ANSWER:B
EXPLANATION:Number Systems Recall’s = 0.20220222022220… from the previous section. Notice that it is non- terminating and non-recurring. Therefore, from the property above, it is irrational number.

ANSWER:D
EXPLANATION:The sum of the zeroes of the polynomial $$3 x^2-8 x+6$$.
Basic Concept
– The sum of the zeroes of a quadratic polynomial $$a x^2+b x+c$$ is $$-\frac{b}{a}$$
How to solve
Identify the coefficients of the polynomial and use the formula for the sum of the zeroes.
Step 1
Identify the coefficients of the polynomial.
$$
\begin{aligned}
a & =3 \\
b & =-8 \\
c & =6
\end{aligned}
$$

Step 2
Use the formula for the sum of the zeroes is $$-\frac{b}{a}$$

Solution
The sum of the zeroes of the polynomial $$3 x^2-8 x+6$$ is $$\frac{8}{3}$$.

ANSWER:A
EXPLANATION:The relationship between the pair of linear equations $$2 x+3 y-9=0$$ and $$4 x+6 y-18=0$$.
How to solve
Compare the coefficients of the two equations to determine their relationship.
Step 1
Compare the coefficients of the two equations.
$$
\begin{aligned}
& \frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2} \\
& \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2} \\
& \frac{c_1}{c_2}=\frac{-9}{-18}=\frac{1}{2}
\end{aligned}
$$
Step 2
Since $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$, the two equations represent the same line.
Step 3
Therefore, the pair of linear equations are coincident and consistent.
Solution
The pair of linear equations $$2 x+3 y-9=0$$ and $$4 x+6 y-18=0$$ are coincident and consistent.

ANSWER:C
EXPLANATION:$$\begin{aligned}
& 21,18,15,………. \\
& a=21 \\
& d=18-21=-3 \\
& n=35 \\a_n
& \Rightarrow a + (n-1)d\\
& \Rightarrow 21+(n-1) d \\
& \Rightarrow 21+(35-1)-3 \\
& \Rightarrow 21 + (34)(-3)\\
& \Rightarrow21-102 \\
& \Rightarrow-81
\end{aligned}$$

ANSWER:C
EXPLANATION:cot $$30^{\circ}$$ = √3 = $$\tan 60^{\circ}$$

ANSWER:B
EXPLANATION:The distance of a point from the x-axis is called its y-coordinate or ordinate.

ANSWER:C
EXPLANATION: $$ \begin{gathered}
7=7 \times 1 \\
28=7 \times 2 \times2\\
H C F(7,28)=7 \
\end{gathered}$$

ANSWER:B
EXPLANATION:Let the probability of raining tomorrow be $$p(A)$$ and the probability of not raining be $$p(\bar{A})$$
We know that,
$$
p(A)+p(\bar{A})=1
$$
Therefore,
$$
\begin{aligned}
& 0.85+p(\bar{A})=1 \\
& p(\bar{A})=1-0.85=0.15
\end{aligned}
$$
The probability that it will not rain tomorrow is 0.15 .

ANSWER:B
EXPLANATION:$$\frac{1}{3} \pi r^{2} h$$

ANSWER:C
EXPLANATION:We are given the quadratic equation of the form $$x^2+5 x+5=0$$.
Compare it with the standard form of the quadratic equation $$a x^2+b x+c=0$$, we get the values of $$\mathrm{a}, \mathrm{b}$$ and c given by
$$
a=1, b=5, c=5
$$

To determine the value of the discriminant we need to substitute the above values of $$a, b$$ and $$c$$ in the formulae of the discriminant given by $$D=b^2-4 a c$$.

We get,
$$
D=5^2-4(1)(5)
$$

Evaluate the square term,
$$
D=25-4(1)(5)
$$

$$
D=25-20
$$

$$
D=5
$$

ANSWER:True
EXPLANATION:Prime Factors of 5313 are:
3, 7, 11, 23.

ANSWER:B
EXPLANATION:01 Identify the number of terms and the first and last term
we have to find the sum of the first 100 positive integers. So, the number of terms, $$n$$ = 100 . The first term, $$a$$=1 and the last term $$a_n$$ = 100 .

02 Apply the formula for the sum of an arithmetic series
$$
S_n=\frac{n}{2}\left(a+a_n\right)
$$
formula.
03 Substitute the values into the formula
Substitute the values we found in Step 1 into the formula:
$$
S_{100}=\frac{100}{2}(1+100)
$$
04 Simplify the expression
First, add the first term and the last term:
$$
S_{100}=\frac{100}{2}(101)
$$
Then, simplify the fraction:
$$
S_{100}=50(101)
$$
05 Calculate the sum
Finally, multiply the remaining numbers to find the sum of the first 100 positive integers:
$$
S_{100}=5050
$$
So, the sum of the first 100 positive integers is 5050.

