The document Mathematics: CBSE Sample Question Paper (2020-21) (Standard) - 2 Notes | EduRev is a part of the Class 10 Course CBSE Sample Papers For Class 10.

All you need of Class 10 at this link: Class 10

**Class X****Mathematics – Standard****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 two sections - I and II.

2. Section I has 16 questions of 1 mark each. Internal choice is provided in 5 questions.

3. Section II has 4 questions on case study. Each case study has 5 case-based sub-parts. An examinee is to attempt any 4 out of 5 sub-parts.

**Part – B : **

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

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

3. Section V, 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**

**Sections - I**

**Section I has 16 questions of 1 mark each. Internal choice is provided in 5 questions.**

**Q.1. Calculate the largest number which divides 70 and 125, leaves remainders 5 and 8, respectively. (1 Mark)****OR****If p is a prime number, then find LCM of p, p ^{2} and p^{3}. (1 Mark)**

OR

p

p = p × 1

p

Required LCM = p × p × p = p

OR

A number is chosen at random from the numbers – 3, – 2, – 1, 0, 1, 2, 3. What will be the probability that square of this number is less than or equal to 1. (1 Mark)

No. of all possible outcomes = 7

No. of favourable outcomes = 3

The numbers whose square is ≤ 1 = – 1, 0, 1

∴ Required probability = 3 / 7

Here, a = −2 and b = +5

∴ α + β = −2 + 5 = 3 and

αβ = −2 × 5 = −10

So, required polynomial is x

If we multiply this polynomial by any real number, let 5 and 2, we get 5x

6x – 3y + 10 = 0

2x – y + 9 = 0

Represents what kind of lines.

So, the system of linear equations is inconsistent (no solution) and graph will be a pair of parallel lines.

2x

3x

x(5 + 2√6) + 3 = 0

It is not of the form of ax

So it is not a quadratic equation.

Thus, the 30th term is t

∴ ∠AOB + ∠COD = 180°

125° + ∠COD = 180°

∠COD = 180° – 125° = 55°**OR**

∠TOQ = 180° – 70° = 110°

(angle of supplementary)

Then, ∠TRQ = 1 / 2 ∠TOQ

(angle at the circumference of the circle by same arc)

= 1 / 2 x 110° = 55°.**Q.9. If the circumference of a circle and the perimeter of a square are equal, then find the relation between area of circle and area of square. (1 Mark)****Ans. **According to question,

Circumference of a circle = Perimeter of square

Let ‘r’ and ‘a’ be the radius of circle and side of square respectively.

2πr = 4a

22 / 7 r = 2a

11r = 7a

r = 7a / 11 ..(i)

Area of circle, A_{1} = πr^{2}

From equation (i), we have

A_{1} = π(7a / 11)^{2}

= 22 / 7(49a^{2} / 121)

= 14a^{2} / 11 ...(ii)

Area of circle, A_{2} =a^{2} ...(iii)

From equation (ii) and (iii), we have

A_{1} > A_{2}

Area of circle is greater than the area of square.**Q.10. During conversion of a solid from one shape to another, what will be the volume of new shape? (1 Mark)****Ans. **During reshaping a solid, the volume of new solid will be equal to old one or remains unaltered.**Q.11. In the figure of DABC, the points D and E are onthe sides CA, CB respectively such that DE || AB, AD = 2x, DC = x + 3, BE = 2x – 1 and CE = x. Then, the value of x is ...................... . (1 Mark)****Ans. **

Using basic proportionality theorem

or, 5x = 3 or, x = 3 / 5**Detailed Answer :**

In AB C, DE || AB

Then, CD / CA = CE / CB

or,

or,

or,

or, (x + 3)(3x – 1) = x(3x + 3)

or, 3x^{2} – x + 9x – 3 = 3x^{2} + 3x

or, 8x – 3 = 3x

or, 8x – 3x = 3

or, 5x = 3

∴ x = 3 / 5**Q.12. Find the value of sin ^{2}60° + 2tan 45° – cos^{2}30°. (1 Mark)**sin

OR

If sin A = 3 / 4, then find value of sec A. (1 Mark)

Ans.

