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**Class X****Mathematics – Basic****Sample Question Paper 2020-21****Max. Marks : 80****Duration : 3 hrs.**

**General Instructions : **

1. This question paper contains two parts A and B.

2. Both Part A and Part B have internal choices.**Part – A**

1. It consists of two sections - I and II.

2. Section I has 16 question. Internal choice is provided in 5 questions.

3. Section II has four case study-based questions. Each case study has 5 case-based sub-parts. An examinee is to attempt any 4 out of 5 sub-parts.**Part – B**

1. Question No 21 to 26 are Very short answer Type questions of 2 marks each.

2. Question No 27 to 33 are Short Answer Type questions of 3 marks each.

3. Question No 34 to 36 are Long Answer Type questions of 5 marks each.

4. Internal choice is provided in 2 questions of 2 marks, 2 questions of 3 marks and 1 question of 5 marks.

**Part - A**

**Section - I**

**Q.1. If two positive integers a and b are written as a = x ^{3}y^{3} and b = x^{4}y^{3}; x, y are prime numbers, then find the value of HCF (a, b). (1 Mark)**

b = x

Therefore,

HCF of a and b = x

OR

Given that one of the zeroes of the cubic polynomial ax

Let f(x) = ax

αβ + βγ+ γα = c / a

One root is zero (given) so, a = 0. βγ = c / a.

a

= 4 + (125 – 1)(– 5)

= 4 – 620

∴ a

∴

√100 = 10

= 0

Given, radius of circle, r = OC = 8 cm

Diameter of the circle = AC = 2 × OC = 2 × 8 = 16 cm which is equal to the diagonal of a square.

Let side of square be a cm

Using Pythagoras theorem in ΔABC,

AB^{2} + BC^{2} = AC^{2}

a^{2} + a^{2} = 16^{2}

2a^{2} = 256

a^{2} = 128 cm^{2}**Q.7. Write the first four terms of an A.P., whose first term is –2 and the common difference is –2. (1 Mark)****Ans. **In the given A.P., a = – 2, d = – 2,

t_{n} = a + (n - 1)d

t_{1} = (-2) + (1 - 1) (-2) = -2

t_{2} = (-2) + (2 - 1) (-2) = -4

t_{3} = (-2) + (3 - 1) (-2) = -6

t_{4} = (-2) + (4 - 1) (-2) = -8**Q.8. If 4 tan θ = 3, then find the. (1 Mark)****ORIf tan (3x + 30°) = 1, then find the value of x. (1 Mark)**

⇒ tan θ = 3 / 4

tan (3x + 30°) = 1 = tan 45°

or, 3x + 30° = 45° or, x = 5°

Ans.

Product of zeroes, α × β = (c / a)

Therefore, for 4x

Product of zeroes = 0 / 4 = 0.

OR

The mean of n quantities x

∵

∴

or, Mean = 4.6

OR

If the distance between the points (p, –1) and (4, –5) is √41 , then find the value of p. (1 Mark)

a

∴ for 4x – 2y = 12 and – kx + 4y = – 24

4 / -k = -2 / 4 = -12 / 24

∴ 2k = 16 or k = 8

By distance formula, distance between (x

(x

∴

∴ p

p

(p – 9)(p + 1) = 0

∴ p = 9 or p = – 1

So, KP / PM = KQ / QN (By BPT)

or, KP / PM = KQ / KN - KQ

or, 4 / 13 = KQ / 20.4 - KQ

or, 4 × 20.4 – 4 KQ = 13 KQ

or, 17 KQ = 4 × 20.4

∴ KQ == 4.8 cm.

OR

What is the name of a line which intersects a circle at two distinct points? (1 Mark)

∴ r

∴ Circumference of the inner circle, 2πr

∴ r

∴ Width of the ring = r

= 14 – 10.5 cm = 3.5 cm.

A line intersecting the circle at two distinct points is called a secant.

**Section - II**

**Q.17. The given figure, shows the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated at A (3, 1), B (6, 4) and C (8, 6) respectively.**

**(i) Find the distance between Ashima and Bharti. (1 Mark)****(a) 2√3 units ****(b) 3√2 units ****(c) √2 units ****(d) √3 units****Ans. **(b)**Solution. **Distance between Ashima and Bharti

√18 = 3√2 units

**(ii) Find the distance between Bharti and Camella. (1 Mark)****(a) 2√2 units ****(b) √2 units ****(c) √3 units ****(d) 3√2 units****Ans. **(a)**Solution. **Distance between Bharti and Camella.

√8 = 2√2 units

**(iii) At how much distance Ashima is seated from Camella? (1 Mark)****(a) 2√5 units ****(b) 5√2 units ****(c) 2 units ****(d) 5 units****Ans. **(b)**Solution. **Distance between Ashima and Camella.

