# Mathematics Unsolved Sample Paper Solutions CBSE Class 10

## Section - A

1. (b) (2, 5)

**Explanation:** Given A(0, 3) and B(6, 9) and let (x, y) be the coordinates of P. Since the line is trisected then AP : PB = 1 : 2

Using section formula,

So, coordinates of P are (2, 5).

2.(c) A = 52.5°, = 7.5°

**Explanation:** Given, tan (A + B) = √3=tan60°

⇒ A + B = 60° ...(i)

and tan (A – B) = 1 = tan 45°

or A – B = 45° ...(ii)

Adding (i) and (ii), we get 2A = 105°

A = 52.5°

Put A = 52.5° in equation (i), we get

52.5º + B = 60°

B = 60° – 52.5°

⇒ B = 7.5°

Hence A = 52.5° and B = 7.5°.

3. (b) – 4

**Explanation:**Given : n

^{th}term of an A.P.,

_{n}= 7 – 4n

_{n + 1}= 7 – 4(n + 1)

= 7 – 4n – 4 = 3 – 4n

So, common difference,

_{n + 1}– a

_{n}

= (3 – 4n) – (7 – 4n) = – 4

4. (c) 0.7

**Explanation:** Given, P (winning) = 0·3

So, P(losing) = 1 – 0·3 = 0·7

5. (a) 5.5

**Explanation:**

Class |
Frequency f_{i} |
x_{i} |
f_{i}x_{i} |

1—3 | 9 | 2 | 18 |

3—5 | 22 | 4 | 88 |

5—7 | 27 | 6 | 162 |

7—10 | 17 | 8·5 | 144.5 |

Σf_{i} = 75 |
Σf_{i}x_{i} = 412·5 |

_{i}= 75

_{i}x

_{i}= 412·5

We know,

Mean of given distribution is 5·5.

6.

(d) 30°

Explanation: In ΔABC, tan 60° =(h/x)

⇒ √3 = (h/x)

h = x√3 …(i)

In **Δ**ADB, tan θ =(h/3x)

⇒ tan θ = (x√3x/3x)

⇒ tan θ = (1/√3)

⇒ tan θ = tan 30°

⇒ θ = 30°

7. (b) 4

**Explanation:** A composite number is a natural number which has more than two factors. Here, factors of 4 = 1, 2, 4.

∴ 4 is a smallest composite number.

∴ 4 is a smallest composite number.

8. (c) 10 m

**Explanation:** Let, DC be the tower of height 15m. and AB be the height of pole

Now, in ΔAED, tan 30° =(DE/AE)

⇒(1/√3)=(DE/AE)

⇒ AE = √3 DE …(i)

In **Δ**BCD, tan 60° =(DC/BC)

⇒ √3 =(15/BC)

⇒ BC =(15/√3)

⇒ BC = 5 √3 …(ii)

As, AE = BC

∵ √3 DE = 5 √3 [using (i) and (ii)]

⇒ DE = 5 m

Height of pole, AB = EC = DC – DE

= 15 – 5 = 10 m.

Thus, height of the pole is 10m.

9. (a) 32

**Explanation:** Given : OA = OB = 10.5 cm

and ∠AOB = 60°

Perimeter of the sector = OA + OB + AB

10. (b) 13.5 m

**Explanation:** In ∆ABC,

11. (a) 3 median–2 mean

**Explanation:** The relationship between three measures of central tendencies is given by

Mode = 3 median – 2 mean.

12. (b) 2

**Explanation:** A circle can have maximum two parallel tangents.

13. (b) 30°

**Explanation:** Given,

⇒ α = 60°

and cos β = 0

cos β = cos 90°

⇒ β = 90°

Now, β – α = 90° – 60° = 30°

14. (b) 2r cm

**Explanation:** Here, maximum diameter of sphere is equal to the diameter of cylinder.

·.· Radius of cylinder = r

∴ Diameter of cylinder = Diameter of sphere = 2r

15. (a) 18

**Explanation:** Prime factors of 36 = 3 × 3 × 2 × 2

Prime factors of 54 = 3 × 3 × 2 × 3

Hence HCF (35, 54) = 3 × 3 × 2 = 18

16. (c) k ≠ 10

**Explanation:** Given system of equations are,

3x – y – 5 = 0

and 6x – 2y – k = 0

For the system to have no solution,

17. (a) (7/2)

**Explanation:** Here,

18. (d) no solution

Explanation: Given, equations are x + 2y + 5 = 0 and – 3x – 6y + 1 = 0

_{1}= 1, b

_{1}= 2, c

_{1}= 5

_{2}= – 3, b

_{2}= – 6, c

_{2}= 1

Hence, the pair of equations has no solution.

19. (c) Assertion is true but the Reason is false.

