CBSE Sample Papers

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

CBSE Sample Papers for Class 10 Maths Standard Set 4 for Practice

Time: 3 Hours
Maximum Marks: 80

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 of 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. It consists of three sections III, IV and V.
2. In section III, Question Nos. 21 to 26 are Very Short Answer Type questions of 2 marks each.
3. In section IV, Question Nos. 27 to 33 are Short Answer Type questions of 3 marks each.
4. In section V, Question Nos. 34 to 36 are Long Answer Type questions of 5 marks each.
5. Internal choice is provided in 2 questions of 2 marks, 2 questions of 3 marks and 1 question of 5 marks.

Part – A
Section-I

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

Question 1.
Find the distance between the lines 2x + 4 = 0 and x – 5 = 0.
OR
Find the distance of P (3, -2) from the y-axis.
Solution:
7 units OR 3 units

Question 2.
Find the value of k for which the given system has unique solution. 2x + 3y – 5 = 0, kx – 6y – 8 = 0
Solution:
k ≠ – 4

Question 3.
If cosec \(\theta=\frac{5}{3} \), then find the value of cos θ + tan θ.
Solution:
\(\frac{31}{20}\)

Question 4.
The given figure is a sector of circle of radius 10.5 cm. Find the perimeter of the sector.  [Take π = \(\frac{22}{7}\) ]

Solution:
32 cm

Question 5.
If in an equilateral triangle, the length of the median is √3 cm, then find the length of the side of equilateral triangle.
OR
In an equilateral triangle of side 3 √3 cm, find the length of the altitude.
Solution:
2 cm OR 4.5 cm

Question 6.
Find the nature of roots of ax² + bx + c = 0, a > 0, b = 0, c > 0.
Solution:
no real roots

Question 7.
Is x = – 3 a solution of x² + 6x + 9 = 0?
OR
What will be the nature of roots of quadratic equation 2x² + 4x -7 = 0?
Solution:
Yes OR Real and unequal

Question 8.
In the given figure, ABC is circumscribing a circle. Find the length of BC.

Solution:
10cm

Question 9.
In the given figure, O is the centre of the circle, PA and PB are tangents to the circle. Find the measure of ∠AQB.

OR
In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60°, find ∠PRQ.

Solution:
70° OR 120°

Question 10.
ΔABC is isosceles in which ∠C = 90°. If AC = 6 cm, then find AB².
Solution:
72 cm²

Question 11.
To divide line segment AB in the ratio m : n, a ray AX is drawn so that ∠B AX is an acute angle and then points are marked on ray AX at equal distance. Find the minimum number of these points.
Solution:
m + n

Question 12.
If a, p are the zeroes of the polynomial 2y² + 1y + 5, write the value of a + p 4- a,p.
Solution:
-1

Question 13.
Form a quadratic polynomial, the sum and product of whose zeroes are (-3) and 2 respectively.
Solution:
x² + 3x + 2

Question 14.
For which values of k will the pair of equations kx + 3y = k – 3 and 12x + ky = k have no solution?
Solution:
k = – 6

Question 15.
Find the number of spherical lead shots each 4.2 cm in diameter can be obtained from a rectangular solid lead with dimensions 66 cm, 42 cm and 21 cm.
Solution:
1500

Question 16.
Find the probability of
(a) an impossible event.
(b) a sure event.
OR
Find the probability that a number selected at random from the number 1, 2, 3, …, 35 is a multiple of 7.
Solution:
(a) 0
(b) 1
OR
\(\frac{1}{7}\)

Section-ll

Case Study based questions are compulsory. Attempt any four sub-parts of each question. Each sub-part carries 1 mark.

Case Study Based-1

Question 17.
Traffic Lights

Traffic Lights (or traffic signals) are lights used to control movement of traffics. They are placed on roads at intersections and crossings. The different colours of light tell drivers what to do. The traffic lights at different road crossings change after every 48 sec, 72 sec and 108 sec respectively.

(a) 108 can be expressed as a product of its primes as ……….
(i) 2³ x 3²
(ii) 2³ x 3³
(iii) 2² x 3²
(iv) 2² x 3³
Solution:
(iv) 2² x 3³

(b) The HCF of 48, 72, 108 is ………..
(i) 18
(ii) 16
(iii) 12
(iv) 10
Solution:
(iii) 12

(c) The LCM of 48, 72, 108 is ……….
(i) 520
(ii) 432
(iii) 396
(iv) 420
Solution:
(ii) 432

(d) If all the traffic lights change simultaneously at 8:20:00 hrs, they will again change simultaneously at
(i) 8 : 27 : 12 hrs
(ii) 8 : 32 : 24 hrs
(iii) 8 : 40 : 08 hrs
(iv) 8 : 24 : 24 hrs
Solution:
(i) 8 : 27 : 12 hrs

(e) The [HCF x LCM] for the numbers 48, 72, 108 is ……..
(i) 2472
(ii) 3680
(iii) 4090
(iv) 5184
Solution:
(iv) 5184

Case Study Based-2

Question 18.
Pollution —A Major Problem
One of the major serious problems that the world is facing today is the environmental pollution. Common types of pollution include light, noise, water and air pollution.

In a school, students thoughts of planting trees in and around the school to reduce noise pollution and air pollution.

Condition I: It was decided that the number of trees that each section of each class will plant be the same as the class in which they are studying, e.g. a section of class I will plant 1 tree a section of class II will plant 2 trees and so on a section of class XII will plant 12 trees. Condition II: It was decided that the number of trees that each section of each class will plant be the double of the class in which they are studying, e.g. a section of class I will plant 2 trees, a section of class II will plant 4 trees and so on a section of class XII will plant 24 trees.

Refer to Condition I

(a) The AP formed by sequence e. number of plants by students is ………
(i) 0, 1, 2, 3, …, 12
(ii) 1, 2, 3, 4, …, 12
(iii) 0, 1, 2, 3, …, 15
(iv) 1,2, 3, 4, …, 15
Solution:
(ii) 1, 2, 3, 4, …, 12

(b) If there are two sections of each class, how many trees will be planted by the students?
(i) 126
(ii) 152
(iii) 156
(iv) 184
Solution:
(iii) 156

(c) If there are three sections of each class, how many trees will be planted by the students?
(i) 234
(ii) 260
(iii) 310
(iv)   326
Solution:
(i) 234

Refer to Condition II

(d) If there are two sections of each class, how many trees will be planted by the students?
(i) 422
(ii) 312
(iii) 360
(iv) 540
Solution:
(ii) 312

(e) If there are three sections of each class, how many trees will be planted by the students?
(i) 468
(ii) 590
(iii) 710
(iv)  620
Solution:
(i) 468

Case Study Based-3

Question 19.
Student-Teacher Ratio

Student-teacher ratio expresses the relationship between the number of students enrolled in a school and the number of teachers in that school. It is important for a number of reasons. For example, it can be an indicator of the amount of individual attention any child is likely to receive, keeping in mind that not all class size are going to be the same.

The following distribution gives the state-wise student-teacher ratio in higher secondary schools of India (28 states and 7 UTs only).

(a) The mode of the above data is ………
(i) 25.5
(ii) 30.6
(iii) 35.2
(iv) 38.3
Solution:
(ii) 30.6

(b) The mean of the above data is ………….
(i) 29.2
(ii) 30.5
(iii) 38.3
(iv) 40.1
Solution:
(i) 29.2

(c) The modal class is ………..
(i) 20-25
(ii) 40 – 45
(iii) 30 -35
(iv) 50.55
Solution:
(iii) 30 -35

(d) The sum of class marks of 25-30 and 45-50 is
(i) 62
(ii)  70
(iii)  75
(iv) 85
Solution:
(iii)  75

(e) The sum of the upper and lower limits of modal class is
(i) 55
(ii) 65
(iii) 85
(iv) 75
Solution:
(ii) 65

Case Study Based-4

Question 20.
Skysails’ is that genre of engineering science that uses extensive utilization of wind energy to move a vessel in the sea water. The ‘Skysails’ technology allows the towing kite to gain a height of anything between 100 metres to 300 metres. The sailing kite is made in such a way that it can be raised to its proper elevation and then brought back with the help of a ‘telescopic mast’ that enables the kite to be raised properly and effectively.

(a) In the given figures, if sin 8 = cos (30 – 30°), where 0 and 30 – 30° are acute angles, then the value of 9 is ……
(i) 30°
(ii) 60°
(iii) 45°
(iv) None of these.
Solution:
(i) 30°

(b) What should be the length of the rope of the kite sail in order to pull the ship at the angle θ (calculated in part (a) and be at a vertical height of 200 m?
(i) 300 m
(ii) 400 m
(iii) 500 m
(iv) 600 m
Solution:
(ii) 400 m

(c) If BC = 15 m, θ = 30°, then AB is ………..
(i) \(2 \sqrt{3} \mathrm{~m}\)
(ii) 15 m
(iii) 24 m
(iv)\(5 \sqrt{3} \mathrm{~m}\)
Solution:
(iv)\(5 \sqrt{3} \mathrm{~m}\)

(d) Suppose AB = BC = 12 m, then 0 = ………
(i) 0°
(ii) 30°
(iii) 45°
(iv) 60°
Solution:
(iii) 45°

(e) Given that BC = 6 m and 0 = 45°. The values of AB and AC are respectively………..
(i) AB = 4 m, AC = \(4 \sqrt{2} \mathrm{~m}\)
(ii)  AB = 7 m, AC =\(7 \sqrt{5} \mathrm{~m}\)
(iii) AB = 9 m, AC = \(9 \sqrt{3} \mathrm{~m}\)
(iv) AB = 6 m, AC =\(6 \sqrt{2} \mathrm{~m}\)
Solution:
(iv) AB = 6 m, AC =\(6 \sqrt{2} \mathrm{~m}\)

Part-B
Section-III

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

Question 21.
Is 7 ^x 11 ^x 13 + 11 a composite number? Justify your answer.
Solution:
Yes

Question 22.
Check whether (1,2), (3, 4), (1, 4), (2, 8) are the vertices of a square.
OR
Determine if the points (1,5), (2, 3) and (-2, -11) are collinear.
Solution:
No OR No

Question 23.
Find all zeroes of the polynomial 2x³ + x² – 6x – 3, if two of its zeroes are \(-\sqrt{3} \text { and } \sqrt{3}\)
Solution:
\(-\sqrt{3}, \sqrt{3},-\frac{1}{2}\)

Question 24.
Draw a line-segment PQ = 8.4 cm by using ruler and compass only. Find a point R on PQ such that \(\frac{3}{4}\)
Solution:
Point ‘R’ is 3.6 cm away from ‘P’.