√50 = 5√2 units

**(iv) Write the formula for distance between : (x _{1}, y_{1}) and (x_{2}, y_{2}) (1 Mark)**

(a)

(b)

(c)

(d) (x_{1} + x_{2})^{2} + (y_{1} - y_{2})

**(v) Write the formula for mid-points of two points : (1 Mark)****(a)****(b)****(c)****(d)****Ans. **(b)**Solution.**

Mid points =** **.**Q.18. There is a small island in the middle of 100 m wide river and a tall tree stands on the island. A and B are points directly opposite to each other on two banks and in line with the tree.**

**(i) It the angles of elevation of the top of the tree from A and B are respectively 45° and 60°. Find the height of tree. (1 Mark)****(a) 50 (3 − √3) ****(b) 50 (3 − √3) ****(c) 50 (3 + √3) ****(d) 50 (√3 +3)****Ans. **(a)**Solution. **In ΔAPD,

tan 45° = PD / AD

1 = PD / x

PD = x ...(i)

In ΔBPD,

tan 60° = PD / BD

√3 = PD / 100 − x

√3 (100 – x) = PD ...(ii)

Equating (i) and (ii)

x = √3 (100 – x)

x (√3 +1) = 100√3

= 50 (3 − √3) m

Height of tree, PD = x = 50 (3 − √3) m

**(ii) Find the distance from A to tree. (1 Mark)****(a) 50 ( 3 − √3) m ****(b) 51 (3 − √2) m ****(c) 50 (2 − √3) m ****(d) 25 (2 − √3) m.****Ans. **(a)**Solution. **AD = x = 50 (3 − **√**3) m.

**(iii) Find the distance from B to tree. (1 Mark)****(a) 50 (3 − √1) m ****(b) 25 (3 − √1) m ****(c) 50 (3 + √1) m ****(d) 25 (3 − √1) m****Ans. **(a)**Solution. **BD = 100 – x

= 100 – 50 (3 − **√**3)

= 50 (2 − 3 + **√**3)

= 50 (**√**3 − 1) m.

**(iv) The value of tan 45° is equal to : (1 Mark)****(a) 0(b) 1(c) 2(d) √3 / 2**

**(v) The value of tan 60° is equal to : (1 Mark)****(a) 1 / √3(b) √3(c) 1 / √2(d) √3 / 2**

**(i) Find the circumference of circular table cover. (1 Mark)****(a) 200.50 cm ****(b) 201.2 cm ****(c) 201.14 cm ****(d) 220.15 cm****Ans. **(c)**Solution.** Circumference of table Cover = 2πr.

= 2 x 22 / 7 x 32 cm

= 201.14 cm.

**(ii) Find the diameter of circle having radius 32 cm. (1 Mark)****(a) 60 cm ****(b) 65 cm ****(c) 62 cm ****(d) 64 cm****Ans. **(d)**Solution. **Diameter = 2 × radius

= 2 × 32 cm

= 64 cm.

**(iii) Find the area of the circle (see the figure). (1 Mark)****(a) 3281.25 cm ^{2} **

= 22 / 7 x (32)

= 3218.28 cm

**(iv) Find the area of equilateral triangle ABC. (see the figure). (1 Mark)****(a) 786 √3 cm ^{2} **

Now, ∠BOM = 1 / 2 × 120° = 60°

So, from ΔBOM, we have

OM / OB = cos 60° = 1 / 2

i.e. ⇒ OM = 16 cm. (∴ OB = 32 m)

Also, BM / OB = sin 60° = √3 / 2

i.e., BM = 16√3 cm

BC = 2BM = 32√3 cm.

Hence, area of ΔBOC = 1 / 2 BC.OM.

= 1 / 2 × 32√3 × 16

Area of ΔABC = 3 × area of ΔBOC

= 3 x 1 / 2 × 32√3 × 16

= 768√3 cm^{2}.

**(v) Find the area of shaded region (designed). (1 Mark)****(a) 1900 cm ^{2} **

= Area of circle – Area of ΔABC

= (3218.28 – 768√3) cm

= 1888.104 cm

**(i) Find the diameter of each conical depression. (1 Mark)****(a) 1 cm ****(b) 1.5 cm. ****(c) 2.0 cm ****(d) 2.5 cm****Ans. **(a)**Solution. **Diameter of each conical depression = 2r

= 2 × 0.5 cm

= 1 cm.