ANSWER:similar
EXPLANATION:All circles are similar, but not all circles are congruent

ANSWER:two
EXPLANATION: The maximum number of tangents that can be drawn to a circle from a point outside the circle is two

ANSWER:$$S_n=\frac{n}{2}[2 a+(n-1) d$$
EXPLANATION: $$\text { or } S_n=\frac{n}{2}[a+l] ;[\text { Where } l=[a+(n-1) d]] \text { or } S_n=\frac{n}{2}[2 a+(n-1) d]$$

ANSWER:False
EXPLANATION:y-2x=0

2-2(0)=0

2-0=0

2=0

ANSWER:False
EXPLANATION:the ratio of sine and cosine is always less than 1. So, the value of $$\sin A$$ or $$\cos A$$ never exceeds the value 1.

ANSWER:$$Cos^2A=1$$
EXPLANATION:$$Sin^2A+Cos^2A=1$$

ANSWER:6.833
EXPLANATION:sum of first 6 prime number/total number of terms
2+3+5+7+11+13/6=6.833

ANSWER:Class Mark = (Upper Class Limit + Lower Class Limit) / 2
EXPLANATION:The formula for calculating the class mark of grouped data is: Class Mark = (Upper Class Limit + Lower Class Limit) / 2
Upper Class Limit: The highest value within a particular class interval.
Lower Class Limit: The lowest value within a particular class interval.
Example:
If a class interval is “20 – 30”, the class mark would be calculated as:
(30 + 20) / 2 = 25

ANSWER:mode = 15.99
EXPLANATION:Concept :

The relationship connecting three measures of central tendency is given by

3 × Median = 2 × Mean + Mode

Solution :

Step 1

Write down mean and median

Here it is given that ,

Mean = 15.6

Median = 15.73

Step 2

Calculate mode of the grouped frequency distribution

3 × Median = 2 × Mean + Mode

⇒ ( 3 × 15.73 ) = ( 2 × 15.6 ) + Mode

⇒ 47.19 = 31.2 + Mode

⇒ Mode = 47.19 – 31.2

⇒ Mode = 15.99

Hence mode = 15.99

ANSWER:
EXPLANATION:Given,
$$
\begin{aligned}
& 3 x-5 y-4=0\\
& 2 y+7=9 x\\
=>& 3 x-5 y=4(I) \\
=>& 2 y+7=9 x(II)
\end{aligned}
$$
$$\qquad$$
$$\qquad$$
From (ii), $$x=\frac{2 y+7}{9}$$ $$\qquad (III)$$
substitute $$x=\frac{2 y+7}{9}$$ in (i), then
$$
\begin{aligned}
& 3\left(\frac{2 y+7}{9}\right)-5 y=4 \\
& \Rightarrow \frac{2 y+7}{3}-5 y=4 \\
& \Rightarrow \frac{2 y+7-15 y}{3}=4 \\
& \Rightarrow 7-13 y=12 \\
& \Rightarrow y=-\frac{5}{13}
\end{aligned}
$$

Put $$y=-\frac{5}{13}$$ in (iii), then
$$
\begin{aligned}
& x=\frac{2\left(-\frac{5}{13}\right)+7}{9}=\frac{-10+91}{9 \times 13} \\
& \Rightarrow x=\frac{9}{13}
\end{aligned}
$$

Hence, $$x=\frac{9}{13}$$ and $$y=-\frac{5}{13}$$ is the solution for the given pair of linear equations.

ANSWER:
EXPLANATION:$$
\begin{aligned}
& 2 x^2+x-6=0 \\
& \Rightarrow 2 x^2+4 x-3 x-6=0 \\
& \Rightarrow 2 x(x+2)-3(x+2)=0 \\
& \Rightarrow(x+2)(2 x-3)=0 \\
& \Rightarrow x+2=0 \text { or } 2 x-3=0 \\
& \Rightarrow x=-2 \text { or } x=\frac{3}{2}
\end{aligned}
$$
The roots of the given equation are $$-2, \frac{3}{2}$$.

