Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 7 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Basic Set 7 with Solutions
Time: 4 Hours
Maximum Marks: 80
TERM – 1
Time: 2 Hours
Maximum Marks: 40
General Instructions:
Read the following instructions very carefully and strictly follow them:
- This question paper contains 50 questions out of which 40 questions are to be attempted. All questions carry equal marks.
- This question paper consists of three Section-A, B and C.
- Section A contains 20 questions. Attempt any 16 questions from Q. No. 1 to 20.
- Section B contains 20 questions. Attempt any 16 questions from Q. No. 21 to 40.
- Section C contains of two Case Studies containing 5 questions in each case. Attempt any 4 questions from Q. No. 41 to 45 and another 4 from Q. No. 46 to 50.
- There in only one correct option for every multiple choice question (MCQ). Marks will not be awarded for answering more than one option.
- There is no negative marking.
Section – A
Question 1.
HCF of 92 and 152 is: 1
(a) 4
(b) 19
(c) 23
(d) 57
Answer:
(a) 4
Explanation: Prime factorisation of 92
= 2 × 2 × 23
= 22 × 23
Prime factorisation of 152
= 2 × 2 × 2 × 19
= 23 × 19
Common factors of 92 and 152 = 22
= 2 × 2
= 4
Question 2.
In ΔABC, DE || BC, AD = 4 cm, DB = 6 cm and AE = 5 cm. The length of EC is: 1
(a) 7 cm
(b) 6.5 cm
(c) 7.5 cm
(d) 8 cm
Answer:
(c) 7.5 cm
Explanation: Given that in ABC, DE || BC, then
\(\frac{A D}{D B}=\frac{A E}{E C}\)
[By basic proportionality theorem]
Given, AD = 4 cm, DB = 6 cm, AE = 5 cm
By BPT,
\(\frac{4}{6}=\frac{5}{E C}\)
EC = \(\frac{30}{4}\)
= 7.5 cm
Question 3.
The value of k, for which the pair of linear equations x + y – 4 = 0, 2x + ky – 3 = 0 have no solution, is: 1
(a) 0
(b) 2
(c) 6
(d) 8
Answer:
(b) 2
Explanation: For no solution, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Here,
a1 = 1, b1 = 1, c1 = -4 and a2 = 2, b2 = k,
c2 = -3
Now, \(\frac{1}{2}=\frac{1}{k} \neq \frac{-4}{-3}\)
⇒ k = 2
Question 4.
The value of (tan2 45° – cos2 60°) is:
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{3}{2}\)
(d) \(\frac{3}{4}\)
Answer:
(d) \(\frac{3}{4}\)
Explanation: (tan2 45°- cos2 60°) = (1 – (\(\frac{1}{2}\))2)
[∵ tan45°= 1 and cos60°= \(\frac{1}{2}\)]
= (1 – (\(\frac{1}{4}\))
= (\(\frac{4-1}{4}\) = 1 – (\(\frac{3}{2}\)
Question 5.
A point (x, 1) is equidistant from (0, 0) and (2, 0). The value of x is: 1
(a) 1
(b) 0
(c) 2
(d) 1
Answer:
(a) 1
Explanation: Let A(x, 1), O(0,0) and B(2,0) be three points.
Given that A is equidistant from O and B.
x2 + 1 = (2 -x)2 + 1
x2 + 1 = 4 + x2 – 4x + 1
4x = 4
x = 1
Question 6.
Two coins are tossed together. The probability of getting exactly one head is:
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{3}{4}\)
(d) 1
Answer:
(b) \(\frac{1}{2}\)
Explanation: If two coins are tossed simultaneously we obtain any one of the following as an outcome.
HH, HT, TH,TT
Total number of elementary events = 4
One head is obtained if any one of the following elementary events occurs: HT, TH
Favourable number of elementary events = 2
Hence, required probability = \(\frac{2}{4}\) = \(\frac{1}{2}\).
Question 7.
A circular arc of length 22 cm subtends an angle θ at the centre of the circle of radius 21 cm. The value of θ is:
(a) 90°
(b) 50°
(c) 60°
(d) 30°
Answer:
(c) 60°
Explanation: Given arc length = 22 cm,
radius = 21 cm, angle subtended by arc = θ
We know that,
Length of an arc of sector angle
θ = \(\frac{\theta}{360}\) × 2πr
⇒ 22 = \(\frac{\theta}{360}\) × 2 × \(\frac{22}{7}\) × 21
⇒ θ = 60°
Question 8.