Question 25.
If a cos θ – b sin θ = c, prove that a sin θ + b cos θ = \(\pm \sqrt{a^{2}+b^{2}-c^{2}}\)
OR
If sin θ+ sin² θ=1, then prove that cos² θ + cos⁴ θ = 1.

Question 26.
Prove that the lengths of tangents drawn from an external point to a circle are equal.

Section-IV

Question 27.
Point P divides the line segment joining the points A(2, 1) and B(5, -8) such that \(\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{1}{3}\). If P lies on the line 2x -y + k O, find the value of k.
Solution:
-8

Question 28.
In the given figure. if LM || CB and LN || CD, prove that \(\frac{\mathrm{AM} {\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)

OR
In the given figure, DE || AC and DF || AE. Prove that \(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\)

Question 29.
In the given figure, DE JI AC and DF  AE. Prove that \(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\)

Solution:
7.868 cm²

Question 30.
The area of an equilateral triangle is \(49 \sqrt{3} \mathrm{~cm}^{2}\). Taking each angular point as centre, a circle is described with radius equal to half the length of the side of the triangle as shown in figure. Find the area of the triangle not included in the circles.

In the given figure, prove that AD = BE if ∠A = ∠B and DE || AB.

Question 31.
The weight of tea in 70 packets are as follows:

Determine the modal weight.
Solution:
201.7 kg

Question 32.
Check whether the pair of equations 5x – y = 1 and are consistent. If so, solve them graphically.
Solution:
Consistent, x = 2, y = 3

Question 33.
Find the mean and median for the following data:

Solution:
Mean = 26.4; Median = 27.2

Section-V

Question 34.
Two hoardings on cleanliness are kept on two poles of equal heights standing opposite to each other on either side of the road, which is 80 m wide. From a point between them on the road, the angle of elevation of the top of the poles are 60° and 30° respectively. Find the height of the pole and the distance of the point from the poles.
OR
A tree breaks down due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with it. The distance from the foot of the tree to the point where the top touches the ground is 8 metres. Find the height of the tree before it was broken.
Solution:
20√3 m; 20 m and 60 m OR 13.86 m

Question 35.
A well of diameter 3 m and 14 m deep is dug. The earth, taken out of it, has been evenly spread all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
\(\frac{9}{8} \mathrm{~m}\)

Question 36.
A part of monthly expenditure of a family is constant and the remaining varies with the price of wheat. When the price of wheat is ₹ 800 per quintals, the total monthly expenditure is ₹ 5400 and when it is ₹ 1000 per quintal, the total monthly expenditure ₹ 6000. Find the total monthly expenditure of the family when the cost of wheat is ₹ 950 per quintal. Assuming the consumption of wheat to be the same.
Solution:
₹ 5850

Students can access the CBSE Sample Papers for Class 10 Maths Standard with Solutions and marking scheme Set 3 will help students in understanding the difficulty level of the exam.

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

Time: 3 Hours
Maximum Marks: 80

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 of 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. It consists of three sections III, IV and V.
2. In section III, Question Nos. 21 to 26 are Very Short Answer Type questions of 2 marks each.
3. In section IV, Question Nos. 27 to 33 are Short Answer Type questions of 3 marks each.
4. In section V, Question Nos. 34 to 36 are Long Answer Type questions of 5 marks each.
5. Internal choice is provided in 2 questions of 2 marks, 2 questions of 3 marks and 1 question of 5 marks.

Part – A
Section-I

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

Question 1.
If one of the zeroes of the quadratic polynomial x¹ + 3x + k is 2, then find the value of k
Solution :
Since 2 is a zero of p(x) = x² + 3x + k, then p(x) =0
⇒(2)² + 3 x 2 + k = 0
⇒ 4 + 6 + k = 0
⇒ k = -10

Question 2.
Find the total number of factors of a prime number.
OR
Find the HCF and the LCM of 12, 21, 15.
Solution :
We know that prime number is a number which has exactly two factors, i.e., 1 and the number itself. So, the total number of factors of a prime number is 2.
OR
12 = 2 x 2 x 3 = 2² x 3¹
15 = 3 x 5 = 3¹ x 5¹ 21 = 3 x 7 = 3¹ x 7¹ HCF(12, 15,21)= 3
LCM(12, 15, 21) = 2² x 3¹ x 5¹ x 7¹ = 4 x 3 x 5 x 7 = 420

Question 3.
Find the value of k for which the system of equations x + y – 4 = 0 and 2x + ky = 3, has no solution.
Solution :
For the equations, x + y -4 = 0 and 2x + ky = 3
⇒ 2x + ky  –  3 = 0

Question 4.
Find the value of x for which 2x, (x + 10) and (3x + 2) are the three consecutive terms of an AP.
OR
The first term of an AP is p and the common difference is q. Find its 10th term.
Solution :
2x (x + 10) and (3x + 2) are three consecutive terms of an AP.
Then, (x+10)-2x= (3x + 2)-(x+10)
⇒ x + 10 – 2x = 3x + 2 – x – 10
⇒ -x + 10 = 2x- 8
⇒ 3x = 18
⇒ x = 6
a_n = a + (n – 1 )d  ⇒ a_m =p + (10 – 1 )q =p + 9q
a₁₀ = p + 9q

Question 5.
In figure, ΔABC is circumscribing a circle. Find the length of BC.

Solution :
Line segments AB, BC and AC are tangents of the circle.
AP = AR = 4 cm     …(i)
[Tangents drawn from the same point to a circle are equal]
Similarly, BP = BQ = 3 cm ……………… (ii)
and   CQ = CR     ………….. (iii)
Since, AC =11 cm  ⇒ AR + CR = 11 cm
⇒ 4 cm + CR = 11 cm [from (i)]
⇒ CR = 7 cm ……………… (iv)
BC = BQ + CQ = 3 cm + 7 cm  [From (ii),(iii) and (iv)]
= 10 cm

Question 6.
If in the given figure, DE || BC, then find EC.

Solution :

Question 7.
Find the value of  \(\left(\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}\right)\)
Solution :
\(\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}=\sin ^{2} \theta+\frac{1}{\sec ^{2} \theta}=\sin ^{2} \theta+\cos ^{2} \theta=1\)

Question 8.
Find the value of (1 + tan² θ) (1 – sin θ) (1 + sin θ).
Solution :
(1 + tan² θ)(1 – sin θ)(1 + sin θ) = (1 + tan² θ)(1 – sin² θ)
\(=\sec ^{2} \theta \cos ^{2} \theta=\frac{1}{\cos ^{2} \theta} \times \cos ^{2} \theta=1\)

Question 9.
Two cones have their heights in the ratio 1 : 3 and radii in the ratio 3:1. What is the ratio of their volumes? 1
Solution :
Let h₁ and h₂ be the heights of two cones and r₁ and r₂ by the radii

Question 10.
If ABC is an equilateral triangle of side 2a, then find the length of one of its altitudes.
OR
In the given figure, MN || QR. If PM = x cm, MQ = 10 cm, PN = (x – 2) cm, NR = 6 cm, then find the

Solution :
In ∠ADB and ∠ADC,
AB = AC [Sides of equilateral triangle ABC]
AD = AD [Common]
∠ADB = ∠ADC [90°, AD ⊥ BC]
ΔADB ≅ ΔADC
ABC is an equilateral triangle such that each side is 2a and AD ⊥ BC.

Question 11.
Base of an isosceles triangle is \(\frac{2}{3}\) times its congruent sides. Perimeter of the triangle is 32 cm.
Formulate this problem as a pair of equations.
Solution :
Let the congruent side of isosceles triangle be x cm and its base be y cm.
Then \(y=\frac{2}{3} x\)
2x – 3y = 0 ……………. (i)
Also,   x + x + y=32  ⇒ 2x + y = 32 ………….. (ii)

Question 12.
Check whether x(x + 2) – 3 = (x+ 4)x is a quadratic equation.
Solution :
Since  x(x + 2) – 3 = x(x + 4)
⇒ x² + 2x – 3 = x² + 4x
⇒ 2x + 3 = 0
This is linear equation not a quadratic equation.

Question 13.
Is x = -2 a solution of 3x² + 13x + 14 = 0?
OR
State whether the equation (x + 1) (x – 2) + x = 0 has two distinct real roots or not. Justify your answer.
Solution :
Putting the value of x in the quadratic equation,
LHS = 3x² + 13x + 14 = 3 (-2)² + 13 (-2) + 14
= 12 – 26 + 14 = 0 = RHS
Hence, x = -2 is a solution.
OR
We have (x + 1) (x – 2) + x = 0
⇒ x²-x-2 + x= 0  ⇒ x²-2 = 0
D = b² – 4ac = 0 – 4(1) (-2) = 8 > 0
∴ Given equation has two distinct real roots.

Question 14.
Two concentric circles of radii a and b (a> b) are given. Find the length of the chord of the larger circle which touches the smaller circle.
OR
In the given figure, CP and CQ are tangents from an external point C to a circle with centre O. AB is another tangent which touches the circle at R. If CP = 11 cm and BR = 4 cm, find the length of BC.

Solution :

OR
In the given figure, CP = CQ  [tangents drawn from an external point are equal]
So,  CP = CQ = 11 cm
Also,   BR = BQ   [tangents drawn from an external point are equal]
So,  BR = BQ = 4 cm
∴ Now, BC = CQ – BQ
= (11 -4) cm = 7 cm

Question 15.
Draw a line segment of length 6 cm. Using compasses and ruler, find a point P on it which divides it in the ratio 3:4.
Solution :

Steps of construction:
1. Draw a line segment AB = 6
2. Draw any ray AX making an acute angle XAB with AB.
3. Along AX mark 7 (3 + 4) points A₁ A₂, A₃, A₄,……………, A₇ at equal distances such that
AA₁ = A₁A₂ = A₆A₇
4. Join A₇B
5. From A₃, draw A₃P parallel to A₇B (by making an angle equal to ∠AA₇B at A₃) to meet AB at point P.
Then  AP : PB = 3 : 4.