**Explanation:** Suppose the given system of equations has infinitely many solutions if

⇒ 3a = a + b

⇒ 2a – b = 0

Also clearly a = 4 and a + b = 12

⇒ b = 8

∴ 2a – b = 8 – 8 = 0

So, the assertion is true.

On the other hand, a pair of equations have a unique solution, if

So, reason is false. Hence correct option is (c).

20. (d) Assertion is false but the Reason is true.

**Explanation:**The highest power of x in the polynomial 4x

^{3}– x

^{2}+ 5x

^{4}+ 3x – 2 is 4.

Therefore, the degree of the polynomial p(x) is 4.

Hence, assertion is false but reason is true.

**(Any two)**

## Section - B

^{2}– 7x – 3

Put p(x) = 0

^{2}– 7x – 3 = 0

^{2}– 9x + 2x – 3 = 0

⇒ 3x(2x – 3) + 1(2x – 3) = 0

⇒ (3x + 1) (2x – 3) = 0

⇒ (3x + 1) = 0 or (2x – 3) = 0

**Hence Verified.**

22. Given A = (1, 4), B = (– 1, 2) and P = (x, y)

Also, AP = BP

on squaring both sides,

^{2}+ (4 – y)

^{2}= (x + 1)

^{2}+ (y – 2)

^{2}

^{2}– 2x + 16 + y

^{2}– 8y = x

^{2}+ 1 + 2x + y

^{2 + 4 – 4y }

– 2x – 8y + 16 = 2x – 4y + 4

⇒ – 2x – 2x – 8y + 4y = 4 – 16

⇒ – 4x – 4y = – 12

⇒ – 4(x + y) = – 12

⇒ x + y = 3. **Ans.**

23. Given : Sample space = {1, 2, 3, 4, 5, 6}.

(i) Prime numbers are 2, 3, 5.

∴ P(Prime number) =(3/6)=(1/2) **Ans.**

(ii) Numbers divisible by 2 are 2, 4, 6.

∴ P(number divisible by 2) =(3/6)=(1/2) **Ans.**

**OR**

Total number of possible outcomes = 1, 2, 3, 4, 5, 6.

Let E be the event of getting an even number.

Then, the favourable outcomes are 2, 4, 6.

Hence, the total number of favourable outcomes = 3

∴ P(E) =(3/6)=(1/2)** Ans.**

24. Consider part of curve between x = – 2 and x = 2

The curve cuts X-axis at x = – 1 and x = 1

Hence, required zeroes are – 1 and 1. **Ans.**

**OR**

Given equations are, x – y = – 1 ...(i)

and 3x + 2y = 12 ...(ii)

Multiplying equation (i) with 2, we get

2x – 2y = – 2 ...(iii)

Adding equations (ii) and (iii), we get

5x = 10

⇒ x = 2

Putting x = 2 in equation (i), we get

2 – y = – 1

⇒ – y = – 3

⇒ y = 3.

Hence, the values of x and y are 2 and 3 respectively. **Ans.**

25. Since, jacks, queens and kings of red colour are removed. Then,

Total number of possible outcomes = 52 – 6 = 46

_{1}be the event of getting a black king.

∴ Favourable outcomes = king of spade and king of club.

No. of favourable outcomes = 2

_{1}) =(2/46)=(1/23).

**Ans.**

_{2}be the event of getting a card of red colour.

∴ Favourable outcomes = 10 cards of heart and 10 cards of diamond.

No. of favouable outcomes = 20

_{2}) =(20/46)=(10/23).

**Ans.**

_{3}be the event of getting a card of black colour

∴ Favourable outcomes = 13 cards of spade and 13 cards of club.

No. of favourable outcomes = 26

P(E_{3}) =(26/46)=(13/23) **Ans.**

## Section - C

26. Let O be the centre of the given circle. Suppose the tangent at P meets BC at Q. Join BP.

To Prove : BQ = QC

Proof : Since, tangent at any point of circle is perpendicular to radius through the point of contact.

∴ ∠ABC = 90°

In ∆ABC, ∠1 + ∠5 = 90° [y angle sum property, ∠ABC = 90°]

and, ∠3 = ∠1

[Angle between tangent and the chord equals angle made by the chord in alternate segment]

Now, ∠3 + ∠5 = 90° ...(i)

Also, ∠APB = 90° [Angle in semi-circle]

∴ ∠3 + ∠4 = 90° [∠APB + ∠BPC = 180°, linear pair] ...(ii)

From (i) and (ii), we get

∠3 + ∠5 = ∠3 + ∠4

or ∠5 = ∠4

∴ PQ = QC [Sides opposite to equal angles are equal]

Also, QP = QB

[Tangents drawn from an internal point to a circle are equal]

∴ QB = QC **Hence Proved.**

27.