ANSWER
EXPLANATION:Solution
Step 1: Given trigonometric ratio is:
$$\sec \theta=\frac{13}{12}$$,
where, $$\sec \theta=\frac{\text { Hypotenuse }}{\text { Base }}$$
Step 2: Find the perpendicular, using the
Pythagoras theorem.
According to Pythagoras theorem,
$$(\text { Hypotenuse })^2=(\text { Perpendicular })^2+(\text { Base })^2$$
Put Hypotenuse $$=13$$ and Base $$=12$$ in the equation.
$$13^2=$$ Perpendicular $$^2+12^2$$
Perpendicular $${ }^2+144=169$$
Perpendicular $${ }^2+144-144=169-144$$
Perpendicular $${ }^2=25$$
Perpendicular $$=\sqrt{25}$$
Perpendicular $$=5$$
Step 3: Find other trigonometric ratios.
$$
\begin{aligned}
& \sin (\theta)=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{5}{13} \\
& \cos (\theta)=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{12}{13} \\
& \tan (\theta)=\frac{\text { Perpendicular }}{\text { Base }}=\frac{5}{12} \\
& \operatorname{cosec}(\theta)=\frac{\text { Hypotenuse }}{\text { Perpendicular }}=\frac{13}{5} \\
& \cot (\theta)=\frac{\text { Base }}{\text { Perpendicular }}=\frac{12}{5}
\end{aligned}
$$

So, when the value of $$\sec (\theta)=\frac{13}{12}$$, the values of other trigonometric ratios are $$\sin (\theta)=\frac{5}{13}, \cos (\theta)=\frac{12}{13}, \tan (\theta)=\frac{5}{12}, \cos e$$ $$c(\theta)=\frac{13}{5}, \cot (\theta)=\frac{12}{5}$$

ANSWER:
EXPLANATION:Solution
Radius $$=r=3 \mathrm{~cm}$$
Volume of the sphere $$=\frac{4}{3} \pi r^3=\frac{4}{3} \times \frac{22}{7} \times 3^3=113.14 \mathrm{~cm}^3$$

ANSWER:
EXPLANATION:C.S.A= 2πrh

= 2 x 22/7 x 2 x 7

= 2 x 22 x 2

= 88 Cm.

ANSWER:
EXPLANATION:Solution
Evaluate the value of $$y$$
The distance formula is
$$
d=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}
$$
.
Here,
$$
\begin{aligned}
& d=10 \text { units } \\
& \left(x_1, x_2\right)=(2,10) \text { and }\left(y_1, y_2\right)=(-3, y) .
\end{aligned}
$$

Substituting in the formula,
$$
\begin{aligned}
& \Rightarrow 10=\sqrt{(10-2)^2+(y+3)^2} \\
& \Rightarrow 10=\sqrt{64+y^2+9+6 y}\left[(\mathbf{a}+\mathbf{b})^2=\mathbf{a}^2+\mathbf{2 a b}+\mathbf{b}^2\right]
\end{aligned}
$$

Now, take square roots on both the sides
$$
\begin{aligned}
& \Rightarrow 100=73+6 y+y^2 \\
& \Rightarrow y^2+6 y-27=0 \\
& \Rightarrow y^2+9 y-3 y-27=0 \\
& \Rightarrow y(y+9)-3(y+9)=0 \\
& \Rightarrow(y+9)(y-3)=0 \\
& \Rightarrow y=3,-9
\end{aligned}
$$

Hence, the value of $$y$$ is 3 or -9 .

ANSWER:
EXPLANATION:Let the points $$(5,-2),(6,4)$$, and $$(7,-2)$$ are representing the vertices $$A, B$$, and C of the given triangle respectively.
$$
\begin{aligned}
\ \mathrm{AB}=\sqrt{(5-6)^2+(-2-4)^2}\\=\sqrt{(-1)^2+(-6)^2}\\=\sqrt{1+36}=\sqrt{37} \\
\mathrm{BC}=\sqrt{(6-7)^2+(4-(-2))^2}\\=\sqrt{(-1)^2+(6)^2}\\=\sqrt{1+36}=\sqrt{37} \\
\ \mathrm{CA}=\sqrt{(5-7)^2+(-2-(-2))^2}\\=\sqrt{(-2)^2+0^2}=2
\end{aligned}
$$

Therefore, $$\mathrm{AB}=\mathrm{BC}$$
As two sides are equal in length, therefore, ABCis an isosceles triangle.

