Students can access the CBSE Sample Papers for Class 10 Maths with Solutions and marking scheme Term 2 Set 2 will help students in understanding the difficulty level of the exam.

## CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 2 with Solutions

Time Allowed: 2 Hours

Maximum Marks: 40

General Instructions:

- The question paper consists of 14 questions divided into 3 sections A, B, C.
- All questions are compulsory.
- Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
- Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
- Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one question. It contains two case study based questions

SECTION – A

(12 Marks)

Question 1.

If x = \(\frac{2}{3}\) and x = -3 are the roots of quadratic equation ax^{2} + 7x + b = 0, find the values of a and b.

OR

Solve for x:

\(\sqrt{2 x+9}+x=13\)** (2)
**Answer:

Since, \(\frac{2}{3}\) and -3 are the roots of given quadratic equation,

Hence, the respective values of a and b are 3 and -6.

OR

We have,

⇒ (x – 20) (x – 8) = 0

⇒ x = 20, 8

Hence, the values of x are 8 and 20.

Question 2.

The mean of eight numbers is 23. One more number is added to it. The new mean reduced to 22. Find the number added. **(2)
**Answer:

Given: Mean of eight numbers = 23

Sum of eight numbers = 23 x 8

= 184

Let the newly added number be x.

Now, there are nine numbers in the set of data.

So, Sum of nine numbers = 184 + x

∴ Mean of nine numbers = \(\frac{184+x}{9}\)

But, mean of nine numbers = 22 [Given]

Hence, the number odded is 14.

Question 3.

In the given figure, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 40°. If AT is a tangent to the circle at point A, find ∠BAT. **(2)
**

Answer:

Given: ∠ACB = 40°

Since, angle in a semi-circle is a right angle

∴ ∠ABC = 90°

So, in ΔABC

∠BAC = 180° – ∠ABC – ∠ACB

= 180° – 90° – 40°

= 50°

Now,

∠OAT = 90° [∵ Tangent! Radius]

⇒ ∠OAB + ∠BAT = 90°

⇒ 50° + ∠BAT = 90°

⇒ ∠BAT = 90° – 50° = 40°

Question 4.

Given below is a cumulative frequency distribution tabLe. Find the missing vaLues a, b, c and d.

Class interval | Frequency | Cumulative Frequency |

0 – 10 | 17 | 17 |

10 – 20 | a | 45 |

20 – 30 | 32 | 77 |

30 – 40 | b | 101 |

40 – 50 | 19 | c |

50 – 60 | d | 130 |

Answer:

Class interval | Frequency | Cumulative Frequency |

0 – 10 | 17 | 17 |

10 – 20 | a = 45 – 17 = 28 | 45 |

20 – 30 | 32 | 77 |

30 – 40 | b = 101 – 77 = 24 | 101 |

40 – 50 | 19 | c = 19 + 101 = 120 |

50 – 60 | d= 130 – 120 = 10 | 130 |

**Concept Applied
**The cumulative frequency of a doss is equaL to the sum of the frequency of the class arid the cumulative frequency of the preceding class.

Question 5.

The 19^{th} term of an A.P. is equal to three times its sixth term. If its 9^{th} term is 19, find the A.P. **(2)
**Answer:

Let a be the first term and d be the common difference of the A.P.

Now, a

_{19}= 3a

_{6}

⇒ a + 18 d = 3 (a + 5 d)

⇒ a + 18 d = 3 a + 15 d

⇒ 2a – 3d = 0 ………………………….. (i)

Also, a

_{9}= 19

⇒ a + 8d = 19 ………………………….. (ii)

Solving equations (i) and (ii), we get

a = 3, d = 2

So, the A.P. is 3, 5, 7, 9 …

Question 6.

If the total surface area of a solid hemisphere is 462 cm^{2}, find its volume. \(\left[\text { Use } \pi=\frac{22}{7}\right]\)

OR

Rakesh, a student of cLass X, kept together two cubes, each of side 5 cm, to form a cuboid. Find the surface area of the cuboid so formed. **(2)
**Answer:

Let the radius of the hemisphere be r cm.

