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CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions

February 17, 2022 by Bhagya

Students can access the CBSE Sample Papers for Class 10 Maths with Solutions and marking scheme Term 2 Set 4 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 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.
Which term of the A.P. 3, 15, 27, 39, …. will be 132 more than its 54th term?
OR
Find the 15th term from the end of the A.P. 3, 8,13 253. (2)
Answer:
Given A.P. is 3,15, 27, 39…….
Here, a = 3, d = 15 – 3 = 12
Let nth term be the required term of the A.P.
∴ an = 132 + 054
⇒ a + (n – 1)d = 132 + 0 +53d
⇒ (n – 1)12 = 132 + 53 x 12 = 768
⇒ \(n-1=\frac{768}{12}=64\)
⇒ n = 65
Hence, 65th term is the required term of the A.P.
OR
Given A.P. is 3, 8,13, 253.
Now,
15th term from the end of the A.P. 3,8,13,… 253 = 15th term from the
beginning of the A.P. 253, 248 13, 8, 3.
For the A.P. 253, 248 13, 8,3
a = 253, d = 248-253 = -5
∴ 15th term, a15 = a + 14d
= 253 + 14 x (-5)
= 253 – 70
= 183
Hence, the required term is 183.

Related Theory
Alternatively, nth term from the end of an A.P. is given as an = 1 – (n – 1) d
where, 1 is the last term and d is the common difference of the A.P.

Question 2.
The volume of a right circular cylinder with its height equal to the radius, is 25\(\frac{1}{7}\)cm3. Find the height of the cylinder. (2)
Answer:
Given: Radius of cylinder (r)
= Height of cylinder (h)
Also,
Volume of cylinder = 25\(\frac{1}{7}\)cm3CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 4
Hence, height of the cylinder is 2 cm.

Question 3.
Mrs. Bansal recorded the marks obtained by her students in the subject of mathematics, in the form of the following table. (3)

Marks obtained 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100
Number of  students 3 10 25 7 5

Answer:

Class Cumulative Frequency
Less than 20 3
Less than 40 3 + 10 = 13
Less than 60 3 + 10 + 25 = 38
Less than 80 3 + 10 + 25 + 7 = 45
Less than 100 3 + 10 + 25 + 7 + 5 = 50

Question 4.
Find the nature of roots of the quadratic equation 4x2 – 12x + 9 = 0. (2)
Answer:
Given equation is 4x2 – 12x + 9 = 0
∴ o = 4, b = -12, c = 9
We know,
Discriminant = b2 – 4ac
= (-12)2 – 4 x 4 x 9
= 144 – 144
= 0
Hence, the nature of roots is real and equal.

Question 5.
Draw a line segment AB of length 7.6 cm and divide it in the ratio 5:8. (2)
Answer:
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 5
Steps of construction:

  • Draw a Line segment AB of length 7.6 cm.
  • From point A, draw a ray AX such that ∠BAX in acute.
  • Mark (5 + 8) i.e. 13 points, namely, A1, A2, A3 A12, A13 on AX at equaL distances.
  • Join A13B.
  • From point A5, draw a Line A5P, parallel to A13B, intersecting line segment AB at point P.

Thus, P is the required point which divides AB in the ratio 5:8.