ANSWER:
EXPLANATION:Let the polynomial be $$a x^2+b x+c$$, and its zeroes be $$a$$ and $$\beta$$.
$$
\begin{aligned}
& \alpha+\beta=\sqrt{2} \\
& \alpha \beta=\frac{1}{3}=\frac{c}{a}
\end{aligned}
$$
$$
\begin{aligned}
& a+\beta=\frac{3 \sqrt{2}}{3}=\frac{-b}{a} \\
& a=3, b=-3 \sqrt{2}, c=1
\end{aligned}
$$
Quadratic polynomial is $$3 x^2-3 \sqrt{2} x+1$$.

ANSWER:
EXPLANATION:From the figure, it can be observed that points $$X, Y, Z$$ are dividing the line segment in a ratio $$1: 3,1: 1,3: 1$$ respectively.
Using Sectional Formula, we get,
Coordinates of $$\mathrm{X}=\frac{1 \times 2+3 \times(-2)}{1+3}, \frac{1 \times 8+3 \times 2}{1+3}$$ $$=\left(-1, \frac{7}{2}\right)$$
Coordinates of $$\left.Y=\frac{2-2}{2}, \frac{2+8}{2}=(0,5)\right)$$
Coordinates of $$Z=\left(\frac{3 \times 2+1 \times(-2)}{1+3}, \frac{3 \times 8+1 \times 2}{1+3}\right)$$
$$
=\left(1, \frac{13}{2}\right)
$$

ANSWER:
EXPLANATION:Let the radius of the circle be $$r$$.
Circumference $$=22 \mathrm{~cm}$$
$$
\begin{aligned}
& 2 \pi r=22 \\
& 2 \times \frac{22}{7} \times r=22 \\
& r=22 \times \frac{7}{22} \times \frac{1}{2} \\
& =r=\frac{7}{2}
\end{aligned}
$$

Here $$\theta=90^{\circ}$$
Area of the $$\left(\frac{1}{4}\right)^{t h}$$ quadrant of the circle $$=\frac{\theta}{360^{\circ}} \times \pi r^2$$
$$
\begin{aligned}
& =\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times\left(\frac{7}{2}\right)^2 \mathrm{~cm}^2 \\
& =\frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \mathrm{~cm}^2 \\
& =\frac{77}{8} \mathrm{~cm}^2
\end{aligned}
$$

ANSWER:
EXPLANATION:

Step 1: Draw appropriate diagram and use the concept of congruence:
As we know that by theorem the tangents from the external point are equal
In $$\triangle O A P, \triangle O A S$$
So, $$A P=A S$$
$$\Rightarrow O A=O A$$ (It is the common side)
$$\Rightarrow O P=O S$$ (They are the radii of the circle)
So, by $$S S S$$ congruency $$\triangle O A P \cong \triangle O A S$$
Thus, $$\angle P O A=\angle A O S$$
This shows that $$\angle 7=\angle 8$$
In this same manner, the angles which are equal are
$$
\begin{aligned}
& \Rightarrow \angle 4=\angle 3 \\
& \Rightarrow \angle 2=\angle 1 \\
& \Rightarrow \angle 6=\angle 5
\end{aligned}
$$

Step 2: Use the concept of complete angle
On adding these angles we get,
$$
\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=
$$
$$360^{\circ} \quad$$ [Complete angle]
On rearranging,
$$
\begin{aligned}
& (\angle 7+\angle 8)+(\angle 2+\angle 1)+(\angle 4+\angle 3)+(\angle 6+ \\
& \angle 5)=360^{\circ} \\
& \Rightarrow 2 \angle 1+2 \angle 8+2 \angle 5+2 \angle 4=360^{\circ}
\end{aligned}
$$

On taking out 2 as common we get,
$$
\Rightarrow(\angle 1+\angle 8)+(\angle 5+\angle 4)=180^{\circ}
$$

Thus, $$\angle B O C+\angle D O A=180^{\circ}$$
Similarly, $$\angle A O B+\angle D O C=180^{\circ}$$
Hence, the opposite sides of any quadrilateral which is circumscribing a given circle will subtend supplementary angles at the center of the circle.