Given: TSA of hemisphere = 462 cm

^{2}

⇒ 3πr

^{2}= 462

**Related Theory**

TSA of hemisphere = CSA of hemisphere + Area of base of hemisphere)

= 2πr

^{2}+ πr

^{2}

= 3πr

^{2}

OR

When two cubes of edge 5 cm each are joined together side by side, a cuboid is formed whose dimensions are 10 cm x 5 cm x 5 cm.

SECTION – B

(12 Marks)

Question 7.

The tops of two poles of heights 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with the horizontal, find the length of the wire. **(3)
**Answer:

Let AB and CD be the two poles.

∴ AB = 20 m, CM = 14 m, and ∠ACM = 30°

So, AM = AB – MB

= AB – CD [∵ MB = CD]

= 20 – 14

= 6 m

Also, the wire joining the tops of two poles is AC.

So, in ΔAMC

sin 30° \(\frac{\mathrm{AM}}{\mathrm{AC}}\)

⇒ \(\frac{1}{2}=\frac{6}{A C}\)

⇒ AC = 12

Hence, the Length of the wire is 12 m.

Question 8.

Find two consecutive positive integers, sum of whose squares is 365. **(3)
**Answer:

Let the two consecutive positive integers be x and (x + 1).

Then, according to the question

x

^{2}+ (x + 1)

^{2}= 365

⇒ 2x

^{2}+ 2x + 1 = 365

⇒ 2x

^{2}+ 2x – 364 =0

⇒ x

^{2}+ x – 182 = 0

⇒ (x + 14) (x – 13) = 0

⇒ x = -14, 13

Since, x is a positive integer,

x = 13

So, the two consecutive positive integers are 13 and 14.

Question 9.

Draw a circle of radius 2 cm with centre O and take a point P outside the circle such that OP = 6.5 cm. From P, construct two tangents to the circle. **(3)
**Answer:

Steps of construction:

- Draw a circLe with centre 0 and radius 2 cm.
- Take a point P outside the circLe and join it to centre O such that OP = 6.5 cm.
- Draw perpendicuLar bisector of OP to get its mid-point M.
- Taking M as centre and OM = MP as radius, draw a circLe, intersecting the circLe drawn in step (1), at points A and B.
- Join PA and PB.

Thus, PA and PB are the required tangents to the circLe with centre O.

Question 10.

Find the sum of all two-digit numbers which when divided by 7, leaves 2 as a remainder.

OR

Mala saved ₹ 10 in the first week of a year and then increased her weekly savings by ₹ 2. If in the nth week, her weekly savings is ₹ 68, then find the value of n. Also, find her weekly savings in 52^{nd} week of the year. **(3)
**Answer:

List of two-digit numbers which Leave remainder 2, when divided by 7 is:

(14 + 2), (21 + 2), (28 + 2) (91 + 2)

i.e. 16, 23, 30 ….. 93

This List of numbers forms an A.P. with a = 16, d = 7 and L = 93.

Let there be n terms in the A.P.

Then, L = a + (n – 1 )d

⇒ 93 = 16 + (n – 1) x 7

⇒ 7(n – 1) = 77

⇒ n – 1 = 11

⇒ n = 12

OR

Since, savings done by MaLa every week increases uniformLy by ₹ 2, so her savings form an A.P. with a = 10, d = 2 and a

_{n}= 68.

We know,

a_{n} = a + (n – 1)d

⇒ 68 = 10 + (n – 1) x 2

⇒ 2(n – 1) = 58

⇒ n – 1 = 29

⇒ n = 30

So, the vaLue of n is 30.

Now,

a_{52} = a + 51d

= 10 + 51 x 2

= 10 + 102

= 112

Hence, her savings in the 52^{nd} week is ₹112.

**Caution**

In word problems, learn to differentiate savings in the nth week and total savings in n weeks. Otherwise, error could occur.

SECTION – C

(16 Marks)

Question 11.