A quadratic polynomial having sum and product of its zeroes as 5 and 0 respectively, is:
(a) x2 + 5x
(b) 2x(x – 5)
(c) 5x2 – 1
(d) x2 – 5x + 5
Answer:
None of the option is correct.
Explanation: Given, sum of zeroes = 5 and product of zeroes = 0
Quadratic polynomial = x2 – (sum of zeroes)x + product of zeroes
Required polynomial = x2 – 5x + 0 = x2 – 5x
Question 9.
If P(E) = 0.65, then the value of P(not E) is: 1
(a) 1.65
(b) 0.25
(c) 0.65
(d) 0.35
Answer:
(d) 0.35
Explanation: Given, P(E) = 0.65,
then P(notE) = 1 – 0.65 = 0.35 [∵ P(E) + P(not E) = 1]
Question 10.
Zeroes of a quadratic polynomial x2 – 5x + 6 are: 1
(a) -5, 1
(b) 5, 1
(c) 2, 3
(d) -2, -3
Answer:
(c) 2, 3
Explanation: We have,
x2 – 5x + 6
= x2 – (3 + 2)x + 6
= x2 – 3x -2x + 6
= x(x – 3) – 2(x – 3)
= (x – 3)(x – 2)
For zeroes, (x – 3)(x – 2) = 0
x – 3 = 0 or x – 2 = 0
⇒ x = 3, 2
Question 11.
Perimeter of a rectangle whose length (l) is 4 cm more than twice its breadth (b) is 14 cm.
The pair of linear equations representing the above information is:
(a) l + 4 = 2b 2(l + b) = 14
(b) l – b =4 2(l + b) = 14
(c) l = 2b + 4 l + b = 14
(d) l = 2b + 4 2(l + b) = 14
Answer:
(d) l = 2b + 4; 2(l + b) = 14
Explanation: Given, length of a rectangle is 4 more than twice its breadth
⇒ 1 = 4+ 2b
Perimeter of rectangle = 2(l + b) = 14
Question 12.
The ratio in which the point (4, 0) divides the line segment joining the points (4, 6) and (4,-8) is:
(a) 1 : 2
(b) 3 :4
(c) 4 : 3
(d) 1 : 1
Answer:
(b) 3 : 4
Explanation: Let A (4, 0) divide the line segment joining the points B(4, 6) and C(4, – 8) in the ratio m : n.
Thus, the point A is
[latex]\frac{m \times 4+n \times 4}{m+n}, \frac{m \times(-8)+n \times 6}{m+n}[/latex]
Thus we have
\(\frac{m \times 4+n \times 4}{m+n}\) = 4 and \(\frac{m \times(-8)+n \times 6}{m+n}\) = 0
\(\frac{4(m+n)}{m+n}\) = 4 and -8m + 6n = 0
4 = 4 and 8m = 6n
⇒ \(\frac{m}{n}=\frac{6}{8}=\frac{3}{4}\)
Therefore, (4, 0) divides (4, 6) and (4, -8) in the ratio 3 : 4
Question 13.
Which of the following is not defined?
(a) sec 0°
(b) cosec 90°
(c) tan 90°
(d) cot 90°
Answer:
(c) tan 90°
Explanation: tan 90° = \(\frac{\sin 90^{\circ}}{\cos 90^{\circ}}=\frac{1}{0}\)
which is not defined [sin 90° = 1 and cos 90° = 0]
Question 14.
In the given figure, a circle is touching a semi-circle at C and its diameter AB at O. if AB = 28 cm, what is the radius of the inner circle?
(a) 14 cm
(b) 28 cm
(c) 7 cm
(d) \(\frac{7}{2}\) cm
Answer:
(c) 7 cm
Explanation:
Here, in figure, given that AB = 28 cm.
So, AO = \(\frac{1}{2}\)× AB
= \(\frac{1}{2}\) × 28cm
= 14cm
Now. AO = CO = BO
= 14cm [Radii of same circle]
OC is diameter of inner circle, so radius of inner circle = \(\frac{1}{2}\) × CO = \(\frac{1}{2}\) × 14 cm = 7 cm
Question 15.
The vertices of a triangle OAB are 0(0, 0), A(4, 0) and B(0, 6). The median AD is drawn on OB. The length AD is:
(a) \(\sqrt{52}\) units
(b) 5 units
(c) 25 units
(d) 10 units
Answer:
(b) 5 units
Explanation:
Given that AD, is median.