ANSWER:
EXPLANATION:We know that, the lengths of tangents drawn from an external point to a circle are equal.
$$
\begin{aligned}
& \mathrm{TP}=\mathrm{TQ} \\
& \text { In } \triangle \mathrm{TPQ} \\
& \mathrm{TP}=\mathrm{TQ} \\
& \Rightarrow \angle \mathrm{TQP}=\angle \mathrm{TPQ} \ldots \ldots \text { (1) (In a triangle, equal sides ho } \\
& \text { to them) } \\
& \angle \mathrm{TQP}+\angle \mathrm{TPQ}+\angle \mathrm{PTQ}=180^{\circ} \text { (Angle sum property) } \\
& 2 \angle \mathrm{TPQ}+\angle \mathrm{PTQ}=180^{\circ} \text { (Using (1)) } \\
& \Rightarrow \angle \mathrm{PTQ}=180^{\circ}-2 \angle \mathrm{TPQ} \ldots \ldots \text { (1) }
\end{aligned}
$$

We know that, a tangent to a circle is perpendicular to the radius through the point of contact.
OP $$\perp$$ PT
$$
\begin{aligned}
& \angle \mathrm{OPT}=90^{\circ} \\
& \Rightarrow \angle \mathrm{OPQ}+\angle \mathrm{TPQ}=90^{\circ} \\
& \Rightarrow \angle \mathrm{OPQ}=90^{\circ}-\angle \mathrm{TPQ} \\
& \Rightarrow 2 \angle \mathrm{OPQ}=2\left(90^{\circ}-\angle \mathrm{TPQ}\right)=180^{\circ}-2 \angle \\
& \mathrm{TPQ} \ldots \ldots \text { (2) }
\end{aligned}
$$

From (1) and (2), we get
$$
\angle P T Q=2 \angle O P Q
$$

ANSWER:
EXPLANATION:Let’s draw a parallelogram ABCD as per the given question.If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

This is referred to as the AA similarity criterion for two triangles.

In ΔABE and ΔCFB,

∠BAE = ∠FCB (opposite angles of a parallelogram)

∠AEB = ∠FBC [AE || BC and EB is a transversal, alternate interior angles]

Thus, ΔABE ~ ΔCFB (AA criterion)

ANSWER:B
EXPLANATION:Step 1. Explaining the diagram.

Let $$A B C D$$ be quadrilateral where $$A C$$ and $$B D$$ intersects each other at $$O$$ such that, $$\frac{A O}{B O}=\frac{C O}{D O}$$
Step 2. Showing $$A B C D$$ is trapezium
Construction-From the point $$O$$, draw a line $$E O$$ touching $$A D$$ at $$E$$ in such a way that,
$$E O\|D C\| A B$$
In $$\triangle D A B, E O \| A B$$
By using Basic Proportionality Theorem
$$
\frac{D E}{E A}=\frac{D O}{O B} .
$$

Also, given,
$$\frac{A O}{B O}=\frac{C O}{D O}$$
$$\Rightarrow \frac{A O}{C O}=\frac{B O}{D O} \quad[$$ applying alternendo]
$$\Rightarrow \frac{C O}{A O}=\frac{D O}{O B} \quad[$$ [applying invertend
$$o]$$
$$
\Rightarrow \frac{D O}{O B}=\frac{C O}{A O} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots(i i)
$$

From equation (i) and (ii),
We have
$$
\frac{D E}{E A}=\frac{C O}{A O}
$$

Therefore, By applying converse of Basic Proportionality Theorem,
$$E O \| D C$$
Also $$E O\|A B \Rightarrow A B\| D C$$.
Hence, quadrilateral $$A B C D$$ is a trapezium with $$A B \| C D$$.

ANSWER:
EXPLANATION:Let us assume that $$6+\sqrt{ } 2$$ is a rational number.
So it can be written in the form $$\frac{a}{b}$$
$$
6+\sqrt{ } 2=\frac{a}{b}
$$

Here $$a$$ and $$b$$ are coprime numbers and $$b \neq 0$$
$$
6+\sqrt{ } 2=\frac{a}{b}
$$

By solving the equation we get,
$$
\begin{aligned}
& \Rightarrow \sqrt{ } 2=\frac{a}{b}-6 \\
& \Rightarrow \sqrt{ } 2=\frac{a-6 b}{b}
\end{aligned}
$$

This shows $$\frac{a-6 b}{b}$$ is a rational number.
But we know that $$\sqrt{ } 2$$ is an irrational number, it contradicts our assumption.
Our assumption $$6+\sqrt{ } 2$$ is a rational number is incorrect.
Therefore, $$6+\sqrt{ } 2$$ is an irrational number
Hence, it is proved that $$6+\sqrt{ } 2$$ is an irrational number.