(√3 / 2)

= 2

∴ sec A = 4 / √7

Side of square = diameter of circle = 8 cm

∴ Radius of circle, r = 8 / 2 = 4 cm

Area of circle = πr

= π × 4 × 4 = 16πcm

πr

∴ r = 7 m

2πrh = 264

or, 2 x 22 / 7 x 7 x h = 264

or, h = 6 m

∴ h / 2r = 6 / 14 = 3 / 7

Hence, h : d = 3 : 7

OR

Write the co-ordinates of a point P on x-axis which is equidistant from the points A(– 2, 0) and B (6, 0). (1 Mark)

or 32 + k

k = ± 4

Using distance formula :

Point on x-axis is (2, 0)

Let (x, 0) be equidistant from (– 2, 0) and (6, 0) by distant formula :

On squaring both sides,

(x + 2)

⇒ x

⇒ 4x + 12x = 36 – 4

⇒ 16x = 32

⇒ x = 2

Hence, the required point P(x, 0) = (2, 0)

Find the sum of lower limits of median class and modal class.

The modal class is the class having the maximum frequency.

The maximum frequency 20 belongs to class (15 – 20).

Here, n = 66

So, n / 2 = 66 / 2 = 33

33 lies in the class 10–15.

Therefore, 10 – 15 is the median class.

So, sum of lower limits of (15 – 20) and (10 – 15) is (15 + 10) = 25.

**Section-II**

**Case study based questions are compulsory. Attempt any four sub parts of each question. Each subpart carries 1 mark**

**Q.17. ****Case Study based - 1 :**

In a room, 4 friends are seated at the points A, B, C and D as shown in figure. Reeta and Meeta walk into the room and after observing for a few minutes Reeta asks Meeta.__Asks Meeta__

**(a) What is the position of A? (1 Mark)****(i) (4, 3) (ii) (3, 3) (iii) (3, 4) (iv) None of these**

A = (3, 4)

**(b) What is the middle position of B and C? (1 Mark)****(i) (15 / 2, 11 / 2)****(ii) (2 / 15, 11 / 2)****(iii) (1 / 2, 1 / 2)****(iv) None of these****Ans. **(i)**Solution. **Mid point of B and C :

[∵ Co-ordinates of B = (6, 7) and C = (9 , 4)]

**(c) What is the position of D? (1 Mark)****(i) (6, 0) ****(ii) (0, 6) ****(iii) (6, 1) ****(iv) (1, 6)****Ans. **(iii)**Solution. **Point D lies at x = 6 and y = 1

D = (6, 1)

**(d) What is the distance between A and B? (1 Mark)****(i) 3√2 ****(ii) 2√3 ****(iii) 2√2 ****(iv) 3√3****Ans.** (i)**Solution. **Since, A = (3, 4) and B = (6, 7)

Using distance formula

= √18 = 3√2 unit

**(e) What is the equation of line CD? (1 Mark)****(i) x – y – 5 = 0 ****(ii) x + y – 5 = 0 ****(iii) x + y + 5 = 0 ****(iv) x – y + 5 = 0****Ans. **(i)**Solution.** Equation of line CD = Equation of line through C(9, 4) and D(6, 1)

i.e.,

or, y – 4 = (x – 9)

or, x – y – 5 = 0**Q.18. Case Study based - 2 : **

Seema placed a lighedbulb at point O on the ceiling and directly below it placed a table. Now, she put a cardboard of shape ABCD between table and lighted bulb. Then a shadow of ABCD is casted on the table as A'B'C'D' (see figure). Quadrilateral A'B'C'D' in an enlargement of ABCD with scale factor 1 : 2, Also, AB = 1.5 cm, BC = 25 cm, CD = 2.4 cm and AD = 2.1 cm; ∠A = 105°, ∠B = 100°, ∠C = 70° and ∠D = 85°.

**(a) What is the measurment of angle A'? (1 Mark)****(i) 105° ****(ii) 100° ****(iii) 70° ****(iv) 80°****Ans. **(i)**Solution.** Quadrilateral A'B'C'D' is similar to ABCD.

∴ ∠A' = ∠A

⇒ ∠A' = 105°

**(b) What is the length of A'B'? (1 Mark)****(i) 1.5 cm ****(ii) 3 cm ****(iii) 5 cm ****(iv) 2.5 cm****Ans. **(ii)**Solution. **Given scale factor is 1 : 2

∴ A'B' = 2 AB

⇒ A'B' = 2 × 1.5 = 3 cm

**(c) What is the sum of angles of quadrilateral A'B'C'D'? (1 Mark)****(i) 180° ****(ii) 360° ****(iii) 270° ****(iv) None of these****Ans. **(ii)**Solution. **Sum of the angles of quadrilateral A'B'C'D' is 360°

**(d) What is the ratio of sides A'B' and A'D'? (1 Mark)****(i) 5 : 7 ****(ii) 7 : 5 ****(iii) 1 : 1 ****(iv) 1 : 2****Ans. **(i)**Solution. **A'B' = 3 cm and A'D' = 2 AD