So, D is the mid-point of OB.
Coordinates of D = (\(\frac{0+0}{2}\), \(\frac{0+6}{2}\)) = (0, 3)
Using distance formula:
Length of AD = \(\sqrt{(4-0)^2+(0-3)^2}\)
= \(\sqrt{4^2+3^2}\)
= \(\sqrt{16+9}\)
= \(\sqrt{25}\)
= 5 units
Question 16.
In a right-angled triangle PQR, ∠Q = 90°. If ∠P = 45°, then value of tan P – cos2 R is:
(a) 0
(b) 1
(c) \(\frac{1}{2}\)
(d) \(\frac{3}{2}\)
Answer:
(c) \(\frac{1}{2}\)
Explanation:
In right-angled triangle PQR,
∠Q = 90° and ∠P = 45°
Now,
∠P +∠Q + ∠R = 180°
[Angle sum property of triangle]
45° + 90° + ∠R =180°
⇒ 135° + ∠R =180°
⇒ ∠R =180°- 135°
= 45°
Now, tan P – cos2R = tan 45° – (cos 45°)2
= 1 – (\(\frac{1}{\sqrt{2}}\))2
1 – \(\frac{1}{2}\) = \(\frac{1}{2}\) [∵ tan45°=0 and cos45°= \(\frac{1}{\sqrt{2}}\)]
Question 17.
If tan θ = \(\frac{3}{2}\), then the value of sec θ is:
(a) \(\frac{\sqrt{13}}{3}\)
(b) \(\frac{\sqrt{5}}{3}\)
(c) \(\sqrt{\frac{13}{3}}\)
(d) \(\frac{3}{\sqrt{13}}\)
Answer:
(a) \(\frac{\sqrt{13}}{3}\)
Explanation: Given, tanθ = \(\frac{2}{3}\)
We have identity, sec2θ – tan2θ = 1
Putting value of tan θ, we get
sec2θ – (\(\frac{2}{3}\))2 = 1
⇒ sec2θ – \(\frac{4}{9}\) = 1
⇒ sec2θ = 1 + \(\frac{4}{9}\) = \(\frac{13}{9}\)
⇒ secθ = \(\frac{\sqrt{13}}{3}\)
Section – B
Question 18.
The perimeter of the sector of a circle of radius 14 cm and central angle 45° is: 1
(a) 11 cm
(b) 22 cm
(c) 28 cm
(d) 39 cm
Answer:
(d) 39 cm
Explanation: Given, radius = 14 cm,
angle subtended by arc = 45°
∵ Length of an arc of sector angle
θ = \(\frac{\theta}{360}\) × 2πr
= \(\frac{45}{360}\) × 2 × \(\frac{22}{7}\) × 14
= 11 cm
Now, perimeter of sector = s + r+ r = s + 2r
= 11 cm + 2 × 14 cm
= 11 cm + 28 cm
= 39 cm
Question 19.
A bag contains 16 red balls, 8 green balls and 6 blue balls. One ball is drawn at random. The probability that it is blue ball is:
(a) \(\frac{1}{6}\)
(b) \(\frac{1}{5}\)
(c) \(\frac{1}{30}\)
(d) \(\frac{5}{6}\)
Answer:
(b) \(\frac{1}{5}\)
Explanation: Total number of balls
= 16 + 8 + 6 = 30
Number of blue balls = 6
So, the probability of getting a blue ball = \(\frac{\text { Number of blue balls }}{\text { Total number of balls }}\)
= \(\frac{6}{30}\)
= \(\frac{1}{5}\)
Question 20.
If sin θ – cos θ = 0, then the value of θ is: 1
(a) 30°
(b) 45°
(c) 90°
(d) 0°
Answer:
(b) 45°
Explanation: Given,
sin θ – cos θ = 0
⇒ sin θ = cos θ
⇒ \(\frac{\sin \theta}{\cos \theta}\) = 1
⇒ tan θ = tan 45° [tan 45° = 1]
⇒ θ = 45°
Question 21.
The probability of happening of an event is 0.02. The probability of not happening of the event is:
(a) 0.02
(b) 0.80
(c) 0.98
(d) \(\frac{49}{100}\)
Answer:
(c) 0.98
Explanation: Given, Probability of happening an event
= 0.02
Then, Probability of not happening of the event
= 1 – 0.02 = 0.98 [∵ P(E) + P(not E) = 1]
Question 22.