**(ii) Find the volume of the cuboid. (1 Mark)****(a) 520 cm ^{3} **

= 15 cm × 10 cm × 3.5 cm

= 525 cm

**(iii) Write the formula to find out the volume of conical depression. (1 Mark)****(a) 1 / 2 πr ^{2}h**

**(iv) Find the volume of wood taken out to make four cavities. (1 Mark)****(a) 30.5 cm ^{3} **

= 1.47 cm

**(v) Find the volume of the wood in the entire stand. (1 Mark)****(a) 523.53 cm ^{3} **

= Volume of cuboid – volume of 4 cavities

= 525 cm

= 523.53 cm

**Part – B**

**All questions are compulsory. In case of internal choices, attempt anyone.**

**Q.21. Find the zeroes of the quadratic polynomial x ^{2} + 6x + 8. (2 Mark)ORIf zeroes of the polynomial x^{2} + 4x + 2a are a and 2 / α, then find the value of a. (2 Mark)**

As, p(x) = 0

x

x

x(x + 4) + 2(x + 4) = 0

(x + 4) (x + 2) = 0

∴ x = – 4 or – 2

∴ Roots are – 4 or – 2.

Product of (zeroes) roots = c / a

= 2a / 1 = α. 2 / α

or, 2a = 2

∴ a = 1

OR

Find whether the lines represented by 2x + y = 3 and 4x + 2y = 6 are parallel, coincident or intersecting. (2 Mark)

In the given equation,

2x

a = 2, b = k, c = 3

∴ Discriminant = k

= k

∴ k

or k = ± √24 = ± 2√6

Here, a

and a

If a

then the lines are coincident.

Clearly, 2 / 4 = 1 / 2 = -3 / -6 = 1 / 2

Hence, lines are coincident.

Let there be n terms in the A.P. Then, a

a

∴ a + (n − 1)d = 497

∴ 56 + (n − 1) × 7 = 497

[∵ a = 56 and d = 7]

∴ 7n + 49 = 497

∴ 7n = 448 or n = 64

Let the ratio in which y-axis divides AB line segment be k : 1

By section formula

∴

∴ 3k – 6 = 0 or k = 2

∴ k : 1 = 2 : 1

and 4k +7 = y (k + 1)

(substitute value of k)

4(2) + 7 = y(2 + 1) 15

= 3y or y = 5

∴ (0, y) = (0, 5)

Thus, the line segment joining the points A(– 6, 7) and B(3, 4) is divided by the y-axis in the ratio 2 : 1 internally and the coordinates of the point of division is (0, 5).

Let a triangle ABC with each side equal to 2a.

∠A = ∠B = ∠C = 60°

Draw AD perpendicular to BC

ΔBDA ≌ ΔCDA by RHS

BD = CD

∠BAD = ∠CAD = 30° by CPCT

In ΔBDA, cosec 30° = AB / BD = 2a / a = 2

and cos 60° = BD / AB = a / 2a = 1 / 2

Mean = 30

Median = 25

Empirical formula is given by :

3 Median = Mode + 2 Mean

or, 3 × 25 = Mode + 2 × 30

or, 75 = Mode + 60

or, 75 – 60 = Mode

or, 15 = Mode

or, Mode = 15.

**All questions are compulsory. In case of internal choices, attempt anyone.**

**Q.27. Three sets of English, Hindi and Sociology books dealing with cleanliness have to be stacked in such a way that all the books are stored topic wise and the height of each stack is the same. The number of English books is 96, the number of Hindi books is 240 and the number of sociology books is 336. (3 Mark) (i) Assuming that the books are of the same thickness, determine the number of stacks of English, Hindi and Sociology books. (ii) Which mathematical concept is used in the problem ?**

(2m – 1)x + 3y – 5 = 0

and 3x + (n – 1)y – 2 = 0

We have, By prime factorisation

96 = 2

240 = 2

and 336 = 2

∴ HCF of 96, 240 and 336 is 2

So, there must be 48 books in each stack.

Number of stacks of English books

= 96 / 48 = 2

Number of stacks of Hindi books

240 / 48 = 5

Number of stacks of Sociology books

336 / 48 = 7

We have, for equation

(2m – 1)x + 3y – 5 = 0 ...(i)

a

and for equation

3x + (n – 1)y – 2 = 0 ...(ii)

a

For a pair of linear equations to have infinite number of solutions

a

Or 2m - 1 / 3 = 3 / n - 1 = 5 / 2

or 2(2m –1) = 15 and 5(n – 1) = 6

Hence, m = 17 / 2 and n = 11 / 5

OR

A father’s age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father. (3 Mark)

∴ a

∴ a + 6d = a + 4d + 12

6d – 4d = 12

2d = 12 or d = 6,

As 3

∴ a

∴ a + 2 (6) = 16

a + 12 = 16

a = 4

∴ AP = 4 , 10 , 16.........

Let the age of father be x years and sum of the ages of his children be y years.