= 2 × 2.1 = 4.2 cm

∴ A'B' / A'D' = 3 / 4.2 = 30 / 42

= 5 / 7 = 5 : 7

**(e) What is the sum of angles of C' and D'? (1 Mark)****(i) 105° ****(ii) 100° ****(iii) 155° ****(iv) 140°****Ans. **(iii)**Solution. **C' = ∠C = 70°

and ∠D' = ∠D = 85°

∴ ∠C' + ∠D' = 70° + 85° = 155°**Q.19. Case Study based - 3 : **

An electrician has to repaired and electric fault on the pole of height 5 cm. She needs to reach a point 1.3 m below the top of the pole to undertake the repair work (see figure)

**(a) What is the length of BD? (1 Mark)(i) 1.3 m (ii) 5 m (iii) 3.7 m (iv) None of these**

So, BD = AD – AB = (5 – 1.3) m = 3.7 m

**(b) What should be the length of Ladder, when inclined at an angle of 60° to the harizontal? (1 Mark)**

(ii) 3.7 / √3 m

**(iii) 3.7 m (iv) 7.4 m**

∴ sin 60° = BD / BC

⇒ √3 / 2 = 3.7 / BC

⇒ BC =

⇒ BC = 4.28 m (approx.)

**(c) How far from the foot of pole should she place the foot of the ladder? (1 Mark)****(i) 3.7 ****(ii) 2.14 ****(iii) 1 / √3****(iv) None of these****Ans. **(ii)**Solution. **In DBDC,

∴ cot 60° = DC / BD

⇒ 1 / √3 = DC / 3.7

⇒ DC = 3.7 / √3

⇒ DC = 2.14 m (approx.)

**(d) If the horizontal angle is changed to 30°, then what should be the length of the ladder? (1 Mark)****(i) 7.4 m ****(ii) 3.7 m ****(iii) 1.3 m ****(iv) 5 m****Ans. **(i)**Solution. **In ΔBDC,

∴ sin 60° = BD / BC

⇒ 1 / 2 = 3.7 / BC

⇒ BC = 3.7 × 2 = 7.4 m

**(e) What is the value of ∠B? (1 Mark)****(i) 60° ****(ii) 90° ****(iii) 30° ****(iv) 180°****Ans. **(iii)**Solution. **In ΔADC, angle D is 90°.

So, by angle sum property.

∠B + ∠D + ∠C = 180°

or, ∠B = 180° – (90° + 60°)

= 30°**Q.20. Case Study based - 4 :**

The weights (in kg) of 50 wrestlers are recorded in the following table :

**(a) What is the upper limit of modal class. (1 Mark)****(i) 120 ****(ii) 130 ****(iii) 100 ****(iv) 150****Ans. **(ii)**Solution. **Modal Class = 120 – 130

Upper Unit = 130

**(b) What is the mode of the given data (1 Mark)(i) 21 (ii) 50 (iii) 25 (iv) 80**

**(c) How many wrestlers weights have more than 120 kg weight? (1 Mark)****(i) 32 ****(ii) 50 ****(iii) 16 ****(iv) 21****Ans. **(i)**Solution. **No. of wrestlers with more than 120 kg weight = 21 + 8 + 3 = 32

**(d) What is the class mark for class 130 – 140? (1 Mark)(i) 105 (ii) 125 (iii) 135 (iv) 145Ans. **(iii)

Solution.

= 270 / 2 = 135

**(e) Which method is more suitable to find the mean of the above data? (1 Mark)**

(i) Direct method

(ii) Assumed mean method

(iii) Step-Deviation method

(iv) None of these**Ans. **(iii)

**Part - B**

**Section - III**

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

**Q.21. Write the denominator of the rational number 257 / 500 in the form 2 ^{m} × 5^{n}, where m and n are non-negative integers. Hence write its decimal expansion without actual division. (2 Mark)**

= 2

Decimal expansion =

= 0.514

∴ AC = CB

(By isosceles triangle property)

But, AD = BE (Given) ...(i)

Or, AC – AD = BC – BE

∴ CD = CE ...(ii)

Dividing equation (ii) by (i),

CD / AD = CE / BE

By converse of BPT,

DE || AB.