Question 16.
The radius of a circle is 5 cm. Find the circumference of the circle whose area is 49 times the area of given circle.
Solution :
The area of the given circle = πr² = π(5)² = 25π sq. cm
Area of the other circle = 49 × 25π
Let the radius of this circle be R Then
Then πR² = 49 x 25 x π  ⇒ R² = (7)² x (5)²
R = 7 x 5 = 35 cm
The required circumference = 2πR = 2 x \(\frac{22}{7}\) x 35 = 220 cm

Section-II

Case Study based questions are compulsory. Attempt any four sub-parts of each question. Each sub-part carries 1 mark.

Case Study Based-1

Question 17.
Treasure Island Shikha and Sanjana are playing a board game of

Treasure Island.
Shikha and Sanjana are playing a board game of Treasure Island.

Refer to Skull Rock and Cave of Death
(a) The mid-point of the segment joining A(3, 5) and C(5, 3) is……..
(i) (2,3)
(ii) (3,5)
(iii) (4,3)

Refer to Three Palms
(b) The distance of point D(6, 4) from origin is
(i) \(5 \sqrt{7}\)units
(ii) \(7 \sqrt{5}\) units
(iii) \(4 \sqrt{10}\) units
(iv) \(2 \sqrt{23}\) units

Refer to 4-Cross Cliffs and Three Palms
(c) The distance between the points B(2,3)and D(6, 4) is
(i) \(\sqrt{13} \text { units }\)units
(ii) 4 units
(iii) \(5 \sqrt{3}\)units
(iv) \(\sqrt{17}\)units

(d) The coordinate of the point which divides the join of B(2,3) and D(6,4) in the ratio 2 : 3 is
(i) \(\left(\frac{13}{4}, \frac{19}{4}\right)\)
(ii) \(\left(\frac{18}{5}, \frac{17}{5}\right)\)
(iii) \(\left(\frac{11}{8}, \frac{13}{8}\right)\)
(iv) \(\left(\frac{19}{3}, \frac{16}{3}\right)\)

(e) If \(P\left(\frac{x}{3}, 4\right)\) is the mid-point of the line segment joining the points C(5,3) and
A(3, 5), then
(i) 8
(ii) 10
(iii) 12
(iv) 16
Solution :

Case Study Based-2

Question 18.
What are you Smoking?
Given below is Air Quality Index of different localities of Delhi on 27th December 2019 by Times of India Newspaper on 28th. December 2019

The data recorded from above AQI (Air Quality Index) is given below:

(a) The sum of the lower limits of the median class and modal class is ……………
(i) 650
(ii) 660
(iii) 750
(iv) 710

(b) The modal class is …………
(i) 310-320
(ii) 330-340
(iii) 360-370
(iv) 380-390

(c) The upper limit of the median class is ………
(i) 330
(ii) 350
(iii) 360
(iv) 390

(d) The difference of the upper limit of the median class and the lower limit of the modal class is ……..
(i) 0
(ii) 1
(iii) 2
(iv) 3

(e) The mean AQI is ……..
(i) 335.8
(ii) 354.7
(iii) 360.4
(iv) 395.9
Solution :

We have \(n=34, \frac{n}{2}=\frac{34}{2}=17\)

∴ The sum of the lower limits of the median class and modal class = 350 + 360 = 710 So, option
(iv) is the correct answer.

(b) Here, the maximum frequency is 7 and the corresponding class is 360-370.
So, modal class is 360-370. So, option (iii) is the correct answer.

(c) The median class is 350-360 which has upper limit 360. So, option (iii) is the correct answer.

(d) The difference between upper limit of the median class and the lower limit of the modal class = 360 – 360 = 0. So, option (i) is the correct answer.

(e)

\(\text { Mean }=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{12060}{34}=354.7(\text { approx. })\)
So, option (ii) is the correct answer.

Case Study Based-3

Question 19.
Skipping Rope
Skipping rope is a good exercise. It bums calories, makes bones strong and improves heart health.

During skipping, when rope goes up and down it makes the shape of parabolas (graphs of quadratic polynomials). Observe the following skipping pictures.

Refer Picture 1
(a) The graph of polynomial p(x) represented by Picture 1 is shown below. The number of zeroes of the polynomial is

(i) 0
(ii) 1
(iii) 2
(iv) 4

Refer Picture 2
(b) The graph of polynomial p(x) represented by Picture 2 is shown below. Which of the following has negative (-) sign?

(i) a
(ii) b
(iii) c
(iv) All of there

(c) If \(\frac{1}{4}\) and 1 are the sum and product of zeroes of a polynomial whose graph is represented by Picture (3), the quadratic polynomial is
(i) \(k\left(x^{2}-\frac{1}{4} x-1\right)\)
(ii) \(k\left(\frac{1}{4} x^{2}-x-1\right)\)
(iii) \(k\left(x^{2}+\frac{1}{4} x+1\right)\)
(iv) \(k\left(\frac{1}{4} x^{2}+x+1\right)\)

(d) Let the Picture (1) represent the quadratic polynomial f(x) = x² – 8x + k whose sum of the
squares of zeroes is 40, The value of k is
(i) 8
(ii) 10
(iii) 12
(iv) 20

(e) Let the Picture (3) represent the quadratic polynomial f(x) = x² + 7x + 10. Then its zeroes are
(i) -1, -5
(ii) -2, -5
(iii) 1, 5
(iv) 2, 5
Solution :
(a) Since the graph y p(x) cuts the x-axis at two different points, so the polynomial has two zeroes. So, option (iii) is the correct answer.

(b) The parabola)’ = ax² + bx + c open downwards. Therefore, a <O. The vertex \(\left(\frac{-b}{2 a}, \frac{-\mathrm{D}}{4 a}\right)\) of the parabola is on OX¹.
\( \quad \frac{-b}{2 a}<0 \quad \Rightarrow b<0\)
Parabola y =p(x) cuts y-axis at P(O, c) which lies on OY’. Therefore e <O.
Hence, O, b <O and c< O.
So, option (iv) is the correct answer.

(c) We have sum \(\frac{1}{4}\) and product
\(f(x)=k\left(x^{2}-\frac{1}{4} x-1\right)\)
So, option (iii) is the correct answer.

(d)

(e) We have
x² + 7x + 10 =x² + 5x + 2x + 10 = x(x + 5) + 2(x + 5) = (x + 2)(x + 5)
Now, when x + 2 = 0 or x + 5 = 0, i.e., whenx = -2 andx = -5
Therefore, the zeroes of x² + lx + 10 are -2 and -5

Question 20.

Fair Play
Garima has two children, Tapan and Maya. Every Sunday is a game night in the family. Tonight Garima has planned for a game with three cubes, one purple and two yellow. She placed the three cubes in a bag and called for her children.
Garima: Do you want to play a game of probability?
Maya: What is probability?
Garima: Let me ask you something before I answer you. Can you predict what is in this bag?
Tapan: I cannot guess that!
Maya: I am 100% sure it is a toy!
Garima: I am glad you think that Maya. Just now you used the concept of probability.
Whether an event can happen or not, can’t be predicted with total certainty. But we can always predict how- likely or unlikely it is for an event to happen.
And for predicting that, we use a concept called probability.
\(\text { Probability (an event to happen) }=\frac{\text { Number of ways event can happen }}{\text { Total number of ways all events can happen }}\)
Placing the bag of cubes in the centre, Garima explained the rules of the game to the children. Garima: Without looking, the first player will pick out a cube from the bag and then the second player will also pick out one cube without looking. If the two cubes picked out were the same colour, then the first person will win the game. If the boxes picked out are of two differently coloured cubes, then the second player will be the winner.

(a) In the first round, Maya pulled out a cube, which was yellow. What is the probability that Tapan will win the game?
(i) \(\frac{1}{2}\)
(ii) \(\frac{1}{3}\)
(iii) \(\frac{2}{3}\)
(iv) 0

(b) in the second round, Tapan started by picking out a purple cube. What is the probability for Tapan to win the round?
(i) 1
(ii) \(\frac{1}{3}\)
(iii) 0
(iv) \(\frac{2}{3}\)

(c) In the third round, Maya pulled out a cube. The probability that the pulled out cube is not of yellow colour is
(i) 1
(ii) \(\frac{1}{2}\)
(iii) \(\frac{1}{3}\)
(iv) \(\frac{2}{3}\)

(d) In the fourth round, Tapan pulled out a cube. The probability that the pulled out cube is either purple or yellow is
(i) 1
(ii) \(\frac{1}{2}\)
(iii) \(\frac{1}{3}\)
(iv) \(\frac{2}{3}\)

(e) In the last round, Maya pulled out a cube. The probability that the pulled out cube is of green colour is
(i) 1
(ii) \(\frac{1}{2}\)
(iii) \(\frac{1}{3}\)
(iv) 0
Solution :
(a) After taking out one yellow cube, the bag is left with 1 yellow cube whose probability of pulling out \(\frac{1}{2}\) So, option (i) is the correct answer.

(b) After taking out one purple cube, the bag has no purple cube. So, the probability for Tapan to win the round is
\(\frac{0}{3}\) = 0
So, option (iii) is the correct answer.

(c) The number of yellow cube is 1. So, the probability of pulling out a cube not of yellow colour is \(\frac{1}{3}\)
So, option (iii) is the correct answer.

(d) The number of purple and yellow cubes =1 + 2 = 3.
∴ The required probability =\(\frac{3}{3}\) = 1
So, option (i) is the correct answer.

(e) The number of green cube = 0
.’. The required probability = \(\frac{0}{3}\) = 0
So, option (iv) is the correct answer.

Part-B
Section-III

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

Question 21.
Write a rational number between \(\sqrt{2} \text { and } \sqrt{3}\)
Solution :
We have  √2 =1.4142135…. and √3 = 1.7320508…
Since every terminating decimal or repeating decimal represents a rational number.
So, 1.666666… is a rational number between \(\sqrt{2} \text { and } \sqrt{3}\)
Also 1.5, 1.6, 1.7 are rational numbers between \(\sqrt{2} \text { and } \sqrt{3}\)

Question 22.
Find the ratio in which the point (-3, p) divides the line segment joining the points (-5,-4) and (-2,3).
Hence find the value of p.
OR
If the point C(-1, 2) divides internally the line segment joining A(2, 5) and B(x, y) in the ratio 3 : 4, find the coordinates of B.
Solution :

Question 23.
Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial f(x)=ax²+bx+c,a≠0,c≠0.
Solution :
Let the zeroes of the polynornial f(x) = ax² + bx + c be α and β.
Then \(\alpha+\beta=\frac{-b}{a} \text { and } \alpha \beta=\frac{c}{a}\)
Now, the zeroes of the required polynomial are reciprocals of a and p.
∴ The required quadratic polynomial is given by

Question 24.
Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle.
Solution :

Steps of construction:
1. Draw a line segment AB of length 7 cm.
2. With A as centre, draw a circle of radius 3 cm.
3. With B as centre, draw a circle of radius 2 cm.
4. Draw the perpendicular bisector of AB. Let P be the mid-point of segment AB.
5. With P as centre and radius PA draw a circle which intersects the circle with centre A at M and the circle with centre B at R and S.
6. Join BM and BN. Also join AR and AS. Then, BM, BN, AR and AS are required tangents.