Two concentric circles are centred at O. The area of shaded region, if outer and inner radii are 14 cm and 7 cm respectively, is:
(a) 462 cm2
(b) 154 cm2
(c) 231 cm2
(d) 308 cm2
Answer:
(a) 462 cm2
Explanation: Area of shaded region
= Area of outer circle – Area of inner circle
= π × (14)2 – π × (7)2
[∵ Area of circle = πr2, where r is radius of circle]
Area of shaded region
= π × (142 – 72) cm2
= π × (196 – 49) cm2
= \(\frac{22}{7}\) × 147 cm2
= 462 cm2
Question 23.
\(\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}\) can be simplified to get:
(a) 2 cos2 θ
(b) sec2θ
(c) \(\frac{2}{\sin ^2 \theta}\)θ
(d) 2 sec2 θ
Answer:
(d) 2 sec2 θ
Explanation:
\(\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}\) = \(\frac{1-\sin \theta+1+\sin \theta}{(1+\sin \theta)(1-\sin \theta)}\)
= \(\frac{2}{1-\sin ^2 \theta}\)
= \(\frac{2}{\cos ^2 \theta}\) [∵ 1 – sin2 θ = cos2 θ] =
= 2 sec2 θ
Question 24.
The origin divides the line segment AB joining the points A(1, -3) and B(-3, 9) in the ratio: 1
(a) 3 : 1
(b) 1 : 3
(c) 2 : 3
(d) 1 : 1
Answer:
(b) 1 : 3
Explanation: Let O (0, 0) divides the line segment joining the points A (1, -3) and B (-3, 9) in the ratio m : n.
Thus, the coordinates point O are
[latex]\frac{m \times(-3)+n \times 1}{m+n}, \frac{m \times(9)+n \times(-3)}{m+n}[/latex]
Thus, we have
\(\frac{m \times(-3)+n \times 1}{m+n}\) = 0
and \(\frac{m \times(9)+n \times(-3)}{m+n}\) = 0
⇒ -3m + n = 0
and 9m – 3n = 0
⇒ 3m = n
and 9m = 3n
⇒ \(\frac{m}{n}=\frac{1}{3}\) and \(\frac{m}{n}=\frac{3}{9}=\frac{1}{3}\)
Therefore, origin divides line segment AB in the ratio 1 : 3.
Question 25.
The perpendicular bisector of a line segment A(-8, 0) and B(8, 0) passes through a point (0, k). The value of k is: 1
(a) 0 only
(b) 0 or 8 only
(c) any real number
(d) any non-zero real number 1
Answer:
(a) 0 only
Explanation: Given points A (-8, 0) and B (8, 0).
Perpendicular bisector of AB passes through mid-point of AB.
Mid-point of AB = (\(\frac{-8+8}{2}, \frac{0+0}{2}\)) = (0, 0)
It is given that perpendicular bisector passes through (0, k). So, on comparing, k = 0
Question 26.
Which of the following is a correct statement?
(a) Two congruent figures are always similar.
(b) Two similar figures are always congruent.
(c) All rectangles are similar.
(d) The polygons having same number of sides are similar.
Answer:
(a) Two congruent figures are always similar.
Explanation: Two congruent figures are always similar because congruent means exact same. So, if two figures are same then they are similar as well.
Question 27.
The solution of the pair of linear equations x = -5 and y = 6 is: 1
(a) (-5, 6)
(b) (-5, 0)
(c) (0, 6)
(d) (0, 0)
Answer:
(a) (-5, 6)
Explanation: x = -5 represents a line parallel to y-axis and y = 6 represents a line parallel to x-axis
So, point of intersection must have x = -5 and y = 6, both the line intersect at (-5, 6).
Question 28.
A circle of radius 3 units is centered at (0, 0). Which of the following points lie outside the circle? 1
(a) (-1, -1)
(b) (0, 3)
(c) (1, 2)
(d) (3, 1)
Answer:
(d) (3,1)
Explanation: Given radius of circle is 3 units.
As we know that all points on the circumference of circle is at a distance of 3 units.