ANSWER:
EXPLANATION:An AP consists of 50 terms and the $$50^{\text {th }}$$ term is equal to 106.
It is also given that $$a_3=12$$
Using formula $$a_n=a+(n-1) d$$ to find $$n^{\text {th }}$$ term of AP, we get
$$
\begin{aligned}
& a_{50}=a+(50-1) d \text { and } \\
& a_3=a+(3-1) d \\
& \Rightarrow 106=a+(50-1) d \\
& \Rightarrow 106=a+49 d \\
& \Rightarrow 12=a+(3-1) d \\
& \Rightarrow 12=a+2 d
\end{aligned}
$$

These are equations consisting of two variables. Let’s solve them using substitution method.
Using equation $$106=a+49 d$$
We get $$a=106-49 d$$
Putting value of a in the equation $$12=\mathrm{a}+2 \mathrm{~d}$$ we get
$$
\begin{aligned}
& 12=106-49 d+2 d \\
& \Rightarrow 47 d=94 \\
& \Rightarrow d=\frac{94}{47}=2
\end{aligned}
$$

Putting value of $$d$$ in the equation, $$a=106-49 d$$ we get
$$
a=106-49(2)=106-98=8
$$

Therefore, First term, $$a=8$$ and Common difference, $$d=2$$
To find $$29^{\text {th }}$$ term we use formula $$a_n=a+(n-1) d$$ which is used to find $$n^{\text {th }}$$ term of AP, we get
$$
a_{29}=8+(29-1) 2=8+56=64
$$

Therefore, $$29^{\text {th }}$$ term of $$A P$$ is equal to 64.

ANSWER:
EXPLANATION:The first 8 multiples of 8 are
$$
8,16,24,32,40,48,56,64 \ldots . . . . .120
$$

These are in an A.P., having first term as 8 and common difference as 8.
Therefore, $$a=8$$
$$
\begin{aligned}
& d=8 \\
& S_{15}=? \\
& S_n=\frac{n}{2}[2 a+(n-1) d] \\
& \Rightarrow S_{15}=\frac{15}{2}[2(8)+(15-1) 8] \\
& \Rightarrow S_{15}=7.5[6+(14)(8)] \\
& \Rightarrow S_{15}=7.5[16+112] \\
& \Rightarrow S_{15}=7.5(128) \\
& \Rightarrow S_{15}=960
\end{aligned}
$$

ANSWER:(i) a face card
total number of face cards $$=12$$
prbability= $$\frac{12}{52}$$
(ii) a spade.
favourable outcome = 13
prbability=13/52 = $$\frac{1}{4}$$
EXPLANATION:(i) a king of red suit
favourable outcome $$=2$$
prbability=2/52 = $$\frac{1}{26}$$
(ii) a face card
total number of face cards $$=12$$
prbability= $$\frac{12}{52}$$
(iii) a red face card
favourable outcome $$=6$$
prbability $$=6 / 52=\frac{3}{26}$$
(iv) a queen of black suit
favourable outcome $$=2$$
prbability=2/52= $$\frac{1}{26}$$
(v) a jack of hearts
favourable outcome = 1
prbability=1/52
(vi) a spade.
favourable outcome = 13
prbability=13/52 = $$\frac{1}{4}$$

ANSWER:
EXPLANATION:Let speed of train be $$\times \mathrm{km} / \mathrm{hour}$$.
By formula,
$$
\text { Time }=\frac{\text { Distance }}{\text { Speed }}
$$

Given,
If the speed had been $$8 \mathrm{~km} / \mathrm{h}$$ less, then it would have taken 3 hours more to cover the same distance.
$$
\begin{aligned}
& \frac{480}{x-8}-\frac{480}{x}=3 \\
& \Rightarrow \frac{480 x-480(x-8)}{x(x-8)}=3 \\
& \Rightarrow \frac{480 x-480 x+3840}{x^2-8 x}=3 \\
& \Rightarrow 3840=3\left(x^2-8 x\right) \\
& \Rightarrow 3 x^2-24 x=3840 \\
& \Rightarrow 3\left(x^2-8 x\right)=3840 \\
& \Rightarrow x^2-8 x=\frac{3840}{3} \\
& \Rightarrow x^2-8 x=1280 \\
& \Rightarrow x^2-8 x-1280=0
\end{aligned}
$$

Hence, the required equation is $$x^2-8 x-$$ $$1280=0$$.
$$
x^2-8 x-1280=0
$$

Solving for $$x$$ by factorization method, we get,
$$
\begin{aligned}
& x^2-40 x+32 x-1280=0 \\
& x(x-40)+32(x-40)=0 \\
& (x-40)(x+32)=0 \\
& x-40=0 \text { or } x+32=0 \\
& x=40 \text { or } x=-32
\end{aligned}
$$

Speed cannot be negative. Therefore, the value of $$x$$ is $$40 \mathrm{~km} / \mathrm{hr}$$.
The original speed of the train is $$40 \mathrm{~km} / \mathrm{hr}$$.

