- This question paper contains
**50**questions out of which**40**questions are to be attempted. All questions carry**equal**marks. - This question paper contains
**three**Sections:**A**,**B**, and**C.** - Section
**A**has**20**questions. Attempt any**16**questions from Q.No.**1**to**20.** - Section
**B**has**20**questions. Attempt any**16**questions from Q.No.**21**to**40.** - Section
**C**contains**two**Case Studies containing 5 Questions in each case. Attempt any**4**questions from Q.No.**41**to**45**and**4**another from Q.No.**46**to**50.** - There is only
**one**correct option for every multiple choice question (MCQ). Marks will not be awarded for answering more than one option. - There is no negative marking.

(In this section, there are 20 questions. Any 16 are to be attempted).

- 4
- 19
- 23
- 27

Sol. (a);

The prime factorisation of 92 and 152 are:

- 7 cm
- 6.5 cm
- 7.5 cm
- 8 cm

Sol. (c); In $\u25b3$ABC, DE || BC by BPT

- 0
- 2
- 6
- 8

Sol. (b); Since the given pair of linear equations has no solution.

- $\begin{array}{c}1\\ -\\ 2\end{array}$
- $\begin{array}{c}1\\ -\\ 4\end{array}$
- $\begin{array}{c}3\\ -\\ 2\end{array}$
- $\begin{array}{c}3\\ -\\ 4\end{array}$

Sol. (d); We know that,

- 1
- 0
- 2
- $\begin{array}{c}1\\ -\\ 2\end{array}$

Sol. (a); Let P(x, 1) be equidistant from A(0, 0) and B(2, 0)

- $\begin{array}{c}1\\ -\\ 4\end{array}$
- $\begin{array}{c}1\\ -\\ 2\end{array}$
- $\begin{array}{c}3\\ -\\ 4\end{array}$
- 1

Sol. (b); When two coins are tossed together, then the following outcomes are obtained

HH, HT, TH, TT

Exactly one head is obtained, if the events are:

HT, TH

Hence, P(exactly one head) $\begin{array}{cccc}& 2& & 1\\ =& -& =& -\\ & 4& & 2\end{array}$

- 90°
- 50°
- 60°
- 30°

(c); Length of arc (l) = 22 cm

- x
^{2}+ 5x - 2x(x – 5)
- 5x
^{2}- 1 - x
^{2}– 5x + 5

Sol. (b); Let $\alpha $ and $\beta $ be two roots of the given polynomial.

- 1.65
- 0.25
- 0.65
- 0.35

- 4 : 9
- 4 : 3
- 9 : 2
- 9 : 4

Sol. (d); Given that $\u25b3$DEF ~ $\u25b3$PQR and EF : QR = 3 : 2

- –5, 1
- 5, 1
- 2, 3
- –2, –3

Sol. (c); Given quadratic polynomial is

- non-terminating and non-repeating decimal expansion.
- terminating decimal expansion after 2 places of decimals.
- terminating decimal expansion after 3 places of decimals.
- non-terminating but repeated decimal expansion.

Sol.(d); Let $l$ be the length and $b$ be the breadth.

- 1 : 2
- 3 : 4
- 4 : 3
- 1 : 1

(In this section, there are 20 questions. Any 16 are to be attempted).

- 11 cm
- 22 cm
- 28 cm
- 39 cm

Sol. (d);

- 1/2
- 1/5
- 1/30
- 5/6

Sol. (b);

Total number of balls in the bag = 16 + 8 + 6 = 30

Number of blue balls in the bag = 6

- 30°
- 45°
- 90°
- 0°

Sol. (b); Given sin $\mathrm{\theta}$ - cos$\mathrm{\theta}$ = 0

Since, we know that sine and cosine have equal value at 45°.

Thus, $\mathrm{\theta}$ = 45°.

- 0.02
- 0.80
- 0.98
- 49/100

Sol. (c); We know that, sum of the probability of happening and not happening of an event is 1.

- 462 cm
^{2} - 154 cm
^{2} - 231 cm
^{2} - 308 cm
^{2}

(a); Outer radius (R) = 14 cm

Sol. (d);

- 3 : 1
- 1 : 3
- 2 : 3
- 1 : 1

Sol. (b); Let origin divides AB in the ratio k : 1.

- 0 only
- 0 or 8 only
- any real number
- any non-zero real number

Sol. (c); The coordinates of mid-point of line segment joining A(–8, 0) and B(8, 0) =

Since, given that perpendicular bisector AB passes through (0, k).

Thus, the value of k will be any real number as origin is the mid-point of AB.

- Two congruent figures are always similar.
- Two similar figures are always congruent.
- All rectangles are similar.
- The polygons having same number of sides are similar.

Sol. (a);

- Two congruent figures are always similar.
- Two similar figures are not always congruent.
- All rectangles are not similar.
- The polygons having same number of sides are not similar.

Thus, the correct statement is ‘Two congruent figures are always similar.’

- (–5, 6)
- (–5, 0)
- (0, 6)
- (0, 0)

Sol. (a); The graph of linear equations x = –5 and y = 6, intersect each other at (–5, 6).

Thus, the solution of the pair of linear equations x = –5 and y = 6 is (–5, 6).

- (–1, –1)
- (0, 3)
- (1, 2)
- (3, 1)

Sol. (d); Radius of circle = 3 units

- 3
- 9
- 5
- 15

Sol. (b); The given pair of linear equations are

- 0
- 1
- 2
- 4

Sol. (c); HCF of two consecutive even numbers is 2.