Question 25.
The rod AC of a TV disc antenna is fixed at right angles to the wall AB and a rod CD is supporting the disc as shown in figure. If AC = 1.5 m and CD = 3 m, then find
(a) tan θ
(b) sec  θ + cosec θ

If sin θ + cosθ = √3, then prove that tan θ + cot θ = 1.
Solution :

Question 26.
In the given figure, two tangents TP and TQ are drawn to a circle with centre O from an external point T

Prove that: ∠PTQ = ∠OPQ.
Solution :
We know that, tangents drawn from same external point are equal.

Section-IV

Question 27.
Find HCF and LCM of 404 and 96 and verify that HCF x LCM = product of the two given numbers.
Solution :
404 = 2 x 2 x 101 =2² x 101
96 = 2 x 2 x 2 x 2 x 2 x 3 = 2⁵ x3
∴ HCF of 404 and 96 = 2² = 4
LCM of 404 and 96 = 101 x 2⁵ x 3 = 9696 HCF x LCM
= 4 x 9696 = 38784
Also   404 x 96 = 38784
Hence HCF x LCM = Product of 404 and 96.

Question 28.
If the roots of the equation (a – b)x² + (b – c )x + (c – a) = 0 are equal, prove that 2a = b + c.
OR
The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Solution :
For real and equal roots, D = 0 ⇒ (b – c)² – 4(a – b)(c -a) = 0
⇒ b² + c² – 2be – 4ac + 46c + 4a² – 4ab = 0
⇒ 4a² + b² + c² – 4ab + 2 be – 4ac = 0
⇒ (-2a)² + (b)¹ + (c)² + 2(-2 a)b + 2 (b)(c) + 2c(-2a) = 0
⇒ ( – 2a + b + c)² = 0  ⇒ – 2a + b + c = 0
⇒ 2 a = b + c
OR
Let the two consecutive odd numbers be x and x +2
x² + (x + 2)² = 394                                ⇒ x² + x² + 4 + 4x = 394
⇒ 2x² + 4x + 4 = 394                            ⇒ 2x² + 4x – 390 = 0
⇒  x² + 2x – 195 = 0                               ⇒ x² + 15x – 13x – 195 = 0
⇒ x (x + 15) – 13 (x + 15) = 0                ⇒ (x- 13) (x+ 15) = 0
x – 13 = 0                                                  ⇒ or x + 15 = 0
⇒  x= 13                                                   ⇒ or x = – 15 (neglected)

When the first number x = 13, then the second number x + 2 = 13 + 2= 15.

Question 29.
ABCD is a square of side 4 cm. At each comer of the square, a A quarter circle of radius 1 cm, and at the centre, a circle of radius 1 cm, are drawn, as shown in the given figure. Find the area of the shaded region.

Solution :
Area of the shaded portion
= Area of given square ABCD
– 4 x area of each quarter circle
– area of the circle at the centre.

Question 30.
In the given figure, \(\angle \mathrm{D}=\angle \mathrm{E} \text { and } \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) prove that ΔBAC is an isosceles triangle.

OR
Prove that the sum of squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

[ ∴ AB = BC = CD = AD, sides of a rhombus]
Hence, the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Question 31.
In the given figure, PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ, QR and RS.

Solution :
If three or more line segments are perpendiculars to one line, then they are parallel to each other.

Question 32.
The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.
Solution:
Let AB be the tower and angle of elevation from point C = 30°

Question 33.
The median of the following data is 525. Find the values ofx andy, if total frequency is 100.

Solution:

Section – V

Question 34.
A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff of height 6 m. At a point on the plane, the angle of elevation of the bottom and top of the flag­staff are 30° and 45° respectively. Find the height of the tower. (Take 73 = 1.73)
OR
A vertical tower stands on a horizontal plane and is surmounted by a flag-staff of height 5 m. From a point on the ground the angles of elevation of the top and bottom of the flag-staff are 60° and 30° respectively. Find the height of the tower and the distance of the point from base of the tower.
Solution :

Or

Let height of tower (TR) be x m, distance (RP) of a point from the base of tower be y m, height of the flag-staff (QT) be 5 m. Then in the ΔTRP,

Question 35.
A toy is in the form of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of the conical part is 12 cm.

Solution :
Let r be the radius and h the height of the cylindrical part of the toy.
Then, r = 5 cm and h = 13 cm.
Let, r₁ be the radius of the conical part, h₁ its height and 1 its slant height.

Question 36.
A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Solution :
Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h.
∴ Speed of the boat going upstream = (x-y) km/h and
speed of the boat going downstream = (x + y) km/h



On solving (v) and (vi), we get x = 8, y = 3
Hence, speed of the boat in still water = 8 km/h and speed of the stream = 3 km/h.

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

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

Time: 3 Hours
Maximum Marks: 80

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 of 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. It consists of three sections III, IV and V.
2. In section III, Question Nos. 21 to 26 are Very Short Answer Type questions of 2 marks each.
3. In section IV, Question Nos. 27 to 33 are Short Answer Type questions of 3 marks each.
4. In section V, Question Nos. 34 to 36 are Long Answer Type questions of 5 marks each.
5. Internal choice is provided in 2 questions of 2 marks, 2 questions of 3 marks and 1 question of 5 marks.

Part – A
Section-I

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

Question 1.
If xy = 180 and HCF(x, y) = 3, then find the LCM(x, y).
OR
The decimal representation of will terminate after how many decimal places?
Solution :
(LCM)(3)= 180
LCM = 60
OR
Four decimal places.

Question 2.
If the sum of the zeroes of the quadratic polynomial 3×2 – kx + 6 is 3, then find the value of k.
Solution :

Question 3.
For what value of k, the pair of linear equations 3x + y = 3 and 6.x + ky = 8 does not have a solution.
Solution :

Question 4.
If 3 chairs and 1 table costs ₹ 1500 and 6 chairs and 1 table costs ₹ 2400, form linear equations to represent this situation.
Solution :
Let the cost of 1 chair = ₹ x
And the cost of 1 table = ₹ y
3 x + y= 1500
6x + y = 2400

Question 5.
Which term of the AP 27, 24, 21, is zero?
OR
In an Arithmetic Progression, if d = -4, n = 7, an = 4, then find a.
Solution :
a_n = a + (n – 1) d
0= 27 + (n-1)(-3)
30= 3n
n= 10
10th term.
OR
a_n= a + (n — 1 )d
4 = a + 6 X (—4)
a = 28

Question 6.
For what values of k, the equation 9x² + 6kx + 4 = 0 has equal roots?
Solution :
9x² +6kx + 4 = 0
(6k)²-4 x 9 x 4 = 0
36k² = 144
⇒ k² = 4
k = ±2

Question 7.
Find the roots of the equation x² + 1x + 10 = 0.
OR
For what value(s) of ‘a’ quadratic equation 3ax² – 6x + 1 = 0 has no real roots?
Solution :
x² + 7x+ 10= 0
x² + 5x + 2x+ 10= 0
(x + 5)(x + 2) = 0
x = -5, x = -2
OR
3ax² – 6x + 1 = 0
(-6)²– 4(3a)(1) < 0
12a > 36
⇒ a > 3

Question 8.
If PQ = 28 cm, then find the perimeter of ΔPLM.

Answer:
PQ = PT

PL + LQ= PM + MT
PL + LN = PM + MN
Perimeter (ΔPLM) = PL + LM + PM
= PL + LN + MN + PM = 2 (PL + LN) = 2(PL + LQ)
= 2 x 28 = 56 cm

Question 9.
If two tangents inclined at 60° are drawn to a circle of radius 3 cm, then find length of each tangent.
OR
PQ is a tangent to a circle with centre O at point P. If ∠OPQ is an isosceles triangle, then find ∠OQP.
Answer:

Question 10.
In the AABC, D and E are points on side AB and AC respectively such that DE || BC. If
AE = 2 cm, AD = 3 cm and BD = 4.5 cm, then find CE.
Answer:

Question 11.
In the figure, if B₁, B₂, B₃,…. and A₁, A₂, A₃,….. have been marked at equal distances. In what ratio C divides AB?

Answer:
8:5

Question 12.
sin A + cos B =1, A = 30° and B is an acute angle, then find the value of B.
Answer:

Question 13.
If x = 2 sin² θ and y = 2 cos² θ + 1, then find x + y
Answer:
x + y = 2sin²
0 + 2 cos² 9 + 1
= 2(sin² 0 + cos² 0) + 1
= 3

Question 14.
In a circle of diameter 42 cm,if an arc subtends an angle of 60° at the centre where n = then what will be the length of arc?
Answer:

Question 15.
12 solid spheres of the same radii are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. Find the diameter of the each sphere.
Answer:

Question 16.
Find the probability of getting a doublet in a throw of a pair of dice.
OR
Find the probability of getting a black queen when a card is drawn at random from a well- shuffled pack of 52 cards.
Answer:

Section-II

Case Study based questions are compulsory. Attempt any four sub-parts of each question. Each sub-part carries 1 mark.

Case Study Based-1 Sun Room

Question 17.
The diagrams show the plans for a sun room. It will be built onto the wall of a house. The four walls of the sun room are square clear glass panels. The roof is made using four clear glass panels, trapezium in shape, all the same size

• one tinted glass panel, half a regular octagon in shape.