= R.H.S. **Hence Proved.**

L.H.S. = (1 + cot A – cosec A) (1 + tan A + sec A)

= 2

= R.H.S. **Hence Proved.**

28. (i) House = (2, 4), Bank = (5, 8), School = (13, 14) and Office = (13, 21)

∴ Distance between house and office

(ii) Distance covered from house to bank =

∴ Extra distance covered = (5 + 10 + 7) – √410

= (22 −√ 410) units . **Ans.**

29. Given : Quadrilateral ABCD, AC and BD meet at O.

and (AO/ OC) = (BO/ OD)

To prove : ABCD is a trapezium.

Proof : (AO /OC) = (BO/ OD) [Given]

∠1 = ∠2 [Vertically opposite angles]

⇒ ∆AOB ~ ∆COD [SAS criterion]

⇒ ∠3 = ∠4 [Corresponding angles of similar triangles]

⇒ AB || DC [∵ A pair of alternate angles are equal]

⇒ ABCD is a trapezium. **Hence Proved.**

**OR**

Given : ∠A = ∠D = 90°

To prove : AE × EC = BE × ED

Proof : ∠A = ∠D [Each 90°]

∠1 = ∠2 [Vertically opposite angles]

⇒ ∆AEB ~ ∆DEC

⇒ AE × EC = EB × DE. **Hence Proved.**

30. Given : Hexagon ABCDEF in which a circle is circumscribed.

Since tangents to a circle from an exterior point are equal in length.

AP = AU

BP = BQ

CR = CQ

DR = DS

ET = ES

FT = FU

On adding above equations,

(AP + BP) + (CR + DR) + (ET + FT ) = (AU + FU) + (BQ + CQ) + (DS + ES)

⇒ AB + CD + EF = AF + BC + DE. **Hence Proved.**

31. Given, the dimensions of cuboid are 10 cm, 5 cm and 4 cm.

Radius of conical depression = 0·5 cm

Depth of conical depression = 2·1 cm

The edge of cubical depression = 3 cm

^{3}= 200 cm

^{3}

^{2}h

^{3}

^{3}

^{3}

Volume of wood left =

^{3}= 170·8 cm

^{3}.

**Ans.**

## Section - D

32. (i) Given, ABCD is a square of side 14 cm.

^{2}= 196 cm

^{2}

Now, 2 × Diameter of circle = Side of square

⇒ Radius of a circle, r =(7/2)cm

∴ Shaded area = Area of square – 4 × Area of a circle

^{2}= 42 cm

^{2}. Ans.

(ii) Given : Radius of circle, r = 42 cm

Length of arc = 44 cm

33. Given :

^{2}+ x(a + b) = – ab

^{2}+ x(a + b) + ab = 0

^{2}+ ax + bx + ab = 0

⇒ x(x + a) + b(x + a) = 0

⇒ (x + b) (x + a) = 0

⇒ x = – a, – b. **Ans.**

**OR**

(i) Let the fixed charges = ₹ x

Let the charge per day = ₹ y

Latika kept a book for 6 days

She pays fixed charges for 2 days and additional charges for 4 days.

x + 4y = 22 ...(i)

Anand kept book for 4 days.

He pays fixed charges for 2 days and additional charges for 2 days.

x + 2y = 16 ...(ii)

Subtracting equation (ii) from (i), we get

⇒ (x + 4y) – (x + 2y) = 22 – 16

⇒ x + 4y – x – 2y = 6

⇒ 2y = 6

y = 3

Therefore, the charges per day is ₹ 3. **Ans.**

(ii) Put the value of y in equation (i),

x + 4y = 22

⇒ x + 4(3) = 22

⇒ x + 12 = 22

⇒ x = 22 – 12

⇒ x = 10

Thus, fixed charge for first two days is ₹ 10. Ans.

(iii) The graph of the situation is given below :

34. Let BC be the pedestal, DC be the statue and A be the point of observation.

Let BC = h, AB = x.

In ∆ABC, (BC/AB)=tan45°

(h/x)=1

⇒ h = x …(i)

In ∆ABD, (BD/AB)=tan60°

⇒ (h x +1.6/x)=√3

⇒ (h+1.6/√3)=x …(ii)

From equations (i) and (ii), h =(h+1.6/√3)

⇒ h √3 − h = 1·6

⇒ h(√ 3 − 1) = 1·6

⇒ h = 2·1856 ≈ 2·19 m. **Ans.**

35. Let the usual speed be x km/hr.

We know, Time = (Distance/Speed)

So, Usual time taken =(1500/x)hr.