Given, AE = 2 / 3 AB = 2 / 3 x 6 = 4 km

In right angled triangle ADE,

DE^{2} = (3)^{2} + (4)^{2}

or, DE^{2} = 25

∴ DE = 5 km.**Q.23. If sin (A + B) = 1 and sin (A – B) = 1 / 2, 0 ≤ A + B = 90° and A > B, then find A and B. (2 Mark)ORExpress : sin A and tan A in terms of sec A. (2 Mark)**

or, A + B = 90º ...(i)

sin (A – B) = 1 / 2 = sin 30º

or, A – B = 30º ...(ii)

Solving eqns. (i) and (ii),

A = 60º and B = 30º

or,

AE be observer = 1.5 m

BD is the tower = 30 m

∠BAC = θ, BC = 30 – 1.5 = 28.5 m

In ΔBAC, BC / AC = tan θ

⇒ 28.5 / 28.5 = tan θ

⇒ tan θ = 1 = tan 45°

⇒ θ = 45°

Hence, the angle of elevation is 45°.**OR**

Let BD = x m and DC = y m.

From ΔABD, ∠D = 90°

7√3 / x = tan 30°

⇒ 7√3 / x = 1 / √3

⇒ 7√3 x √3

= 21 m

From ΔADC, ∠D = 90°

7√3 / y = tan 60°

⇒ 7√3 = y√3

⇒ y = 7 m,

BC = BD + DC

= 21 + 7 = 28 m.

Hence, the value of BC = 28 m.**Q.25. A child prepares a poster on "save water" on a square sheet whose each side measures 50 cm. At each corner of the sheet, she draws a quadrant of radius 15 cm in which she shows the ways to save water. At the centre, she draws a circle of diameter 21 cm and writes a slogan save water in it. Find the area of the remaining sheet. (2 Mark)****Ans. **Side of square= 50 cm

∴ Area of square = 50 × 50 = 2500 cm^{2}

Radius of quadrant = 15 cm.

Area of 4 quadrants = 4 x 1 / 4 πr^{2} = πr^{2}

= π × 15 × 15

= 22 / 7 x 225

= 707.14 cm^{2}

Area of circle = πr^{2}

= 22 / 7 x (21 / 2)^{2}

= 22 / 7 x 21 / 2 x 21 x 2

= 346.5 cm^{2}

Area of remaining sheet = Area of square – 4(area of quadrant) – Area of circle

= 2500 – 707.14 – 346.5

= 1446.36 sq. cm**Q.26. A teacher took a surprise test of maths. He observes the marks of five students of class. He observes the median is 45.5 and mode is 50.5. So find the mean of the marks of five students using an empirical formula. (2 Mark)ORFind the mode of the following frequency distribution (2 Mark)**

Median = 45.5

3 Median = Mode + 2 Mean

⇒ 3 × 45.5 = 50.5 + 2 Mean

⇒ Mean =

Hence, Mean = 43.

Maximum frequency = 16,

Modal class is 30 – 40

Mode =

= 36

**Section - IV**

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

**Q.27. ****Sum of the ages of a father and the son is 40 years. If father ’s age is three times that of his son, then find their respective ages. (3 Mark)ORA part of monthly hostel charge is fixed and the remaining depends on the number of days one has taken food in the mess. When Swati takes food for 20 days, she has to pay ₹ 3,000 as hostel charges whereas Mansi who takes food for 25 days has to pay ₹ 3,500 as hostel charges. Find the fixed charges and the cost of food per day. (3 Mark)**

Then, x + y = 40 ...(i)

and x = 3y ...(ii)

By solving eqns. (i) and (ii), we get

x = 30 and y = 10

Thus, the ages of father and son are 30 years and 10 years.

Let fixed charge be x and per day food cost be y

Then, x + 20y = 3000 ...(i)

and x + 25y = 3500 ...(ii)

Subtracting (i) from (ii), we get

y = 100 Substituting this value of y in (i), we get

x + 20(100) = 3000 x = 1000

∴ x = 1000 and y = 100

Hence, fixed charge and cost of food per day are ₹ 1,000 and ₹ 100.**Q.28. The ninth term of an A.P. is equal to seven times the second term and twelfth term exceeds five times the third term by 2. Find the first term and the common difference. (3 Mark)ORThe sum of first n terms of three arithmetic progressions are S**

Given, a

or, a + 8d = 7(a + d) ...(i)

a + 8d = 7a + 7d – 6a + d = 0 ...(iii)

and a

Again, a + 11d = 5(a + 2d) + 2 ...(ii)

a + 11d = 5a + 10d + 2 – 4a + d = 2 ...(iv)

Subtracting (iv) from (iii), we get – 2a = – 2

or, a = 1

From (iii),

– 6 + d = 0

d = 6

Hence, first term = 1 and common difference = 6

Since, S

S

and S

or, S

Also,

= n / 2[2n] = n

and

= n(3n - 1) / 2

Now,

= n[4n] / 2

= 2n

Hence Proved.