Refer to Top View
(a) Find the mid-point of the segment joining the points J (6, 17) and I (9, 16).
(i) \(\left(\frac{33}{2}, \frac{15}{2}\right)p\)
(ii) \(\left(\frac{3}{2}, \frac{1}{2}\right)\)
(iii) \(\left(\frac{15}{2}, \frac{33}{2}\right)\)
(iv) \(\left(\frac{1}{2}, \frac{3}{2}\right)\)
Solution:
(iii) \(\left(\frac{15}{2}, \frac{33}{2}\right)\)

Refer to Front View
(b) The distance of the point P from the y-axis is
(i) 4
(ii) 15
(iii) 19
(iv) 25
Solution:
(i) 4

Refer to Front View
(c) The distance between the points A and S is
(i) 4
(ii) 8
(iii) 16
(iv) 20
Solution:
(iii) 16

Refer to Front View
(d) Find the coordinates of the point which divides the line segment joining the points A and B in the ratio 1 : 3 internally.
(i) (8.5, 2.0)
(ii) (2.0, 9.5)
(iii) (3.0, 7.5)
(iv) (2.0, 8.5)
Solution:
(iv) (2.0, 8.5)

Refer to Front View
(e) If a point (x,v) is equidistant from the Q(9. 8) and S( 17, 8), then 1
(i) x + y = 13
(ii) x – 13 = 0
(iii) x – 13 = 0
(iv) x – y – 13
Solution:
(ii) x – 13 = 0

Case Study Based-2

Question 18.
Scale Factor and Similarity
Scale Factor

Solution :
A scale drawing of an object is the same shape as the object but a different size.
The scale of a drawing is a comparison of the length used on a drawing to the length it represents. The scale is written as a ratio.

Similar Figures

The ratio of two corresponding sides in similar figures is called the scale factor.

If one shape can become another using resizing, then the shapes are similar.

Hence, two shapes are similar when one can become the other after a resize, flip, slide or turn.
(a) A model of a boat is made on the scale of 1 : 4. The model is 120 cm long. The full size of the boat has a width of 60 cm. What is the width of the scale model?
(i) 20 cm
(ii) 25 cm
(iii) 15 cm
(iv) 240 cm
Solution :
(iii) 15 cm

(b) What will effect the similarity of any two polygons?
(i) They are flipped horizontally.
(ii) They are dilated by a scale factor.
(iii) They are translated down.
(iv) They are not the mirror image of one another.
Solution :
(iv) They are not the mirror image of one another.

(c) If two similar triangles have a scale factor of a : b, which statement regarding the two triangles is true?
(i) The ratio of their perimeters is 3a : b
(ii) Their altitudes have a ratio a : b
(iii) Their medians have a ratio \(\frac{a}{2}: b\)
(iv) Their angle bisectors have a ratio a²: b²
Solution :
(ii) Their altitudes have a ratio a : b

(d) The shadow of a stick 5 m long is 2 m. At the same time, the shadow of a tree 12.5 m high is

(i) 3 m
(ii) 3.5 m
(iii) 4.5 m
(iv) 5 m
Solution :
(iv) 5 m

(e) Below you see a student’s mathematical model of a farmhouse roof with measurements. The attic floor, ABCD in the model, is a square. The beams that support the roof are the edges of a rectangular prism, EFGHKLMN. E is the middle of AT, F is the middle of BT, G is the middle of CT. and H is the middle of DT. All the edges of the pyramid in the model have length of 12 m.

What is the length of EF, where EF is one of the horizontal edges of the block?
(i) 24 m
(ii) 3 m
(iii) 6 m
(iv) 10 m
Solution :
(iii) 6 m

Case Study Based-3

Question 19.
Applications of Parabolas-Highway Overpasses/Underpasses A highway underpass is parabolic in shape.

Answer:
Parabola: A parabola is the graph that results from p(x) — ax² + bx + c. Parabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the vertex.

(a) If the highway overpass is represented by x² – 2x – 8, then its zeroes are
(i) (2. -4)
(ii) (4, -2)
(iii) (-2, -2)
(iv) (-4, -4)
Answer:
(ii) (4, -2)

(b) The highway overpass is represented graphically. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where the graph of polynomial
(i) intersects x-axis
(ii) intersects y-axis
(iii) intersects y-axis or x-axis
(iv) None of the above
Answer:
(i) intersects x-axis

(c) Graph of a quadratic polynomial is a
(i) straight line.
(ii) Circle
(iii) Parabolla
(iv) ellipse
Answer:
(iii) Parabolla

(d) The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is 1
(i)  x² – 6x + 2
(ii) x² – 36
(iii) x² – 6
(iv) x² – 3
Answer:
(ii) x² – 36

(e) The number of zeroes that polynomial f(x) = (x – 2)² + 4 can have is:
(i) 1
(ii) 2
(iii) 0
(iv) 3
Answer:
(iii) 0

Case Study Based-4

Question 20.

(a) Estimate the mean time taken by a student to finish the race.
(i) 54
(ii) 63
(iii) 43
(iv) 50
Solution :
(iii) 43

(b) What will be the upper limit of the modal class?
(i) 20
(ii) 40
(iii) 60
(iv) 80
Solution :
(iii) 60

(c) The construction of cumulative frequency table is useful in detennining the
(i) mean
(ii) median
(iii) mode
(iv) All of the above
Solution :
(ii) median

(d) The sum of lower limits of median class and modal class is
Solution :
(i) 60
(ii) 100
(iii) 80
(iv) 140
Solution :
(iii) 80

(e) How many students finished the race within 1 minute?
(i) 18
(ii) 37
(iii) 31
(iv) 8
Solution :
(iii) 31

Part-B
Section-III

All questions are compulsory. In case of internal choices, attempt anyone.

Question 21.
3 bells ring at an interval of 4, 7 and 14 minutes. All three bell rang at 6 am, when the three balls will the ring together next?
Solution :
4 = 2 x 2
7 =7 x 1
14 = 2 x 7
LCM = 2 x 2 x 7 = 28
The three bells will ring together again at 6 : 28 am

Question 22.
Find the point on x-axis which is equidistant from the points (2, -2) and (-4, 2).
OR
P(-2, 5) and Q(3, 2) are two points. Find the coordinates of the point R on PQ such that PR = 2QR
Solution:

Question 23.
Find a quadratic polynomial whose zeroes are \(5-3 \sqrt{2} \text { and } 5+3 \sqrt{2}\)
Solution:

Question 24.
Draw a line segment AB of length 9 cm. With A and B as centres, draw circles of radius 5 cm and 3 cm respectively. Construct tangents to each circle from the centre of the other circle.
Solution:

Question 25.
If \(tan \mathrm{A}=\frac{3}{4}\), find the value of \(\frac{1}{\sin A}+\frac{1}{\cos A}\) and 3 cm respectively. Construct tangents to each circle from the centre of the other circle. If \(\sqrt{3} \sin \theta-\cos \theta=0 \text { and } 0^{\circ}<\theta<90^{\circ} \) find the value of θ.
Solution:

Question 26.
In the figure, quadrilateral ABCD is circumscribing a circle with centre O and AD ⊥ AB. If radius of incircle is 10 cm, then find the value of x.

Solution:

Question 27.
Prove that 2 – √3 is irrational, given that √3 is irrational.
Solution:

Question 28.
If one root of the quadratic equation 3x² + px + 4 = 0 is \(\frac{2}{3}\), then find the value of p and the other root of the equation.
OR
The roots α and β of the quadratic equation x² – 5x + 3(k – 1) = 0 are such that α – β = 1. Find the value k.
Solution :

Question 29.
In the figure, ABCD is a square of side 14 cm. Semicircles are drawn with each side of square as diameter. Find the area of the shaded region.

Solution:

Question 30.
The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of the first triangle is 9 cm, find the length of the corresponding side of the second triangle.
OR
In an equilateral triangle ABC, D is a point on side BC such that BD = -BC. Prove that 9 AD² = 7 AB².
Solution:

Question 31.
The median of the following data is 16. Find the missing frequencies a and b, if the total of the frequencies is 70.

Solution:

Question 32.
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°, then find the height of the candle.

Solution:

Let AB = candle
C and D are two coins

Question 33.
The mode of the following data is 67. Find the missing frequency x:
Solution:

Question 34.
The two palm trees are of equal heights and are standing opposite to each other on either side of the 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 the trees.
OR
The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60° respectively. Find the height of the tower, and also the horizontal distance between the building and the tower.
Solution:

Hence, height of the tower = h = 75 m
Distance between the building and the tower = 25 √3 = 43.25 m

Question 35.
Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm at the rate of 0.7 m/sec. By how much will the water rise in the tank in half an hour?
Solution:
For pipe, r = 1cm
Length of water flowing in 1 sec, h = 0.1 m = 70 cm
Cylindrical Tank, R = 40 cm, rise in water level = H
Volume of water flowing in 1 sec = n×h = a × 1 x 1 × 70 = 70a
Volume of water flowing in 60 sec = 70a: x 60
Volume of water flowing in 30 minutes = 70a × 60 ×30
Volume of water in Tank = ar²H = a × 40 × 40 × H
Volume of water in Tank = Volume of water flowing in 30 minutes
a × 40 × 40 × H = 70a x 60 x 30

Question 36.
A motorboat covers a distance of 16 km upstream and 24 km downstream in 6 hours. In the same time, it covers a distance of 12 km upstream and 36 km downstream. Find the speed of the boat in still water and that of the stream.
Solution:
Let speed of the boat in still water = x km/hr,
and Speed of the stream = y km/hr
Downstream speed = (x + y) km/hr
Upstream speed = (x – y) km/hr 24

Thus, speed of the boat in still water = 8 km/hr,
Speed of the stream = 4 km/hr

Students can access the CBSE Sample Papers for Class 10 Maths Standard with Solutions and marking scheme Set 1 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 10 Maths Standard Set 1 with Solutions

Time: 3 Hours
Maximum Marks: 80

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 of 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. It consists of three sections III, IV and V.
2. In section III, Question Nos. 21 to 26 are Very Short Answer Type questions of 2 marks each.
3. In section IV, Question Nos. 27 to 33 are Short Answer Type questions of 3 marks each.
4. In section V, Question Nos. 34 to 36 are Long Answer Type questions of 5 marks each.
5. Internal choice is provided in 2 questions of 2 marks, 2 questions of 3 marks and 1 question of 5 marks.

Part – A
Section-I

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

Question 1.
Find the zeroes of the of the polynomials p(x) = 4x² – 12x + 9
Solution :

Question 2.
In the given figure, DE || BC. Find \(\frac{x}{y}\) DE || BC

Solution :

Question 3.
If 2k + 1, 6, 3& + 1 are in AP, then find the value of k.
OR
If the sum of n terms of an AP is 2n² + 5n, then find the 2^nd term
Solution :
For an AP a, b, c; 2b = a + c ⇒ (2k + 1) + (3k + 1) = 2×6
5k+2=12
⇒ 5k = 10
⇒ k = 2

Question 4.
If x = a, y = b is the solution of the pair of equations x-y-2 and x + y = 4, then find the values of a and b.
Solution :

Question 5.
At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle (see figure). Find the length of the chord CD parallel to XY and at a distance 8 cm from A.

