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

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:

1. The question paper consists of 14 questions divided into 3 sections A, B, C.
2. All questions are compulsory.
3. Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
4. Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
5. 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 ax² + 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, a₁₉ = 3a₆
⇒ a + 18 d = 3 (a + 5 d)
⇒ a + 18 d = 3 a + 15 d
⇒ 2a – 3d = 0 ………………………….. (i)
Also, a₉ = 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 cm², 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 cm²
⇒ 3πr² = 462

Related Theory
TSA of hemisphere = CSA of hemisphere + Area of base of hemisphere)
= 2πr² + πr²
= 3πr²
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
x² + (x + 1)² = 365
⇒ 2x² + 2x + 1 = 365
⇒ 2x² + 2x – 364 =0
⇒ x² + 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,
a₅₂ = 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 f₁ and f₂. (4)

Class Interval	Frequency
0 – 10	f1
10 – 20	5
20 – 30	9
30 – 40	12
40 – 50	f2
50 – 60	3
60 – 70	2
Total	40

Answer:

Class	Frequency	Cumulative Frequency
0 – 10	f1	f1
10 – 20	5	5 + f1
20 – 30	9	14 + f1
30 – 40	12	26 + f1
40 – 50	f2	26 + f1 + f2
50 – 60	3	29 + f1 + f2
60 – 70	2	31 + f1 + f2
	N = 40

We have,
Sum of frequencies = 40
∴ 31 + f₁ + f₂ = 40
⇒ f₁ + f₂ = 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 m² in which 1 m² 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 = 550m²
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