# Board Paper of Class 10 2021 Mathematics (Basic) Term 1 Set 4 - Solutions

#### General Instructions:

- 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.

#### Question 1

## HCF of 92 and 152 is

- 4
- 19
- 23
- 27

Sol. (a);

The prime factorisation of 92 and 152 are:

#### Question 2

## In $\u25b3$ABC, DE || BC, AD = 4 cm, DB = 6 cm and AE = 5 cm. The length of EC is

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

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

#### Question 3

## The value of k, for which the pair of linear equations x + y – 4 = 0, 2x + ky – 3 = 0 have no solution, is

- 0
- 2
- 6
- 8

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

#### Question 4

## The value of (tan^{2} 45^{°} – cos^{2} 60^{°}) is

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

Sol. (d); We know that,

#### Question 5

## A point (x, 1) is equidistant from A(0, 0) and B(2, 0). The value of x is

- 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)

#### Question 6

## Two coins are tossed together. The probability of getting exactly one head is

- $\begin{array}{c}1\\ -\\ 4\end{array}$
- $\begin{array}{c}1\\ -\\ 2\end{array}$
- $\begin{array}{c}3\\ -\\ 4\end{array}$
- 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) $\begin{array}{cccc}& 2& & 1\\ =& -& =& -\\ & 4& & 2\end{array}$

#### Question 7

## A circular arc of length 22 cm subtends an angle $\mathrm{\theta}$ at the centre of the circle of radius 21 cm. The value of $\mathrm{\theta}$ is

- 90°
- 50°
- 60°
- 30°

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

#### Question 8

## A quadratic polynomial having sum and product of its zeroes as 5 and 0 respectively, is

- 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.

#### Question 9

## If P(E) = 0.65, then the value of P(not E) is

- 1.65
- 0.25
- 0.65
- 0.35

Sol. (d); We know that, sum of the probability of happening event (E) and not happening event ($\overline{E}$) is 1.

#### Question 10

## It is given that $\u25b3$DEF ~ $\u25b3$PQR. EF : QR = 3 : 2, then value of ar(DEF): ar(PQR) is

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

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

#### Question 11

## Zeroes of a quadratic polynomial x^{2} – 5x + 6 are

- –5, 1
- 5, 1
- 2, 3
- –2, –3

Sol. (c); Given quadratic polynomial is

#### Question 12

## $\begin{array}{c}57\\ -\\ 300\end{array}$ is a

- 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.

#### Question 13

## Perimeter of a rectangle whose length ($l$) is 4 cm more than twice its breadth (b) is 14 cm. The pair of linear equations representing the above information is

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

#### Question 14

#### Question 15

## The ratio in which the point (4, 0) divides the line segment joining the points (4, 6) and (4, –8) is

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

Sol. (b); Let P(4, 0) divides the line segment joining the points A(4, 6) and B(4, –8) in the ratio k : 1.

#### Question 16

#### Question 17

## In the given figure, a circle is touching a semi-circle at C and its diameter AB at O. If AB = 28 cm, what is the radius of the inner circle?

#### Question 18

## The vertices of a triangle OAB are O(0, 0), A(4, 0) and B(0, 6). The median AD is drawn on OB. The length AD is

#### Question 19

#### Question 20

#### Question 21

## The perimeter of the sector of a circle of radius 14 cm and central angle 45° is

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

Sol. (d);

#### Question 22

## A bag contains 16 red balls, 8 green balls and 6 blue balls. One ball is drawn at random. The probability that it is blue ball is

- 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

#### Question 23

## If sin$\mathrm{\theta}$ - cos$\mathrm{\theta}$ = 0, then the value of $\mathrm{\theta}$ is

- 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°.

#### Question 24

## The probability of happening of an event is 0.02. The probability of not happening of the event is

- 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.

#### Question 25

## Two concentric circles are centred at O. The area of shaded region, if outer and inner radii are 14 cm and 7 cm respectively, is

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

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

#### Question 26

Sol. (d);

#### Question 27

## The origin divides the line segment AB joining the points A(1, -3) and B(-3, 9) in the ratio:

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

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

#### Question 28

## The perpendicular bisector of a line segment A(–8, 0) and B(8, 0) passes through a point (0, k). The value of k is

- 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.

#### Question 29

## Which of the following is a correct statement?

- 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.’

#### Question 30

## The solution of the pair of linear equations x = –5 and y = 6 is

- (–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).

#### Question 31

## A circle of radius 3 units is centered at (0, 0). Which of the following points lie outside the circle?

- (–1, –1)
- (0, 3)
- (1, 2)
- (3, 1)

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

#### Question 32

## The value of k for which the pair of linear equation 3x + 5y = 8 and kx + 15y = 24 has infinitely many solutions, is

- 3
- 9
- 5
- 15

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

#### Question 33

## HCF of two consecutive even number is

- 0
- 1
- 2
- 4

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

#### Question 34

## The zeroes of quadratic polynomial x^{2} + 99x + 127 are

- 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.

#### Question 35

## The mid-point of line segment joining the points (–3, 9) and (–6, –4) is

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

#### Question 36

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

(d); A rational number having denominator, whose prime factors are not in the form 2^{m} × 5^{n}, have non-terminating but repeating decimal expansion.

#### Question 37

## In ΔABC, DE ‖ BC, AD = 2 cm, DB = 3 cm, DE : BC is equal to

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

Sol. (b);

#### Question 38

## The (HCF x LCM) for the numbers 50 and 20 is

- 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.

#### Question 39

## For which natural number n, 6^{n} ends with digit zero?

- 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.

#### Question 40

## (1 + tan^{2} A) (1 + sin A) (1 – sin A) is equal to

- 1
- 0
- 2

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

**Case Study – I**

**(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:

#### Question 41

## The maximum height achieved by ball is

- 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

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

#### Question 42

## The polynomial represented by above graph is

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

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

#### Question 43

## Time taken by ball to reach maximum height is

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

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

#### Question 44

## Number of zeroes of the polynomial whose graph is given, is

- 1
- 2
- 0
- 3

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

Thus, it has two zeros.

#### Question 45

## Zeroes of the polynomial are

- 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.

#### Case Study- II

**(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:

#### Question 46

## Which of the following criteria is not suitable for $\u25b3\mathrm{ABC}$ to be similar to $\u25b3\mathrm{QRP}$?

- SAS
- AAA
- SSS
- RHS

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

#### Question 47

## If each square is of length x unit, then length BC is equal to

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

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

#### Question 48

## Ratio BC : PR is equal to

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

Sol. (c);

#### Question 49

## ar (PQR) : ar (ABC) is equal to

- 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

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

#### Question 50

## Which of the following is not true?

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

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

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