Question 6.
Find the roots of the quadratic equation \(\sqrt{2}\)x2 + 7x + 5\(\sqrt{2}\) = 0.
OR
If α and β are the roots of the equation 2x2 – 11x + 5 = 0, then find the value of (2α + 1) (3β + 4). (2)
Answer:
We have,
⇒ \(\sqrt{2}\)x2 + 7x + 5\(\sqrt{2}\) = 0
⇒ \(\sqrt{2}\)x2 + 2x + 5x + 5\(\sqrt{2}\) = 0
⇒ \(\sqrt{2}\)x (x + \(\sqrt{2}\)) + 5 (x + \(\sqrt{2}\)) = 0
⇒ (\(\sqrt{2}\)x + 5) (x + \(\sqrt{2}\)) = 0
⇒ \(x=-\frac{5}{\sqrt{2}},-\sqrt{2}\)
Flence, the roots of the given quadratic equation are \(-\frac{5}{\sqrt{2}}\) and –\(\sqrt{2}\).
OR
We have,
2x2 – 11x + 5 = 0
⇒ 2x2 – 10x – x + 5 = 0
⇒ 2x(x – 5) – 1(x – 5) = 0
⇒ (2x – 1) (x – 5) = 0
⇒ x = \(\frac{1}{2}\), 5
α = \(\frac{1}{2}\), β = 5
Now, (2α + 1) (3β + 4) = (2 × \(\frac{1}{2}\) + 1) (3 × 5 + 4)
= 2 × 19
= 38

SECTION – B
(12 Marks)

Question 7.
Calculate the mode of the following data:

Marks obtained 0-20 20-40 40-60 60-80 80-100
Number of students 5 9 12 6 3

Answer:
Here, maximum frequency = 12
And, class with maximum frequency is 40 – 60.
∴ Modal class = 40 – 60

We know,
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 6
Hence, mode of the data is 46.67.

Question 8.
The length of shadow of a person is x m long when the angle of elevation of the Sun is 45°. If the length of the shadow increases by (\(\sqrt{3}\) – 1)m, then find the new angle of elevation of the Sun. (3)
Answer:
Let AB be the person, BC be its shadow when angle of elevation of Sun is 45° and BD be its longer shadow.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 7
Now, in ΔABC
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 8
Hence, the new angLe of elevation of the Sun is 30°.

Question 9.
Prove that the lengths of tangents drawn from an external point to a circle are equal. (3)
Answer:
Given: A circle with centre O at which tangents TP and TQ are drawn from an external point T.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 9
To prove: TP = TQ
Construction: Join OP, OQ and OT.
Proof: We know, tangent is perpendicular to radius at the point of contact.
TP ⊥ OP and TQ ⊥ OQ
⇒ ∠OPT = 90°, ∠OQT = 90°
Now, in ΔOPT and AOQT
∠OPT = ∠OQT = 90° [Proved above]
OT = OT [Common side]
OP = OQ [Radii]
∴ ΔOPT ≅ ΔOQT [RHS congrency rule]
∴ PT = QT [c.p.c.t.] Hence, proved.

Question 10.
The marks obtained by 110 students in examination are given below: (3)
Answer:
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 10
Hence, mean marks of the students is 44.82.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 11
Cumulative frequency just greater than 57.5 is 70, which belongs to class 40-50.
∴ Median class = 40 – 50

We know,
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 12
Hence, the median of given frequency distribution is 43.75.

SECTION – C
(16 Marks)

Question 11.

Marks Number of students
30-35 14
35-40 16
40-45 28
45-50 23
50-55 18
55-60 8
60-65 3

Find the mean marks of the students.
OR
Find the median of the following frequency distribution.

Marks Number of students
Below 10 12
Below 20 22
Below 30 35
Below 40 50
Below 50 70
Below 60 86
Below 70 97
Below 80 104
Below 90 109
Below 100 115

Answer:
Let the four consecutive terms in A.P. be (a – 3d), (a – d), (a +„d) and (a + 3d).
∴ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 32
⇒ 4a = 32
⇒ a = 8
Also,
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 13
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 14
For a = 8 and d = 2, the numbers are [8 – 3(2)], (8 – 2), (8 + 2), [8 + 3(2)] i.e. 2, 6, 10, 14..

And, for a = 8 and d = -2, the numbers are [8 – 3(-2)], [8 – (-2)], [8 + (-2)], [8 + 3(-2)] i.e. 14, 10, 6, 2.

Hence, the numbers are 2, 6, 10, 14 or 14, 10, 6, 2.

Question 12.
The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7:15. Find the numbers. (4)
Answer:
Let the radius of other sphere be r.