(i) Find the distance travelled by the balloon during the interval.

(ii) Which mathematical concept is used in the above problem ?

In ΔOLP, tan 60° = PL / OL

⇒

⇒ √3 = 87 / OL

⇒ OL = 87 / √3

In ΔOMQ, tan 30° = QM / OM

⇒

⇒ OM = 87√3

∴ Distance travelled by the balloon,

PQ = LM = OM – OL**(ii) **Height and distance.**Q.30. If (x ^{2} + y^{2})(a^{2} + b^{2}) = (ax + by)^{2}. Prove that x / a + y / b. (3 Mark)**Given, (x

Ans.

⇒ x

⇒ x

⇒ (xb – ya)

[∵ (a - b)

⇒ xb = ya

∴ x / a + y / b

(i) If one ball is drawn at random from the bag, what is the probability that it is not red ?

(ii) If 2 more red balls are put in the bag, the probability of drawing a red ball will be 9 / 8 times the probability of drawing a red ball in the first case. Find the value of x.

Total number of Red balls = x + 2

P(red balls) =

Now, According to the question,

⇒ 180x = 144x + 288

⇒ 36x = 288

⇒ x = 288 / 36 = 8

**Q.32. In the given figure, OP is equal to the diameter of a circle with centre O and PA and PB are tangents. Prove that ABP is an equilateral triangle. (3 Mark)****Ans. ****Construction : **Join A to B.

We have,

OP = diameter

⇒ OQ + QP = diameter

⇒ Radius + QP = diameter

⇒ OQ = PQ = radius

Thus, OP is the hypotenuse of right angled ∆AOP.

So, In ∆AOP, sin θ = AO / OP = 1 / 2

θ = 30°

Hence, ∠APB = 60°

Now, in ∆ABP, AP = PB

So, ∠PAB = ∠PBA = 60°

∴ ∆APB is an equilateral triangle.**Q.33. If bcos θ = a, then prove that cosec θ + cot θ =****(3 Mark)****Ans. **

Given, cos θ = a / b

AC^{2} = AB^{2} – BC^{2}

Hence proved.

**Section - V**

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

**Q.34. Solvewhere a + b ≠ 0. (5 Mark)**

Check whether the equation 5x

Given,

⇒

⇒

⇒

⇒

⇒ x(a + b+ x) = – ab

⇒ x

⇒ (x + a)(x + b) = 0

⇒ x = – a or x = – b

Discriminant ⇒ b

Here, a = 5, b = (– 6) and c = (– 2)

Then, b^{2} – 4ac = (– 6)^{2} – 4 × 5 × – 2

= 36 + 40 = 76 > 0

So the equation has real and two distinct roots.

Again, 5x^{2 }– 6x = 2 (dividing both the sides by 5)

On adding square of the half of coefficient of x

⇒

⇒

⇒**Verification :**

Similarly,

Hence, Verified.**Q.35. In the given figure, D and E trisect BC. Prove that 8AE ^{2} = 3AC^{2} + 5AD^{2}. (5 Mark)**

BE = 2x

and BC = 3x

Now, in ΔABE,

AE

= AB^{2} + 4x^{2}, ...(i)

In ΔABC,

AC^{2} = AB^{2} + BC^{2} = AB^{2} + 9x^{2}

In ΔADB, and AD^{2} = AB^{2} + BD^{2} = AB^{2} + x^{2}

Now, on multiplying (i) by 8,

8AE^{2} = 8AB^{2} + 32x^{2} ...(ii)

and 3AC^{2} + 5AD^{2} = 3(AB^{2} + 9x^{2}) + 5 (AB^{2} + x^{2})

= 3AB^{2} + 27x^{2} + 5AB^{2} + 5x^{2}

= 8AB^{2} + 32x^{2} = 8 (AB^{2} + 4x^{2})

∴ 3AC^{2} + 5AD^{2} = 8AE^{2}. [From eqns. (i)]

Hence proved.**Q.36. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid. (π = 22 / 7) (5 Mark)**

Total volume =

Height of the cylinder(h)= (20 – 7) cm = 13 cm

Radius of circular part(r) = 7 / 2 cm

Volume of solid = Volume of cylinder + 2 × Volume of hemisphere

= 77 / 2(53 / 3)cm

= 680.17 cm

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

116 docs