Increased speed = (x + 100) km/hr

Thus, New time taken =(1500/x+100)

Now, according to the question

⇒ x(x + 100) = 300000

^{2}+ 100x – 300000 = 0

^{2}+ 600x – 500x – 300000 = 0

⇒ x(x + 600) – 500(x + 600) = 0

⇒ (x + 600) (x – 500) = 0

x = 500 and – 600

As speed cannot be negative, so the original speed of the plane is 500 km/hr. **Ans.**

**OR**

Let total time be n minutes.

Since, policeman runs after 1 minute so he will catch the thief in (n –1) minutes.

Total distance covered by thief = 100 m/minute × n minute

= (100 n) m

Now, total distance covered by the policeman

= (100)m + (100 + 10)m + (100 + 10 + 10)m + ......+ (n – 1) terms

i.e., 100 + 110 + 120 + ..... + (n – 1) terms

_{n – 1}=(n-1/2)[2×100+(n-2)10]

Now, (n-1/2)[200+(n-2)×10]=100n.

⇒ (n – 1) (200 + 10n – 20) = 200 n.

^{2}– 10n + 20 – 20n = 200n

^{2}– 30n – 180 = 0

^{2}– 3n – 18 = 0

^{2}– (6 – 3) n – 18 = 0

^{2}– 6n + 3n – 18 = 0

⇒ n(n – 6) + 3(n – 6) = 0

⇒ (n + 3) (n – 6) = 0

∴ n = 6 or n = – 3 (neglect)

Hence, policeman will catch the thief in (6 – 1) i.e., 5 minutes. **Ans.**

## Section - E

36. (i) Each guest will get number of apples =(36/12)=3

Each guest will get no. of bananas =(60/12)=5

Out of 36 apples and 60 bananas each guest will get 3 apples, 5 bananas.** Ans.**

(ii) H.C.F. (36, 42, 60) = 6

So, fruits will be equally distributed among 6 guests. So she could invite max. of 6 guests **Ans.**

(iii) Now Vedika has 30 apples, 60 bananas and 45 mangoes. H.C.F. (30, 45, 60) = 15. **Ans.**

**OR**

Total apples =(36/6)=6, bananas =(60/6)=10 and mangoes = 7. So, total no. of fruits will each guest get 6 + 10 + 7 = 23 fruits. **Ans.**

37. (i) The upper limit of a class and the lower class of its succeeding class differ by 1. **Ans.**

(ii) Here, class intervals are in inclusive from. So, we first convert them in exclusive form. The frequency distribution table in exclusive form is as follows:

Class interval |
Frequency (f _{i}) |
Frequency cumulative c.f. |

199.5 – 209.5 | 4 | 4 |

209.5 – 219.5 | 14 | 18 |

219.5 – 229.5 | 26 | 44 |

229.5 – 239.5 | 10 | 54 |

239.5 – 249.5 | 6 | 60 |

_{i}= N = 60

∴ (N /2) = (60 /2) = 30

Now, the class interval whose cumulative frequency is just greater than 30 is 219.5 – 229.5

∴ Median class is 219.5 – 229.5 **Ans.**

**OR**

= 224.12

∴ Median of the distance travelled is 224.12 km. **Ans.**

(iii) We know, Mode = 3 Median – 2 Mean

= 224.29 km. **Ans.**

38. (i) ∠OCA = 90°

[Since, radius at the point of contact is perpendicular to tangent]

In ΔOAC, ∠OCA + ∠OAC + ∠AOC = 180° (Angle sum property)

⇒ 90° + ∠OAC + ∠AOC = 180°

⇒ ∠OAC + ∠AOC = 90°

⇒ x + y = 90°. **Ans.**

(ii) Since, OB ⊥ PB [Since, radius at the point of contact is perpendicular to tangent]

and ∠PBA = 50° (Given)

∴ ∠OBA = 90° – 50°

= 40°

Also, OA = OB [Radii of circle]

∴ ∠OAB = ∠OBA

= 40° [Angle opposite to equal sides are equal]

**OR**

Here, ∠ABC = 90° (Angle in a semicircle)

So, in ∆ABC, ∠BAC = 180° – 90° – 55° = 35°

Also, ∠OAT = 90°

⇒ ∠BAT + ∠OAB = 90°

⇒ ∠BAT = 90° – 35°

= 55° **Ans.**

(iii) Here, ∠PAB = 72°

∴ ∠OAB = 90° – 72° = 18°

Also, ∠AOB = 132° [Given]

Now, in DOAB, ∠ABC = 180° – 132° – 18°

= 30°. **Ans.**

#### CBSE 36 Sample Question Papers

All Subject Combined

All Subjects Combined for Class 10 Exam 2023

The dot mark ◉ field are mandatory, So please fill them in carefully**To download the Sample Paper (PDF File), Please fill & submit the form below.**