Now, using distance formula,
Distance between (0, 0) and (-1, -1) is,
\(\sqrt{(0+1)^2+(0+1)^2}=\sqrt{1^2+1^2}=\sqrt{2}\) units
Distance between (0, 0) and (0, 3) is,
\(\sqrt{(0-0)^2+(0-3)^2}=\sqrt{3^2}=3\) units
Distance between (0, 0) and (1, 2) is,
\(\sqrt{(0-1)^2+(0-2)^2}=\sqrt{1^2+2^2}=\sqrt{5}\) units
Distance between (0, 0) and (3, 1) is,
\(\sqrt{(0-3)^2+(0-1)^2}=\sqrt{3^2+1^2}=\sqrt{10}\) units
Since, distance between (0, 0) and (3, 1) is greater than 3 units.
Hence, point (3,1) lies outside the circle,
Question 29.
The value of k for which the pair of linear equations 3x + 5y = 8 and kx + 15y = 24 has infinitely many solutions, is: 1
(a) 3
(b) 9
(c) 5
(d) 15
Answer:
(b) 9
Explanation: Given equations are 3x + 5y = 8 and kx + 15y = 24
Equations can be written as 3x + 5y – 8 = 0 and kx + 15y – 24 = 0
For infinitely many solutions, \(\)
Here,
a1 = 3, b1 = 5, c1 = – 8 and a2 = k, b2 = 15,
c2 = -24
Now, \(\frac{3}{k}=\frac{5}{15}=\frac{-8}{-24}\)
⇒ \(\frac{3}{k}=\frac{1}{3}\)
⇒ k = 9
Question 30.
HCF of two consecutive even numbers is: 1
(a) 0
(b) 1
(c) 2
(d) 4
Answer:
(c) 2
Explanation: All even numbers are divisible by 2.
So, HCF of two consecutive even numbers is 2.
Question 31.
The zeroes of quadratic polynomial x2 + 99x + 127 are: 1
(a) both negative
(b) both positive
(c) one positive and one negative
(d) reciprocal of each other
Answer:
(a) both negative
Explanation: Given polynomial is x2 + 99x + 127
On comparing above polynomial with x2 – (sum of roots)x + product of roots, we get:
Sum of roots = – 99
Product of roots = 127
∵ Sum of roots (zeroes) is negative and product of roots (zeroes) is positive, this is possible only when both zeroes will be negative.
∴ Both zeroes are negative.
Question 32.
The mid-point of Line segment joining the points (-3, 9) and (-6, -4) is:
(a) (\(\frac{-3}{2}\), \(\frac{-13}{2}\))
(b) (\(\frac{9}{2}\), \(\frac{-5}{2}\))
(c) (\(\frac{-9}{2}\),\(\frac{5}{2}\))
(d) (\(\frac{9}{2}\), \(\frac{5}{2}\))
Answer:
(c) (\(\frac{-9}{2}\),\(\frac{5}{2}\))
Explanation: The mid-point of line segment joining the points (x1, y1 and (x2, y2) is (\(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\))
So, mid-point of line segment joining the points (-3,9) and (-6,-4) is
(\(\frac{-3-6}{2}, \frac{9-4}{2}\)) = (\(\frac{-9}{2}\),\(\frac{5}{2}\))
Question 33.
The decimal expansion of \(\frac{13}{2 \times 5^2 \times 7}\) is: 1
(a) terminating after 1 decimal place.
(b) non-terminating and non-repeating.
(c) terminating after 2 decimal places.
(d) non-terminating but repeating.
Answer:
Not examinable for 2024 exam.
Question 34.
In ΔABC, DE || BC, AD = 2 cm, DB = 3 cm, DE : BC is equal to:
(a) 2 : 3
(b) 2 : 5
(c) 1 : 2
(d) 3 : 5
Answer:
(b) 2 : 5
Explanation: Given that in ΔABC, DE || BC. then
In ΔADE and ΔABC,
\(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}\)
[By basic proportionality theoremj
And ∠A = ∠A
∴ ΔADC ~ ΔABC [By SAS similarity criterion]
Given, AD = 2 cm, DB = 3 cm;
AB = 2 cm + 3 cm = 5 cm
∴ \(\frac{2}{5}=\frac{D E}{B C}[latex]
[latex]\frac{\mathrm{DE}}{\mathrm{BC}}=\frac{2}{5}[latex]
Question 35.
The (HCF × LCM) for the numbers 50 and 20 is: 1
(a) 1000
(b) 50
(c) 100
(d) 500
Answer:
(a) 1000
Explanation: We know that,
Product of two numbers = HCF of numbers × LCM of numbers
Numbers are 50 and 20, so
HCF × LCM = 50 × 20 = 1000
Question 36.