Solution :
OD² = OM² + DM²
⇒ 5² = 3² + x²
x² = 25 – 9 = 16
x = 4
CD =2 x = 2 × 4 = 8 cm

Question 6.
Find the radius of the largest right circular cone that can be cut out from a cube of edge 4.2 cm.
Solution :
Edge of the cube = 4.2 cm
∴ Radius of the largest right circular cone \(\frac{1}{2}\) (Edge of the Square) \(=\frac{4.2}{2}=2.1 \mathrm{~cm}\)

Question 7.
The circumference of a circle exceeds its diameter by 180 cm. Find its radius
Solution :

Question 8.
A bag contains 40 balls out of which some are red, some are blue and remaining are black. If the probability of drawing a red ball is \(\frac{11}{20} \) and that of blue ball is \(\frac{1}{5} \) , then find the number of black balls.
OR
Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail.
Solution :

Let the number of black balls = x.

Total number of balls = Total possible outcomes = 40

⇒ Number of black balls =10
OR
Number of possible outcomes = 4 as possible outcomes are HH, HT, TH, TT. Favourable outcomes for getting at least one tail are HT, TH, TT No. of favourable outcomes = 3
∴ P(getting at least one tail) =\(\frac{3}{4}\)

Question 9.
If Cos θ \(\frac{7}{8}\) then find the value of \(\frac{(1+\cos \theta)(1-\cos \theta)}{(1-\sin \theta)(1+\sin \theta)}\)
Solution :

Question 10.
Write whether the rational number \(\frac{7}{75}\) will have a terminating decimal expansion or a non-terminating repeating decimal expansion.
OR
Write the HCF of the smallest composite number and the smallest prime number.
Solution :
\(\frac{7}{75}=0.0933333 \ldots \ldots=0.09 \overline{3}\)
So, it is a non-terminating repeating decimal expansion.
OR
The smallest composite number is 4 and the smallest prime number is 2.
The prime factorisation of 4 = 2 x 2 = 2² and the prime factorisation of 2 = 2¹
Now, the HCF of 2 and 4 is the product of smallest power of each common prime factor in . the numbers.
HCF (2, 4) = 2¹ = 2

Question 11.
Find the value(s) of k, if the quadratic equation 3x² – k √3x + 4 = 0 has equal roots.
OR
If -5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, then find the value of k.
Solution :

Question 12.
In the given figure, ABCD is a rectangle. Find the values of x and y.

Solution :

Question 13.
If ax² + bx + c = 0 has equal roots, find the value of
Solution :
For equal roots D = 0
i.e., b² – 4ac =0
⇒ b² = 4ac
\(c=\frac{b^{2}}{4 a}\)

Question 14.
In the given figure, PA and PB are tangents to the circle drawn from an external point P. CD is the third tangent touching the circle at Q. If PA = 15 cm, find the perimeter of APCD.

OR
In the given figure, the quadrilateral PQRS circumscribes a circle with centre O. If ∠POQ = 115°, then find ∠ROS.

Solution :
Since PA and PB are tangents from same external point P.
PA = PB = 15 cm
Now, Perimeter of APCD = PC + CD + DP = PC + CQ + QD + DP
= PC + CA + DB + DP = PA + PB = 15 cm +15 cm = 30 cm
OR
Solution :
Since  ∠POQ = ∠ROS   (Vertically opposite angles)
⇒ ∠ROS = 115°

Question 15.
To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn such that ∠BAX is an acute angle and then at equal distances points are marked on the ray. Find the minimum number of these points.
Solution :
Since 5+ 7 = 12
So, the number of points marked on ray AX = 12.

Question 16.
Find the value of (sin 30° + cos 60°).
Solution :
sin 30° + cos 60° \(=\frac{1}{2}+\frac{1}{2}=\)

Section-II

Question 17.
Case Study Based-1
Medicinal Garden
A medicinal garden is a garden in which different kinds of medicinal plants, like Aloe Vera, Mint, Lemon Balm, etc. are planted with the goal of serving the need of general health maintenance. Observe the following diagram.

Refer to IJKL
(a) The mid-point of the segment joining the points 1(6, 6) and J(6, 18) is
(i) (7,9)
(ii) \(\left(12, \frac{11}{2}\right) \)
(iii) (6, 12)
(iv) (12,24)

Refer to EFGH
(b) The distance between points H(10, 6) and F(14, 18) is
(i) \(8 \sqrt{5}\)unit
(ii) \(4 \sqrt{10}\) unit
(iii) 18 unit
(iv) 24 unit

Refer to ABCD

(c) The coordinates of the points A and B are (22,6) and (22,18) respectively. The x-coordinate of a point R on the line segment AB such that \(\frac{A R}{A B}=\frac{3}{5}\) is…………….
(i) 18
(ii) 24
(iii) 22
(iv) 31

Refer to MQ

(d) The ratio in which the points (20, k) divides the line segment joining the points M(4, 2) and Q(24, 2) is
(i) 4 : 1
(ii) 16 : 15
(iii) 8 : 21
(iv) 10 : 17

Refer to MH and HP

How much longer is HP than MH given that coordinates of H(10, 6), M(4, 2) and P(19,2)
(i) \((\sqrt{95}-2 \sqrt{3}) \text { unit }\)
(ii) \((\sqrt{97}-2 \sqrt{13}) \text { unit }\)
(iii) \((\sqrt{61}-4 \sqrt{5}) \text { unit }\)
(iv) None of there.
Solution:

Question 18.
Case Study Based-2
A Frame House
A frame-house is a house constructed from a wooden skeleton, typically covered with timber board. The concept of similar triangles is used to construct it. Look at the following picture:

Refer to House (i)
(a) The front view of house (7) is shown below in which point P on AB is joined with point Q

If PQ || BC, AP = x m  PB = 10 m. AQ = (x – 2) m, QC = 6 m, then the value of a is
(i) 3m
(ii) 4m
(iii) 5m
(iv) 8 m

(b) The side vies of house (i) is shown below in which point F on AC is joined with point G on DE.

If ACED is a trapezium with AD || CE, F and G are points on non-parallel sides AC and AF
DE respectively such that FG is parallel to AD, then =\(\frac{\mathrm{AF}}{\mathrm{FC}}=\)
(i) \(\frac{\mathrm{DG}}{\mathrm{GE}}\)
(ii) \(\frac{\mathrm{AD}}{\mathrm{CE}}\)
(iii) \(\frac{\mathrm{AF}}{\mathrm{GE}}\)
(iv) \(\frac{\mathrm{DG}}{\mathrm{FC}}\)

(c) The front view of house (ii) is shown below in which point S on PQ is joined with point T on PR.

\(\frac{\mathrm{PS}}{\mathrm{QS}}=\frac{\mathrm{PT}}{\mathrm{TR}} \text { and } \angle \mathrm{PST}=70^{\circ}, \angle \mathrm{QPR}=50^{\circ}\) then the angle ∠QRP =
(i) 70°
(ii) 50°
(iii) 80°
(iv) 60°

(d) Again consider the front view of house (ii). If S and T are points on side PQ and PR respectively such that
ST || QR and PS : SQ = 3 : 1. Also TP = 6.6 m, then PR is

(i) 6.9 m
(ii) 8.8 m
(iii) 10.5m
(iv) 9.4 m

(e) Sneha has also a frame house whose front view is shown below

If MN || AB, BC = 7.5 m, AM = 4 m and MC = 2 m, then length of BN is
(i) 5 m
(ii) 4 m
(iii) 8 m
(iv) 9 m
Solution :




Question 19.
Case Study Based-3

Rainbow is an arch of colours that is visible in the sky, caused by the refraction and dispersion of the sun’s light after rain or other water droplets in the atmosphere. The colours of the rainbow are generally said to be red, orange, yellow, green, blue, indigo and violet.

Each colour of rainbow makes a parabola. We know that for any quadratic polynomial ax² + bx + c, a ≠ 0, the graph of the corresponding equation y = ax² + bx + c has one of the two shapes either open upwards like ∪ or open downwards like ∩ depending on whether a > 0 or a < 0. These curves are called parabolas.

(a) A rainbow is represented by the quadratic polynomial x² + (a + 1 )x + b whose zeroes are 2 and-3. Then
(i) a = -7, b   = -1
(ii)  a = 5, b =-1
(iii)  a – 2, b = – 6
(iv)   a – 0, b = – 6

(b) The polynomial x² – 2x – (7p + 3) represents a rainbow.   If -4 is zero of it, then the value of p is
(i) 1
(ii)  2
(iii) 3
(iv)  4

(c) The graph of a rainbow y=f(x) is shown below.

The number of zeroes of f(x) is
(i) 0
(ii) 1
(iii) 2
(iv) 3

(d) If graph of a rainbow does not intersect the x-axis but intersects y-axis in one point, then number of zeroes of the polynomial is equal to
(i) 0
(ii) 1
(iii) 0 OR 1
(iv) none of these

(e) The representation of a rainbow is a quadratic polynomial. The sum and the product of its zeroes are 3 and -2 respectively. The polynomial is 1
(i) k(x² – 2x – 3), for some real k.
(ii) k(x² – 5x – 9), for some real k.
(iii) k(x² – 3x – 2), for some real k.
(iv) k(x² – 8x + 2), for some real k.
Solution :
(a) x² + (a + 1)x + b =(2)² + (a + 1)2 + b = 0 and (-3)² + (a + l)(-3) + b = 0
⇒ 4 + 2a + 2 + 6=0 and 9 — 3a — 3 + 6 = 0
⇒ 2a + b=-6      … (i)
and  -3a + b=-6   …(ii)
Solving (i) and (ii), we get a = 0 and b = -6
So, option (iv) is correct answer.

(b) p(-4) = 0 ⇒ (-4)² – 2(-4) – (7p + 3) = 0
⇒ 16 + 8 – 7p-3 = 0    ⇒ 21 – 7p = 0
⇒ p = 3
So, option (iii) is correct answer.

(c) ∵ Graph f(x) intersects x-axis at two different points.
∴  Number of zeroes of f(x) = 2.
So, option (iii) is correct answer.