Since, the two given spheres are of same metaL, so their densities must be same.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 15
Now, let the volume of new sphere be R.

Then, volume of new sphere = volume of two spheres
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 16
Hence, the radius of the new sphere is 6 cm.

Concept Applied

Density = \(\frac{\text { Mass }}{\text { Volume }}\) remains same in all shapes and sizes.
OR
Let the number of bricks used to construct the wall be n.

Since, \(\frac{1}{8}\)th of space of wall is covered by mortar.

∴ Remaining space, i.e. 1 – \(\frac{1}{8}\) = \(\frac{7^{\text {th }}}{8}\) of space of wall is occupied by the bricks.

So, n × volume of a brick = \(\frac{7}{8}\) × volume of the wall
⇒ n × 22.5 × 11.25 x 8.75 = \(\frac{7}{8}\) × (270 × 300 × 350)
⇒ \(n=\frac{7 \times 270 \times 300 \times 350}{8 \times 22.5 \times 11.25 \times 8.75}\)
= 11200
Hence, the required number of bricks is 11,200.

Question 13.
Two spheres of same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single bigger sphere. Find the radius of the new sphere.
OR
A mason construct a wall of dimensions 270 cm x 300 cm x 350 cm with bricks, each of size 22.5 cm x 11.25 cm x 8.75 cm. If it is assumed that one-eighth of the space is covered by mortar, then find the total number of bricks used to construct the wall. (4)
Answer:
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 17
Here, AB is the cliff and, CandDarethe two positions of the boat, moving away from the cliff.

Now, in ΔABC,
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 18

Now, distance travelled by the boat during the period of observation
= CD
= BD – BC
= 150 – 50\(\sqrt{3}\)
= 150 – 50 x 1.73
= 150 – 86.5
= 63.5 m

(B) We have, time taken by the boat to cover the distance CD i.e. 63.5 m is 1 min, or, 60 seconds.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 19
Hence, speed of the boat is 1.06 m/sec.

Question 14.
Case study-1
In geology, a cliff is a part of huge rock, which is generally vertical. They are formed by the processes of weathering and erosion with the effect of gravity.

A moving boat is observed from the top of a 150 m high cliff, moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in one minute.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 1

Based on this answer the following questions:

(A) Make a labelled figure on the basis of given information and find the distance travelled by the boat during the period of observation. [Use \(\sqrt{3}\) = 1.73]. (2)

(B) Find the speed of the boat. (2)
Answer:
(A) Given: ∠APB = 50°
We know, tangent is perpendicular to radius at the point of contact.
∴ AP ⊥ OA, BP ⊥ OB
∴ ∠OAP = 90°, ∠OBP = 90°
So, in quadrilateral OAPB,
∠AOB = 360° – ∠OAP
∠APB – ∠OBP
= 360° – 90° – 50° – 90°
= 130°

(B) In ΔOAP,
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 20
= 169
⇒ OP = 13 cm
Similarly, in ΔOBP,
∠OBP = 90°, OB = 3 cm, OP = 13 cm
∴ (PB)2 = (OP)2 – (OB)2
= (13)2 – (3)2 = 169 – 9
= 160
⇒ PB = 4\(\sqrt{10}\)cm

Now,
Area of quadrilateral OAPB
= ar(ΔPOA) + ar(ΔPOB)
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 21

Question 15.
Case study-2
Reshma took one bangle of her mother and one of her size, and put the smaller one within the gap of the bigger one to form two concentric circles. She then took two straws and put them on the bangles as shown in the figure.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 2

Let O be the centre of the two concentric circles, P be the point from where straws PA and PB are drawn to these circles.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 3

If OA = 5 cm, OB = 3 cm and AP = 12 cm, then answer the following questions,

(A) If the angle made between the two straws is 50°, find the angle made between the radii of the two bangles i.e. ∠AOB. (2)

(B) Find the area of the quadrilateral formed by the two straws and the radii of two bangle, Le. area of quadrilateral OAPB. (2)
Answer:

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