After 5 years,

Father’s age = (x + 5) years

Sum of ages of his children = (y + 10) years

According to the given condition,

x = 3y ...(i)

and x + 5 = 2(y + 10)

or, x – 2y = 15 ...(ii)

Solving eqns. (i) and (ii), we get

3y – 2y = 15

⇒ y = 15

Substituting value of y in eq. (i), we get

x = 3 × 15 = 45

Hence, father ’s present age is 45 years.

The distance between PA is equal to PB such that, PA = PB

By distance formula,

∴

– 4x + 25 = 4x + 81

8x = – 56 or x = – 7

So, the point P(–7, 0) is equidistant from the points (2, – 5) and (– 2, 9)

Let ∠OPQ be θ.

∴ ∠TPQ = (90° – θ)

Since TP = TQ (Tangents)

∴ ∠TQP = (90° – θ)

(Opposite angles of equal sides)

Now, ∠TPQ + ∠TQP + ∠PTQ = 180°

⇒ 90° – θ + 90° – θ + ∠PTQ = 180°

⇒ ∠PTQ = 180° – 180° + 2θ

⇒ ∠PTQ = 2θ

⇒ ∠PTQ = 2∠OPQ.**Q.31. If in a triangle ABC right angled at B, AB = 6 units and BC = 8 units, then find the value of sin A.cos C + cos A.sin C. (3 Mark)****Ans. **Here, AC^{2 }= (8)^{2} + (6)^{2} = 100

or, AC = 10 units

(By Pythagoras Theorem)

∴ sin A = 8 / 10, cos A = 6 / 10

and sin C = 6 / 10, cos C = 8 / 10

∴ sin A cos C + cos A sin C

= 8 / 10 x 8 / 10 + 6 / 10 x 6 / 10

= 64 / 100 + 36 / 100

= 100 / 100 = 1**Q.32. Find the area of the minor segment of a circle of radius 42 cm, if length of the corresponding arc is 44 cm. (3 Mark)****Ans. **Here, r = 42 cm

2πrθ / 360° = 44

Area of minor segment = area of sector – area of corresponding triangle

= 21(44 – 21√3) cm^{2}

= 21(44 – 36.37) cm^{2}

= 21 × 7.63 cm^{2}

= 160.23 cm^{2}**Q.33. The table below show the salaries of 280 persons : (3 Mark)****Calculate the median salary of the data.****Ans.**

N / 2 = 280 / 2 = 140.

Median class = 10 – 15.

= 10 + 5 / 133 (140 - 49)

= 10 + 5 x 91 / 133

= 13.42

Hence, Median salary is ₹ 13.42 thousand or ₹ 13420 (approx).**All questions are compulsory. In case of internal choices, attempt anyone.****Q.34. Find the value of a, b and c such that the numbers a, 9, b, 25, c are in A.P (5 Mark)****Ans. **Since a, 9, b, 25, c are in A.P.

Let common difference be d. a + d = 9 ...(i)

a + 3d = 25 ...(ii)

Solving (i) & (ii), we get

2d = 16 or d = 8

and a = 9 – 8 or a = 1

∴ Third term, b = a + 2d

= 1 + 2(8)

= 17

Also, fifth term, c = a + 4d

= 1 + 4(8)

= 33

So the AP is : 1, 9 ,17, 25 ,33**Q.35. Find the coordinates of the point which divide the line segment joining A(0, 5) and B(9, 2) into three equal parts. (5 Mark)ORWhich term of the Arithmetic Progression –7, –12, –17, –22, ... will be – 82 ? Is –100 any term of the A.P. ? **

P divides AB in the ratio of 1 : 2 and Q divides AB in the ratio 2 : 1.

By section formula

∴

x

∴ P(x

Similarly, by section formula

∴ x

∴ Q(x

Given A.P. is

– 7, – 12, – 17, – 22, ......

Here, first term, a

second term, a

third term, a

Common difference, d = a

= – 12 + 7 = – 17 + 12

= – 5 = – 5

∴ d = – 5

Let a

i.e., a

Since, a

∴ – 82 = – 7 + (n – 1)(– 5)

⇒ – 82 = – 7 – 5(n – 1)

⇒ 82 = 5n + 2

⇒ 5n = 80

⇒ n = 16

Hence, 16th term of A.P. is – 82.

Since, these numbers are not factor of 5, hence – 100 will not be a term in the given A.P.

1. Draw a line segment PO = 7 cm.

2. From the point O, draw a circle of radius = 3 cm.

3. Draw a perpendicular bisector of PO. Let M be the mid-point of PO.

4. Taking M as centre and OM as radius draw a circle.

5. Let this circle intersects the given circle at the point Q and R.

6. Join PQ and PR.

On measuring we get

PQ = PR = 6.3 cm

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