For which natural number n, 6n ends with digit zero? 1
(a) 6
(b) 5
(c) 0
(d) None of these
Answer:
(d) None of these
Explanation: We have 6n
For number ends with digit 0, its factor must have 2m × 5n as factor, but 6n =2n × 3n.
In 6n, there are no factors like 2m × 5n, so 6n can never ends with 0.
Question 37.
(1 + tan2 A) (1 + sin A) (1 – sin A) is equal to: 1
(a) [latex]\frac{\cos ^2 A}{\sec ^2 A}\)
(b) 1
(c) 0
(d) 2
Answer:
(b) 1
Explanation: We have
(1 + tan2A)(1 + sinA)(1 – sinA)
= (1 + tan2A) (1 – sin2A)
= sec2A × cos2A = 1
[∵ sec2A – tan2A = 1, cos2A + sin2A = 1]
Section – C
CASE STUDY 1
Sukriti throws a ball upwards, from a rooftop which is 8 m high from ground level. The ball reaches to some maximum height and then returns and hit the ground.
If height of the ball at time t (in sec) is represented by h(m), then equation of its path is given as h = -t2 + 2t + 8.
Based on above information, answer the following question:
Question 38.
The maximum height achieved by ball is:
(a) 7 m
(b) 8 m
(c) 9 m
(d) 10 m
Answer:
(b) 8 m
Explanation:
Given equation of path of the ball is,
h = -t2 + 2t + 8
When ball reaches the ground, then height = 0
Putting h = 0 in equation, we get
0 = -t2 + 2t + 8
t2 – 2t – 8 = 0
t2 – 4t + 2t – 8 = 0
t(t – 4) + 2(t – 4) = 0
(t – 4)(t + 2) =0
t = 4, – 2
Time cannot be negative, so total time required by ball to reach the ground is 4 seconds.
So, time to reach the ball to maximum height is \(\frac{4}{2}\) = 2 seconds.
Therefore, maximum height of the ball
= -22 + 2 × 2 + 8
= -4 + 4 + 8
= 8 m
Question 39.
The polynomial represented by above graph is:
(a) linear polynomial
(b) quadratic polynomial
(c) constant polynomial
(d) cubic polynomial
Answer:
(b) quadratic polynomial
Explanation: The polynomial represented by given graph is quadratic polynomial.
Question 40.
Time taken by ball to reach maximum height is:
(a) 2 sec
(b) 4 sec
(c) 1 sec
(d) 2 min
Answer:
(a) 2 sec.
Explanation: Total time taken by ball to reach the ground is 4 seconds. Therefore, time taken by ball to reach the maximum height is 2 sec.
Question 41.
Number of zeroes of the polynomial whose graph is given, is: 1
(a) 1
(b) 2
(c) 0
(d) 3
Answer:
(b) 2
Explanation: Graph represented is quadratic polynomial, so number of zeroes are 2.
Question 42.
Zeroes of the polynomial are: 1
(a) 4
(b) -2, 4
(c) 2, 4
(d) 0, 4
Answer:
(b) -2. 4
Explanation: im
Given equation of path of the ball is,
h = -t2 + 2t + 8
When ball reaches the ground, then height = 0
Putting h = 0 in equation, we get
0 = -t2 + 2t + 8
t2 – 2t – 8 = 0
t2 – 4t + 2t – 8 = 0
t(t – 4) + 2(t – 4) = 0
(t – 4)(t + 2) =0
t = 4, – 2
Time cannot be negative, so total time required by ball to reach the ground is 4 seconds.
So, time to reach the ball to maximum height is \(\frac{4}{2}\) = 2 seconds.
Therefore, maximum height of the ball
= -22 + 2 × 2 + 8
= -4 + 4 + 8
= 8 m zeroes are -2, 4.
CASE STUDY 2
Quilts are available in various colours and design. Geometric design includes shapes like squares, triangles, rectangles, hexagons etc.
One such design is shown above. Two triangles are highlighted, ΔABC and ΔPQR.
Based on above information, answer the following questions:
Question 43.
Which of the following criteria is not suitable for ΔABC to be similar to ΔQRP?
(a) SAS
(c) SSS
(b) AAA
(d) RHS
Answer:
(d) RHS
Explanation: RHS is not suitable for ΔABC is similar to ΔQRP.
Question 44.