(d) We know that the number of zeroes of a polynomial is equal to number of points of intersection of the graph of polynomial with x-axis.
Since the graph of rainbow does not intersect the x-axis, so it has no zeroes.
So, option (i) is correct answer.

(e) Let the required polynomial be f(x).
Then f(x) = k(x² – 3x – 2) for some real k.
So, option (iii) is correct answer.

Question 20.
Case Study Based-4

(a) The mid-value (class-mark) of 160-170 is ……..
(i) 140
(ii) 145
(iii) 155
(iv) 165

(b) The approximate mean weekly cost-of-living index is 1
(i) 166.4
(ii) 184.5
(iii) 190
(iv) 201.8

(c) The sum of lower and upper limits of modal class is
(i) 290
(ii) 310
(iii) 330
(iv) 350

(d) Mode is the value of the variable which has
(i) maximum frequency
(ii) minimum frequency
(iii) mean frequency
(iv) middle most frequency

(e) The median class of above data is
(i) 150-160
(ii) 160-170
(iii) 170-180
(iv) 190-200

Solution :
(a) Class-mark of 160-170 = \(160-170=\frac{160+170}{2}=\frac{330}{2}=165\)
So, option (iv) is correct answer.

\(\text { Mean }=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{8650}{52}=166.4(\text { approx. })\)
So, option (i) is correct answer.

(c) Maximum frequency is 20
∴ Modal class = 160-170
Lower limit of modal class = 160
Upper limit of modal class =170
Sum of lower and upper limits = 160 + 170 = 330
So, option (iii) is correct answer.

(d) (i) Maximum frequency

n= 52 \(\Rightarrow \frac{n}{2}=26\)
Median class is 160-170.
So, option (ii) is correct answer.

Part-B
Section-III

Question 21.
If two positive integers p and q are written as p = a²b² and q = a³b; a, b are prime numbers, then verify: LCM (p, q) x HCF (p, q) =pq.
Answer:
LCM (p, q) =a³b³ and HCF (p, q) = a²b
LCM (p, q) x HCF (p, q) =a⁵b⁴ = (a²b³) (a³ b) = pq

Question 22.
Draw a line segment of length 6 cm. Using compass and ruler, find a point P on it which divides it in the ratio 3:4.

Solution :
Steps of construction:
1. Draw a line segment AB = 6 cm.
2. Draw an acute angle <BAX.
3. Along AX take 7 points, such that
AA₁ — A₁ A₂ — A₂ A₃ — A₃A₄ — A₄A₅ – A₅A₆ — A₆A₇
4. Join BA₇
5. Through A₃ draw A₃P A₇B which meets AB at P.
6. AP: PB = 3 : 4 and P is the required point.

Question 23.
If 7 sin² A + 3 cos² A = 4, show that tan \(A=\frac{1}{\sqrt{3}}\)
OR
Prove that \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A\)
Solution :

Question 24.
In the given figure, XY and XT’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and XT’ at B. Prove that ∠AOB = 90°.

Solution :

Question 25.
The coordinates of the points P and Q are respectively (4,—3) and (—1, 7). Find the x-coordinate (abscissa) of a point R on the line segment PQ such that \(\frac{P R}{P Q}=\frac{3}{5}\)
OR
Find the ratio in which the point (—3, k) divides the line segment joining the points (—5, —4) and (—2, 3). Also find the value of k.
Solution :

Question 26.
Find k, if the sum of the zeroes of the polynomial x² – (k + 6) x + 2 (2k – 1) is half of their product.
Answer:

Section-IV

Question 27.
In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required if in each room the same number of participants are to be seated and all of them being of the same subject.
Solution :

Question 28.
In the given figure, ABPC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Solution :

Question 29.
Show that in a right triangle, the square of the hypotenuse is equal to the sum of the squares
of the other two sides.
OR
P is the mid-point of side BC of AABC, Q is the mid-point of AP, BQ when produced meets AC at L. Prove that AL = \(\frac{1}{3} \) AC.
Solution :

Question 30.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency k.

Solution :

Question 31.
Using quadratic formula, solve the following equation for X:
abx²+(b²—ac)x—bc=O
OR
The difference of two natural numbers is 5 and the difference of their reciprocals is \(\frac{1}{10}\). Find the numbers.
Solution :

Let the two natural numbers be x and y such that x >y
According to the question,
Difference of numbers,
x – y =  5    ⇒  x = 5 +y ……………(1)
Difference of their reciprocals,
\(\frac{1}{y}-\frac{1}{x}=\frac{1}{10}\)

= (y – 5)(y + 10) = 0  ∴ y = 5  or y = – 10
y is a natural number. ∴ y = 5
Putting the value of)’ in (i), we get
x= 5+5=10
Thus, the required numbers are 10 and 5.

Question 32.
Find the angle of depression from the top of 12m high tower of an object lying at a point 12 m away from the base of the tower.
Solution :
Let AB be the tower of 12 m height and B its base. Let C be a point A 12 m away from base B of tower AB where an object situated.

Question 33.
If the median of the distribution given below is 28.5, find the values of x and y.

Solution :
Here, median = 28.5, n = 60

Section – V

Question 34.
The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distance between the building and the tower.
OR
The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower.
Solution :

Thus, height of the other tower = 10 m and the distance between two towers = BD = 10√3 m.

Question 35.
A hemispherical depression is cut out from one face of a cubical wooden block of edge 21 cm, such that the diameter of the hemisphere is equal to the edge of the cube. Determine the volume and total surface area of the remaining block.
Solution :

Question 36.
A person can row 8 km upstream and 24 km downstream in 4 hours. He can row 12 km downstream and 12 km upstream in 4 hours. Find the speed of the person in still water and also the speed of the current. 5
Solution :
Let the person’s speed of rowing in still water be, x km/h and speed of the current y km/h.
∴ Speed of boat in downstream = (x + y) km/h
and speed of boat in upstream = (x -y) km/h
Since \(\text { Time }=\frac{\text { Distance }}{\text { Speed }}\)

Students can access the CBSE Sample Papers for Class 10 Science with Solutions and marking scheme Set 5 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 10 Science Set 5 for Practice

Time: 3 Hours
Maximum Marks: 80

General Instructions:

(i) The question paper comprises four sections A, B, C and D. There are 36 questions in the question paper. All questions are compulsory.
(ii) Section-A – question no. 1 to 20 – all questions and parts there of are of one mark each.
These questions contain multiple choice questions (MCQs), very short answer questions and assertion – reason type questions. Answers to these should be given in one word or one sentence.
(iii) Section-B – question no. 21 to 26 are short answer type questions, carrying 2 marks each. Answers to these questions should in the range of 30 to 50 words.
(iv) Section-C – question no. 27 to 33 are short answer type questions, carrying 3 marks each.
Answers to these questions should in the range of 50 to 80 words. :
(v) Section—D – question no. – 34 to 36 are long answer type questions carrying 5 marks each. Answer to these questions should be in the range of 80 to 120 words.
(vi) There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
(vii) Wherever necessary, neat and properly labelled diagrams should be drawn.

Section – A

1. If an element X is placed in group 14, what will be the formula and the nature of bonding of  its chloride?

2. Where should an object be placed from a converging lens of focal length 15 cm, so as to obtain the image of same size and real.

3. A solution of a substance ‘X’ is used in white washing.
(i) Name the substance ‘X’ and write its formula.
(ii) Write the reaction of the substance ‘X’ named in (i) above with water.
OR
What happens chemically, when quicklime is added to water filled in a bucket?

4. Three acidic solutions A, B and C have pH = 0, 3 and 5 respectively:
(i) Which solution has the highest concentration of H⁺ ions?
(ii) Which solution has the lowest concentration of H⁺ ions?

5. Generally, when metals are treated with mineral acids, hydrogen gas is liberated, but when metals (except Mn and Mg) are treated with HNO₃, hydrogen is not liberated, why?

6. The volume of glomerular filtrate produced is 18 L but the volume of urine excreted is just 1 – 2 L. Give a suitable reason for this statement.

7. Ozone is deadly poisonous, still it performs an essential function. How?
OR
Write the appropriate names of trophic level ‘Z’ and ‘X’ in the figure given below.

8. Covalent compounds have low melting and boiling point. Why?

9. Give an example of a flower which contains both stamens and carpels.
OR
What are sexually transmitted diseases? Name a STDs which damages the immune system of human body.

10. What is the nature of the image formed by a concave mirror, if the magnification produced by the mirror is +4?    OR
The outer surface of a hollow sphere of aluminium of radius 50 cm is to be used as a mirror. What will be the focal length of this mirror? What type of spherical mirror will it provide?
Answer:
OR 25

11. Let the resistance of an electrical component remain constant, while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
OR
If the charge on an electron is 1.602 x 10^-19 C, find the approximate number of electrons in 1C.
Answer:
OR 6.25 × 10¹⁸

12. Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from the back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of the magnetic field?

13. What will be the amount of energy available to the organisms of secondary consumer trophic level of food chain, if the energy available to producer level is 10000 Joules.

Assertion (A) and Reason (R)
For question numbers 14,15 and 16, two statements are given- one labeled Assertion (A) and the other labeled Reason (R). Select the correct answer to these questions from the codes (i), (ii), (iii) and (iv) as given below:
(i) Both A and R are true, and R is correct explanation of the assertion.
(ii) Both A and R are true, but R is not the correct explanation of the assertion.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Answer:
100 J

14. A. The extent of refraction is different for different medium.
R. Different medium have different refractive index.
Answer:
(i)

15.
R. It is a combination reaction because CO combines with H₂ to form CH₃OH i.e., two substances combine to form a single compound.
Answer:
(i)

16. A. The sex of a child in human beings will be determined by the type of chromosome he/she inherits from the father.
R. A child who inherits ‘X’ chromosome from his father would be a girl (XX), while a child who inherits a ‘Y’ chromosome from the father would be a boy (XY).
OR
A. Two pink coloured flowers on crossing resulted in 1 red, 2 pink and 1 white flower progeny. R. It is due to double fertilisation.
Answer:
(i) OR (iii)

Answer Q. No 17 – 20 contain five sub-parts each. You are expected to answer any four sub¬parts in these questions.

17. Read the following and answer any four questions from 17 (i) to 17 (v) (4 x 1 = 4)
The splitting of a beam of white light into its seven constituent colours, when it passes through a glass prism, is called the dispersion of light.