If median of the following frequency distribution is 32.5, find the missing values f_{1} and f_{2}. **(4)**

Class Interval | Frequency |

0 – 10 | f_{1} |

10 – 20 | 5 |

20 – 30 | 9 |

30 – 40 | 12 |

40 – 50 | f_{2} |

50 – 60 | 3 |

60 – 70 | 2 |

Total | 40 |

Answer:

Class | Frequency | Cumulative Frequency |

0 – 10 | f_{1} |
f_{1} |

10 – 20 | 5 | 5 + f_{1} |

20 – 30 | 9 | 14 + f_{1} |

30 – 40 | 12 | 26 + f_{1} |

40 – 50 | f_{2} |
26 + f_{1} + f_{2} |

50 – 60 | 3 | 29 + f_{1} + f_{2} |

60 – 70 | 2 | 31 + f_{1} + f_{2} |

N = 40 |

We have,

Sum of frequencies = 40

∴ 31 + f_{1} + f_{2} = 40

⇒ f_{1} + f_{2} = 9 ……………………………. (i)

Now,

Median = 32.5 [Given]

∴ Median class = 30 – 40

We know,

Question 12.

If the angles of elevation of the top of the candle from two coins distant a cm and b cm, (a > b), from its base and in the same straight line from it are 30° and 60°, respectively. Find the height of the candle.

OR

Two palm trees of equal heights are standing opposite to each other on either side of a river, which is 80 m wide. From a point O between them on the river, the angles of elevation of the top of the trees are 60° and 30°, respectively. Find the height of the trees and the distances of the point O from base of the trees. **(4)
**Answer:

Let AB be the candle and P and Q be the positions of two coins such that coin Q is nearer to the candle.

∴ ∠APB = 3O°, ∠AQB = 6O° BP = a, BQ = b.

Let height of the candle be h.

Now, in ΔABQ,

**Caution**

in the question, two constants namely. a and b, are given. So, the final answer must contain both the contents. Final answer, with one constant e.g. \(\sqrt{3} b\) or \(\frac{a}{\sqrt{3}}\) is incorrect.

OR

Let AB, CD be the two palm trees of equal heights.

∴ ∠OAB = 60°, ∠OCD = 30°, BD = 80m

Let BO = xm, and AB = CD = hm

Then, OD = BD – OB

= (80 – x) m

Hence, height of the trees is 20\(\sqrt{3}\) m each and distances of the trees from the point O are 20 m and (80 – 20)m, i.e. 60 m.

Question 13.

Case Study-1

A fountain is a decorative reservoir or a structure that jets water into the air for a decorative effect. Nowadays, LEDs of different colours are placed on their bases and by controlling the pressure of water, various dancing fountain shows are organized at different places.

In a society garden, a boundary was constructed around a circular fountain to restrict the children from playing with water.

Suppose the boundary was constructed in the shape of a quadrilateral ABCD, such that the walls AB, BC, CD and DA touches the fountain at points P, Q, R and S respectively.

On the basis of above information, answer the following questions.

(A) If the LED_{s} are placed along the boundaries of the four walls, then prove that AB + CD = BC + AD. **(2)**

(B) Let O be the centre of the fountain and ∠PAS = 90°, prove that ASOP is a square. **(2)
**Answer:

Case Study – 1

(A) We k.now, tangents drawn from an external point to a circle are equal in Length.

∴ AP = AS, BP = BQ, CR = CQ, DR = DS

Adding these equations, we get

AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

Hence, proved.

(B) We have

∠PAS = 90°

Also, AP = AS [From part (A)]

And, OS = OP [Radii]

So, in quadrilateral OPAS, adjacent sides are equal and an angle is a right angle.

∴ OPAS is a square.

Hence, proved.

Question 14.

Case Study-2

A tent is a shelter consisting of sheets of various materials draped over, attached to o frame of poles or attached to a supporting rope.

A student adventure camp of 3 dags was organized by the school for class X students. Their accommodation was organized in tents. A group of four students was to share a tent and they were asked to prepare their own tent. The teacher asked them to prepare a conical tent of radius 7 m and for this purpose, they were provided a canvas of area 551 m^{2} in which 1 m^{2} is used in stitching and wasted.

Based on the given information answer the following questions:

(A) Find the height of the tent. **(2)**

(B) Find the amount of air required by each student in the tent. **(2)
**Answer:

Case Study – 2

(A) Let L h be the slant height and height of the tent, respectively.

Since, 1m2 of canvas is wasted in stitching

∴ Total canvas present in tent

= 551 – 1 = 550m

^{2}

Now, total canvas present in tent

= CSA of tent

= CSAofcone

= πrl

Hence, height of the tent is 24 m.

(B) Amount of air in = Volume of the tent the tent

= Volume of come

Since, each tent is shared by 4 sutdents,

∴ Amount of air required by each tent