If each square is of length x unit, then length BC is equal to: 1
(a) x√2 unit
(b) 2x unit
(c) 2√x unit
(d) x√x unit
Answer:
(a) x√2 unit
Explanation: In ΔABC, AB = AC = x unit.
ΔABC is a right angled triangle, so by Pythago-ras theorem
BC2 = AB2 + AC2 = x2 + x2 = 2x2 [Sides of square are equal in length]
⇒ BC = √2x units
Question 45.
Ratio BC : PR is equal to: 1
(a) 2 : 1
(b) 1 : 4
(c) 1 : 2
(d) 4 : 1
Answer:
(c) 1 : 2
Explanation: From figure, we can see that
PR = 2 × BC
⇒ BC : PR = 1 : 2
Question 46.
ar(PQR): ar(ABC) is equal to: 1
(a) 2 : 1
(b) 1 : 4
(c) 4 : 1
(d) 1 : 8
Answer:
Not examinable for 2024 exam.
Question 47.
Which of the following is not true? 1
(a) ΔTQS ~ ΔPQR
(b) ΔCBA ~ ΔSTQ
(c) ΔBAC ~ ΔPQR
(d) ΔPQR ~ ΔABC
Answer:
(d) ΔPQR ~ ΔABC
Explanation: From given figure, it is clear that ΔTQS ~ ΔPQR, ΔCBA ~ ΔSTQ and ΔBAC ~ ΔPQR.
But ΔPQR is not similar to ΔABC. Here order is not correct.
TERM – 2
Time: 2 Hours
Maximum Marks: 40
General Instructions:
Read the following instructions very carefully and strictly follow them:
- This question paper contains 14 questions. All questions are compulsory.
- This question paper is divided into three sections – Sections A, B and C.
- Section A comprises of 6 questions (Q. No. 1 to 6) of 2 marks each. Internal choice has been provided in two questions.
- Section B comprises of 4 questions (Q. No. 7 to 10) of 3 marks each. Internal choice has been provided in one questions.
- Section C comprises of 4 questions (Q. No. 11 to 14) of 4 marks each. Internal choice has been provided in one questions It also contains two case study based questions.
- Use of calculator is not permitted.
Section – A
Question 1.
In an AP, if a = 50, d = – 4 and Sn = 0, then find the value of n.
OR
Find the sum of the first twelve 2-digit multiples of 7, using an AP. 2
Answer:
Given, a = 50, d = -4 and Sn = 0
We have, Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
Where, n is number of terms.
Putting values of a, d and Sn, we get
0 = \(\frac{n}{2}\)[2 × 50 + (n – 1) × (-4)]
⇒ 4(n – 1) = 100
⇒ n – 1 = \(\frac{100}{4}\) = 25
⇒ n = 25 + 1 = 26
OR
First two-digit multiple of 7 is 14, so first term,
a = 14. Common difference, d = 7
AP is 14, 21, 28, …. up to 12 terms
∵ Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
Sum of 12 terms,
S12 = \(\frac{12}{2}\)[ 2 × 14 + (12 – 1) × 7]
S12 = 6[28 + 11 × 7]
= 6 × [28 + 77]
= 6 × 105 = 630
So, sum of first twelve two-digit multiples of 7 is 630.
Question 2.
Find the nature of the roots of the quadratic equation x2 – 5x + 9 = 0.
OR
Write a quadratic equation with roots – 3 and 5. 2
Answer:
We have x2 – 5x + 9 = 0
For nature of root, we need to find b2 – 4ac
Here b = -5, a = 1 and c = 9.
b2 – 4ac = (-5)2 – 4 × 1 × 9
= 25 – 36 = -11 < 0
So, roots are unequal and imaginary.
OR
(Standard quadratic equation is written as: x2 – (a + b)x + ab = 0,
where, a and b are roots of quadratic equation. Given roots are -3 and 5.
Required quadratic equation:
x2 – (-3 + 5)x + (-3) x 5 = 0
x2 – 2x – 15 = 0
Question 3.
Find the mode of the following frequency distribution:
Class | Frequency |
0 – 20 | 8 |
20 – 40 | 7 |
40 – 60 | 12 |
60 – 80 | 5 |
80 – 100 | 3 |
Answer:
Class | Frequency |
0 – 20 | 8 |
20 – 40 | 7 |
40 – 60 | 12 |
60 – 80 | 5 |
80 – 100 | 3 |
Highest frequency = 12
Modal class = 40 – 60
l = 40, F0 = 7, F1 =12, F2 = 5, h = 20
Question 4.