- both negative
- both positive
- one positive and one negative
- reciprocal of each other

Sol. (a); Let $\mathrm{\alpha}$ and β zeroes of x^{2} + 99x + 127.

Since, sum of zeroes is negative but product of zeroes is positive.

Thus, both the zeroes of the given quadratic polynomial are negative.

Sol. (c); The coordinates of mid-point of line segment joining the points (–3, 9) and (–6, –4)

- terminating after 1 decimal place.
- non-terminating and non-repeating.
- terminating after 2 decimal places.
- non-terminating but repeating.

^{m} × 5^{n}, have non-terminating but repeating decimal expansion.

- 2 : 3
- 2 : 5
- 1 : 2
- 3 : 5

Sol. (b);

- 1000
- 50
- 100
- 500

Sol. (a); We know that,

Product of two numbers is equal to the product of their HCF and LCM.

Thus, HCF x LCM = 50 x 20 = 1000.

- 6
- 5
- 0
- None

Sol. (d);

We observe that for any natural number n, 6^{n} always ends with 6.

Thus, for no natural number n, 6^{n} ends with digit zero.

- 1
- 0
- 2

Sol. (b); (1 + tan^{2} A) (1 + sin A) (1 – sin A)

**(Attempt any 4 questions from Q. No. 41 to 45)**

Sukriti throws a ball upwards, from a rooftop which is 8 m high from ground level. The ball reaches to some maximum height and then returns and hit the ground.

If height of the ball at time t ( in sec) is represented by h (m), then equation of its path is given as $h=\u2013{t}^{2}+2t+8$

Based above information, answer the following:

- 7 m
- 8 m
- 9 m
- 10 m

Sol. (c)

Draw a perpendicular from maximum height on x-axis. It meet x-axis at (1, 0).

$Heighth=\u2013{t}^{2}+2t+8\left(given\right)$

Thus, height of ball at time t = 1 is

$\mathrm{h}=\u2013{\left(1\right)}^{2}+2\times 1+8=\u20131+2+8=9\mathrm{m}.$

- linear polynomial
- quadratic polynomial
- constant polynomial
- cubic polynomial

Sol. (b) The polynomial represented by above graph is quadratic polynomial.

- 2 sec
- 4 sec.
- 1 sec.
- 2 min.

Sol. (c); Time taken by ball to reach maximum height is 1 sec.

- 1
- 2
- 0
- 3

Sol. (b) Since, given graph is of quadratic polynomial.

Thus, it has two zeros.

- 4
- –2, 4
- 2, 4
- 0, 4

Sol. (b); Given polynomial,

$\u2013{t}^{2}+2t+8\phantom{\rule{0ex}{0ex}}=\u2013{t}^{2}+4t\u20132t+8\phantom{\rule{0ex}{0ex}}=\u2013t(t\u20134)\u20132(t\u20134)\phantom{\rule{0ex}{0ex}}=(t\u20134)(\u2013t\u20132)$

On equating to zero.

$(t\u20134)(\u2013t\u20132)=0\phantom{\rule{0ex}{0ex}}\Rightarrow t=4,\u20132.$

Thus, zeroes of polynomial are –2, 4.

**(Attempt any 4 questions from Q. No. 46 to 50)**

Quilts are available in various colours and design. Geometric design include shapes like squares, triangles, rectangles, hexagons etc. One such design shown above. Two triangles are highlighted, $\u25b3ABCand\u25b3PQR$.

Based on above information, answer the following questions:

- SAS
- AAA
- SSS
- RHS

Sol. (d); RHS is not suitable for $\u25b3\mathrm{ABC}$ to be similar to $\u25b3\mathrm{QRP}$.

Sol. (a) In right $\u25b3\mathrm{ABC},\angle \mathrm{A}=90\xb0$

$\therefore \mathrm{BC}\sqrt{{\mathrm{AB}}^{2}+{\mathrm{AC}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\mathrm{x}}^{2}+{\mathrm{x}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{2{\mathrm{x}}^{2}}\phantom{\rule{0ex}{0ex}}=\mathrm{x}\sqrt{2}\mathrm{unit}$

- 2 : 1
- 1 : 4
- 1 : 2
- 4 : 1

Sol. (c);

- 2 : 1
- 1 : 4
- 4 : 1
- 1 : 8

Sol. (c); $\because \u25b3\mathrm{ABC}~\u25b3\mathrm{QPR}$

We know that, ratio of area of two similar triangles is equal to the ratio of square of corresponding sides.

Thus

$\phantom{\rule{0ex}{0ex}}\frac{\mathrm{ar}(\u25b3\mathrm{QPR})}{\mathrm{ar}(\u25b3\mathrm{ABC})}=\frac{{\mathrm{PR}}^{2}}{\mathrm{BC}}=\frac{{2}^{2}}{12}=\frac{4}{1}=\frac{\mathrm{ar}(\u25b3\mathrm{PQR})}{\mathrm{ar}(\u25b3\mathrm{ABC})}[\because \mathrm{ar}(\u25b3\mathrm{PQR})=\mathrm{ar}(\u25b3\mathrm{QPR}\left)\right]$

- $\u25b3\mathrm{TQS}~\u25b3\mathrm{PQR}$
- $\u25b3\mathrm{CBA}~\u25b3\mathrm{STQ}$
- $\u25b3\mathrm{BAC}~\u25b3\mathrm{PQR}$
- $\u25b3\mathrm{PQR}~\u25b3\mathrm{ABC}$

Sol. (d); $\u2206\mathrm{PQR}\nsim \u2206\mathrm{ABC}$

It should be $\u2206\mathrm{QPR}~\u2206\mathrm{ABC}.$