ANSWER:
EXPLANATION:The given quadratic equation is $$\mathrm{k} x(\mathrm{x}-2)+6=0$$.
This equation can be rewritten as $$\mathrm{kx}^2-2 \mathrm{kx}+6=0$$.
For equal roots, it discriminate, $$D=0$$.
$$b^2-4 a c=0$$, where $$a=k, b=-2 k$$ and $$c=6$$
$$
\begin{aligned}
& 4 k^2-24 k=0 \\
& 4 k(k-6)=0 \\
& k=0 \text { or } k=6
\end{aligned}
$$

But $$k$$ cannot be 0 , so the value of $$k$$ is 6 .

ANSWER:
EXPLANATION:Radius of bigger cylinder $$=12 \mathrm{~cm}$$, height of bigger cylinder $$=220 \mathrm{~cm}$$
Radius of smaller cylinder $$=8 \mathrm{~cm}$$, height of smaller cylinder $$=60 \mathrm{~cm}$$
Volume of bigger cylinder
$$
\begin{aligned}
& =\pi r^2 \mathrm{~h} \\
& =\pi \times 12^2 \times 220 \\
& =31680 \pi \mathrm{~cm}^3
\end{aligned}
$$

Volume of smaller cylinder
$$
\begin{aligned}
& =\pi r^2 \mathrm{~h} \\
& =\pi \times 8^2 \times 60 \\
& =3840 \pi \mathrm{~cm}^3 \\
& \text { Total volume } \\
& =31680 \pi+3840 \pi \\
& =35520 \pi \mathrm{~cm}^3
\end{aligned}
$$
$$
\begin{aligned}
& \text { Mass }=\text { Density } \times \text { Volume } \\
& =8 \times 35520 \pi \\
& =892262.4 \mathrm{gm}(1000 \mathrm{gm}=1 \mathrm{~kg}) \\
& =892.3 \mathrm{~kg}
\end{aligned}
$$

ANSWER:
EXPLANATION:Given,
1. Diameter of cylinder = Diameter of cone i.e. $$2 r=1.4 \mathrm{~cm}$$
2. Height of cylinder $$=$$ Height of cone i.e. $$h=2.4 \mathrm{~cm}$$

Let $$l$$ be the slant height of the cone.
From the figure,
$$
\begin{aligned}
l^2= & h^2+r^2 \\
& =(2.4)^2+(0.7)^2 \\
\Rightarrow l & =\sqrt{5.76+0.49} \\
& =\sqrt{6.25} \\
\Rightarrow l & =2.5 \mathrm{~cm}
\end{aligned}
$$

Step 2: Solve for curved surface area of cone
Curved surface area of cone $$=\pi r l$$
$$
\begin{aligned}
& =\frac{22}{7} \times 0.7 \times 2.5 \\
& =5.5 \mathrm{~cm}^2
\end{aligned}
$$

Step 3: Solve for curved surface area of cylinder
Curved surface area of cylinder $$=2 \pi r h$$
$$
\begin{aligned}
& =2 \times \frac{22}{7} \times 0.7 \times 2.4 \\
& =10.56 \mathrm{~cm}^2
\end{aligned}
$$

Step 4: Solve for area of circular base of cylinder
Area of circular base $$=\pi r^2$$
$$
=\frac{22}{7} \times(0.7)^2=1.54 \mathrm{~cm}^2
$$

Step 5: Solve for Total surface area of remaining solid
Total surface area of the remaining solid $$=$$ Curved surface area of cylinder + area of circular base of cylinder + curved surface area of cone
$$
\begin{aligned}
& =10.56+1.54+5.5 \\
& =17.6 \mathrm{~cm}^2 \\
& \quad \approx 18 \mathrm{~cm}^2
\end{aligned}
$$

Hence, total surface area of remaining solid is $$18 \mathrm{~cm}^2$$

ANSWER:
EXPLANATION:Let $$A B$$ be the building of height 7 m and EC be the height of tower.
A is the point from where elevation of tower is $$60^{\circ}$$ and the angle of depression of its foot is $$45^{\circ}$$.
$$
E C=D E+C D
$$

Also, $$C D=A B=7 \mathrm{~m}$$ and $$B C=A D$$
According to question,
In right $$\triangle \mathrm{ABC}$$,
$$\tan 45^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}}$$
$$
\begin{aligned}
& \Rightarrow 1=\frac{7}{B C} \\
& \Rightarrow B C=7 m=A D
\end{aligned}
$$