When a beam of white light enters a prism, it gets refracted and splits into its seven constituent colours, viz. violet, indigo, blue, green, yellow, orange, and red. This splitting of the light ray occurs because of the different angles of bending for each colour. Hence, each colour while passing through the prism bends at different angles with respect to the incident beam. This gives rise to the formation of the coloured spectrum.

(i) What is the cause of dispersion of light by prism?
I. Different colours move with different speed in the prism.
II. Emergent ray bent to different extent towards the base of prism.
III. Different colours move with same speed in the prism but cover different distance.
IV. Emergent ray bent to different extent away from the base of prism.
(a) I only
(b) I and II
(c) III and IV
(d) II and III
Answer:
(b) I and II

(ii) Which colour of white light suffers least deviation when a beam of white light is passed through the prism?
(a) Blue
(b) Red
(c) Violet
(d) Green
Answer:
(b) Red

(iii) Which of the following colours viz., A, B, C and D has more speed in the prism?

(a) A
(b) B
(c) C
(d) D
Answer:
(d) D

(iv) How will you use two identical prisms P₁ and P₂ so that a narrow beam of white light incident on one prism emerges out of the second prism as white light?

Answer:
(c)

(v) Among the seven colours visible due to splitting of white light through prism which colour has shortest wavelength?
(a) Red
(b) Blue
(c) Violet
(d) Yellow
Answer:
(c) Violet

18. Read the following and answer any four questions from 18 (i) to 18 (v) (4 x 1 = 4)

More than a million Americans die of cardiac diseases each year. One of the major causes is high cholesterol levels in the blood. The National Cholesterol Education Program suggests that total blood cholesterol level should be:

Given below are blood report of two persons

(i) Which of the organ can be affected in patient A?
(a) Heart
(b) Kidney
(c) Lungs
(d) Brain
Answer:
(a) Heart

(ii) What information is left out for the blank column?
(a) Total cholesterol
(b) Triglycerides
(c) Low density cholesterol
(d) High density cholesterol
Answer:
(d) High density cholesterol

(iii) A person with high risk category have to be suggested a suitable diet? Which of the following are correct guidelines for the patient
(a) High sugar and starch
(b) Low salt and fats
(c) High proteins
(d) Low sugar and proteins
Answer:
(b) Low salt and fats

(iv) Apart from following a prescribed diet, some other changes should be brought in the lifestyle to avoid aggravation of symptoms in a patient who is already suffering from high blood cholesterol-
A. Yoga and exercise
B. Quitting smoking and alcohol
C. Walking and doing small chores on your own
D. Enjoying loud music
Which of the following is the correct option
(a) A ,C
(b) B,C,D
(c) A,B,C
(d) A, D
Answer:
(c) A,B,C

(v) Which of the following is correct for patient B?
(a) High total cholesterol but triglycerides in normal range
(b) Total cholesterol in normal range but triglycerides are high
(c) Total cholesterol in normal range but low density cholesterol are high
(d) Total cholesterol, triglycerides and low density cholesterol are in normal range
Answer:
(d) Total cholesterol, triglycerides and low density cholesterol are in normal range

19. Read the following and answer any four questions from 19 (i) to 19 (v) (4 x 1 = 4)

Most of the characters or traits of an organism are controlled by the genes. Genes are actually segments of DNA guiding the formation of proteins by the cellular organelles. These proteins may be enzymes, hormones, antibodies, and structural components of different types of tissues. In other words, DNA/ genes are responsible for structure and functions of a living body. Genotype of an individual controls its phenotype.

(i) Select the statements that describe characteristics of genes
I. Genes are specific sequence of bases in a DNA molecule
II. A gene does not code for proteins
III. In individuals of a given species, a specific gene is located on a particular chromosome
IV. Each chromosome has only one gene
(a) I and II
(b) I and III
(c) I and IV
(d) III and IV
Answer:
(b) I and III

(ii) A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. In the progeny, all bore violet flowers, but almost half of them were short. This suggests that the genetic makeup of tall plant can be depicted as
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw
Answer:
(c) TtWW

(iii) Two pea plants one with round green seeds (RRyy) and another with wrinkled yellow (rrYY) seeds produce Fj progeny that have round, yellow (RrYy) seeds. When F₁ plants are selfed, the F₂ progeny will have new combination of characters. Choose the new combination from the following.
I. Round, yellow
II. Round, green
Wrinkled, yellow IV. Wrinkled, green
(a) I and II
(b) I and IV
(c) II and III
(d) I and III
Answer:
(b) I and IV

(iv) A section of DNA providing information for one protein is called—
(a) Nucleus
(b) Chromosomes
(c) Trait
(d) Gene
Answer:
(d) Gene

(v) Which one of the following is present in the nucleus?
(a) Gene
(b) DNA
(c) Chromosomes
(d) All of these
Answer:
(d) All of these

20. Read the following and answer any four questions from 20 (i) to 20 (v) (4 x 1 = 4)
Answer the question numbers 3(a) to 3(d) on the basis of your understanding of the following paragraph and the related studied concepts.

Atomic size refers to radius of atom. The atomic size may be visualised as the distance between the centre of the nucleus and the outermost shell of an atom.

(i) How does atomic size vary along period 2 from left to right and why?
(a) Atomic size increases as atomic number increases
(b) Atomic size increases as number of protons increases
(c) Atomic size decreases as electrons added to same shell and with increase in number of protons, nuclei attracts electrons more
(d) Atomic size decreases as proton has higher positive charge than negative charge on electron and thus protons pulls electrons towards nucleus.
Answer:
(c) Atomic size decreases as electrons added to same shell and with increase in number of protons, nuclei attracts electrons more

(ii) How does atomic size vary in group 1 and group 17 and why?
(a) Atomic size increases because electrons added to the penultimate shell
(b) Atomic size increases because electrons added to next higher energy shell
(c) Atomic size decreases because electrons added to penultimate shell
(d) Atomic size decreases because electrons added to same shell
Answer:
(b) Atomic size increases because electrons added to next higher energy shell

(iii) Which group elements have largest size in periodic table?
(a) Group 1
(b) Group 2
(c) Group 17
(d) Group 18
Answer:
(d) Group 18

(iv) Which element of group 17 is most reactive?
(a) F
(b) Cl
(c) Br
(d) I
Answer:
(a) F

(v) Which of the following has higher distance between the centre of the nucleus and the outermost shell of an atom?
(a) Li
(b) C
(c) Be
(d) I
Answer:
(a) Li

Section-B

21. The resistance of a wire of 0.01 cm radius is 10Ω. If the resistivity of the material of the wire is 50 x 10^-8 Ωm, find the length of the wire.
Answer:
62.8cm

22. Foetus derives its nutrition from the mother.
(i) Identify the tissue used for above purpose. Explain its structure.
(ii) Explain how wastes generated by developing embryo are removed.
OR
Why do we need to adopt contraceptive measures?

23. A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is
(i) pushed into the coil
(ii) held stationary inside the coil?

24. Give reason for the following:
(i) Element carbon forms compounds mainly by covalent bonding.
(ii) Kerosene does not decolourise bromine water while cooking oils do.

25. An alpha particle (positively charged) enters a magnetic field at right angle to it as shown in figure. Explain with the help of relevant rule, the direction of force acting on the alpha particle.

OR
Identify the poles of the magnet in the given figure (i) and (ii).

26. In a test tube A and B shown below, yeast was kept in sugar solution. What products of respiration would you expect in tubes A and B?

Section-C

27. Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 4D. in series with a combination of two resistors (80 each) in parallel and a voltmeter across parallel combination. Each of them dissipates maximum energy and can withstand a maximum power of 16W without melting. Find the maximum current that can flow through the three resistors.
Answer:
1 A

28. Find the current drawn from the battery by the network of four resistors shown in the diagram.

Answer:
0.4A

29. In the electrolysis of water,
(i) Name the gas collected at anode and cathode
(ii) Why is the volume of gas collected at one electrode double than the other?
(iii) What would happen if dil H₂S0₄ is not added to water?

30. A student records the observation to study the rate of respiration in three different people. Study the data collected and answer the questions given below:

(i) Which variable is kept constant?
(ii) Which reading is anomalous?
(iii) Suggest one improvement in this experiment.

31. The electrons in the atoms of four elements A, B, C and D are distributed in three shells having 1, 3, 5 and 7 electrons in outermost shell respectively. State the period in which these elements can be placed in the modem periodic table. Write the electronic configuration of the atoms A and D and the molecular formula of compound formed when A and D will combine.

32. (i) Construct a terrestrial food chain comprising four trophic levels.
(ii) What will happen if we kill all the organisms in one trophic level?
(iii) Calculate the amount of energy available to the organisms at the fourth trophic level if the energy available to the organisms at the second trophic level is 2000 J.
Answer:
(iii) 2 J

33. “pH has a great importance in our daily life” explain by giving three examples.
OR
A compound which is prepared from gypsum has the property of hardening when mixed with a proper quantity of water. Identify the compound and write its chemical formula. Write the chemical equation for its preparation. Mention any one use of the compound.

Section – D

34. You are given balls and stick model of six carbon atoms and fourteen hydrogen atoms and sufficient number of sticks. In how many ways one can join the models of six carbon atoms and fourteen hydrogen atoms to form different molecules of C₆H
OR
(i) Give a chemical test to distinguish between saturated and unsaturated hydrocarbons.
(ii) What is meant by a functional group in an organic compound? Name the functional group present in
(a) CH₃CH₂OH
(b) CH₃COOH
(iii) What is the difference in the molecular formula of any two consecutive members of a homologous series of organic compounds?

35. A student wants to project the image of candle flame on the wall of school laboratory by using a lens:
(i) which type of lens should be used and why?
(ii) at which distance in term of focal length F of the lens should be placed the candle flame so as to get
(a) a magnified and
(b) a diminished image respectively on the wall?
OR
(i) Complete the following ray diagrams:

(ii) A ray of light travelling in air enters obliquely into water. Does the light ray bend towards or away from the normal? Why? Draw a ray diagram to show the refraction of light in this situation.

36. (i) To study the respiration of germinating seeds:

(a) Name two chemicals that are kept in the test tube to absorb carbon dioxide gas released in the conical flask.
(b) Explain why the level of water in the bent tube rises in the set up A
(c) State the observation in set up B:

(ii) What do the following transport:
(a) Xylem
(b) Pulmonary artery
(c) Pulmonary vein
(d) Vena cava