) = 0
i.e., 3 (n – 1) = 21
i.e., n = 8
So, the eighth term is 0.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion

(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

Therefore, the roots of the equation x² - x - 6 = 0 is +3 and -2.

(a) (i) \(\frac{1}{6}\) 
(b) (iii) \(\frac{5}{36}\) 
(c) (i) \(\frac{7}{18}\) 
(d) (i) \(\frac{1}{6}\) 
(e) (ii) \(\frac{11}{36}\)

(i) (b):
x² - 2x -8 =0
\(\Rightarrow\) x² -4x + 2x - 8 = 0 \(\Rightarrow\) x(x - 4) + 2(x - 4) = 0
\(\Rightarrow\)(x - 4)(x + 2) = 0 \(\Rightarrow\)x = 4, x = -2
(ii) (a):
 Intersects x-axis
(iii) (c): 
parabola
(iv) (b): x² - 36 = 0 \(\Rightarrow\) x = 6,-6
 x² – 36
(v) (c):
 f(x) = (x – 2)² + 4 = 0
⇒ x² – 4x + 4 + 4 = 0
⇒ x² – 4x + 8 = 0
Now, D= b² – 4ac = (–4)² – 4.1.8
⇒ D = 16 – 32 = –16 < 0
Hence, the number of zeroes of given polynomial is zero.