Also,
In right $$\triangle \mathrm{ADE}$$,
$$\tan 60^{\circ}=\frac{\mathrm{DE}}{\mathrm{AD}}$$
$$\Rightarrow \sqrt{3}=\frac{D E}{7}$$
$$\Rightarrow D E=7 \sqrt{3} \mathrm{~m}$$
Height of the tower =EC = DE $$+C D$$
$$
=(7 \sqrt{3}+7) m=7(\sqrt{3}+1) m
$$

ANSWER:
EXPLANATION:$$\begin{aligned} & \frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}} \\ = & \frac{5\left(\frac{1}{2}\right)^2+4\left(\frac{2}{\sqrt{3}}\right)^2-(1)^2}{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} \\ = & \frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}} \\ = & \frac{\frac{1}{12}(15+64-12)}{\frac{1+3}{4}} \\ = & \frac{\frac{1}{12} \times 67}{\frac{4}{4}} \\ = & \frac{67}{12}\end{aligned}$$

ANSWER:
EXPLANATION:$$\begin{aligned} & \frac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta} \\ & =\frac{\sin \theta\left(1-2 \sin ^2 \theta\right)}{\cos \theta\left(2 \cos ^2 \theta-1\right)} \\ & =\frac{\tan \theta\left(\sin ^2 \theta+\cos ^2 \theta-2 \sin ^2 \theta\right)}{\left(2 \cos ^2 \theta-\sin ^2 \theta-\cos ^2 \theta\right)} \\ & =\frac{\tan \theta\left(\cos ^2 \theta-\sin ^2 \theta\right)}{\left(\cos ^2 \theta-\sin ^2 \theta\right)} \\ & =\tan \theta\end{aligned}$$

ANSWER:
EXPLANATION:$$D E \| B C$$ (Given)

So, $$\frac{A D}{D B}=\frac{A E}{E C}[$$ Basic proportanality theorem $$]$$
$$
\begin{aligned}
& \frac{D B}{A D}=\frac{E C}{A E}[\text { Taking reciprocal] } \\
& \frac{D B}{A D}+1=\frac{E C}{A E}+1 \text { [Adding } 1 \text { on both sides] } \\
& \Rightarrow \frac{A B}{A D}=\frac{A C}{A E} \\
& \Rightarrow \frac{A D}{A B}=\frac{A E}{A C}
\end{aligned}
$$

ANSWER:
EXPLANATION:Length of the vertical pole $$=6 \mathrm{~m}$$ (Given)
Length of the shadow of the pole $$=4 \mathrm{~m}$$ (Given)
Let Height of tower = h m
Length of shadow of the tower $$=28 \mathrm{~m}$$ (Given)
In $$\triangle A B C$$ and $$\triangle D E F$$,
$$
\begin{aligned}
& \angle C=\angle E(\text { angular elevation }) \\
& \angle B=\angle F=90^{\circ}
\end{aligned}
$$
$$\triangle \mathrm{ABC} \sim \triangle \mathrm{DFE}$$ (By AAA similarity criterion)
$$\frac{A B}{D F}=\frac{B C}{E F}$$ (If two triangles are similar then their corresponding sides are proportional.)
$$
\begin{aligned}
& \frac{6}{h}=\frac{4}{28} \\
& \Rightarrow h=6 \times \frac{28}{4} \\
& \Rightarrow h=6 \times 7 \\
& \Rightarrow h=42 \mathrm{~m}
\end{aligned}
$$

Hence, the height of the tower is 42 m .

ANSWER:B
EXPLANATION:From the table, it can be observed that $$n=60$$
$$
\begin{aligned}
& 45+x+y=60 \\
& x+y=15
\end{aligned}
$$

Median of the data is given as 28.5 which lies in interval 20 – 30.
Therefore, median class $$=20-30$$
Lower limit ( 1 ) of median class $$=20$$
Cumulative frequency ( $$c f$$ ) of class preceding the median class $$=5+x$$
Frequency ( $$f$$ ) of median class $$=20$$
Class size $$(h)=10$$
Median $$=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h$$
$$28.5=20+\left[\frac{\frac{60}{2}-(5+x)}{20}\right] \times 10$$
$$
\begin{aligned}
& 8.5=\left(\frac{25-x}{2}\right) \\
& 17=25-x \\
& x=8
\end{aligned}
$$

From equation (1),
$$
\begin{aligned}
& 8+y=15 \\
& y=7
\end{aligned}
$$

Hence, the values of $$x$$ and $$y$$ are 8 and 7 respectively.