Solve the quadratic equation 2x2 – 5x – 1 = 0 for x. 2
Answer:
Quadratic equation: 2x2 – 5x – 1 = 0.
Here, a = 2, b = -5, and c = -1
Question 5.
In figure, if tangents PA and PB drawn from a point P to a circle with centre O, are inclined to each other at an angle of 70°, then find the measure of ∠POA.
Answer:
Since, O is the centre of the circle and two tangents from point P to the circle are PA and PB.
∴ OA ⊥ AP and OB ⊥ BP
⇒ ∠OAP = ∠OBP = 90°
Now, in quadrilateral PAOB, we have
∠APB + ∠PAO + ∠AOB + ∠PBO = 360°
70° + 90° + ∠AOB + 90° = 360°
250° + ∠AOB = 360°
⇒ ∠AOB = 360° – 250°
∠AOB = 110°.
In ∆OAP and ∆OBP, we have
OP = OP [Common]
∠OAP = ∠OBP [Each 90°]
OA = OB [Radii of the same circle]
∴ ∆OAP ≅ ∆OBP [By RHS congruent rule]
∠POA = ∠POB [By CPCT]
∴ ∠POA = \(\frac{1}{2}\) ∠AOB
= \(\frac{1}{2}\) × 110° = 55°.
Section – B
Question 6.
The frequency distribution given below shows the weight of 40 students of a class. Find the median weight of the students. 3
Weight (in kg) | Number of Students |
40 – 45 | 9 |
45 – 50 | 5 |
50 – 55 | 8 |
55 – 60 | 9 |
60 – 65 | 6 |
65 – 70 | 3 |
Answer:
Weight (in kg) (C.l.) | Number of students (fi) | Cumulative frequency (cf) |
40 – 45 | 9 | 9 |
45 – 50 | 5 | 14 (= 9 + 5) |
50 – 55 | 8 | 22 (= 14 + 8) |
55 – 60 | 9 | 31 (= 22 + 9) |
60 – 65 | 6 | 37 (= 31 + 6) |
65 – 70 | 3 | 40 (= 37 + 3) |
Here Median number: \(\frac{N}{2}=\frac{40}{2}\) = 20
c.f. just > median number (20) = 22
⇒ Median class
= The class interval corresponding to c.f. of 22
⇒ Median class = 50 – 55
Lower Limit of the median class, l = 50
Frequency of the median class, fm = 8
Size of the median class, h = 55 – 50 = 5
Cumulative frequency of the pre-median class, C = 14
We know that,
Median (Me) = l + \(\frac{h}{f_m}\)(\(\frac{N}{2}\) – C)
Median = 50 + \(\frac{5}{8}\) × (20 – 14)
= 50 + \(\frac{5}{8}\) × 6 = 50 + \(\frac{15}{4}\)
= 50 + 3.75 = 53.75
Thus, the required median weight of students is 53.75 kg.
Question 7.
Tn figure, the angles of elevation of the top of a tower AB of height ‘h’m, from two points P and Q at a distance of x m and y m from the base of the tower respectively and in the same straight line with it, are 60° and 30°, respectively, Prove that h2 = xy. 3
Answer:
Given, AP = x m and AQ = y m, ∠AQB = 30° and ∠APB = 60°.
From right ΔABP, we have
\(\frac{\mathrm{BA}}{\mathrm{AQ}}\) = tan 30°
⇒ \(\frac{h}{y}=\frac{1}{\sqrt{3}}\)
⇒ h = \(\frac{y}{\sqrt{3}}\) …………. (i)
From right ΔABP, we have
\(\frac{\mathrm{AB}}{\mathrm{AP}}\) = tan 60°
⇒ \(\frac{h}{x}\) = √3
⇒ h = √3x ……….. (ii)
From (i) and (ii), we get
h2 = \(\frac{y}{\sqrt{3}}\) × √3x
⇒ h2 = xy
Hence, proved
Question 8.
The following table shows the age of patients admitted in a hospital during a particular week:
Age (in years) | Number of Patients |
5 -15 | 5 |
15 – 25 | 12 |
25 – 35 | 20 |
35 – 45 | 24 |
45 – 55 | 15 |
55 – 65 | 4 |
Find the mean age of the patients. 3
Answer:
∵ Mean \((\bar{X})=\frac{\sum f_i x_i}{\sum f_i}\)
= \(\frac{2840}{80}\)
= \(\frac{284}{8}\)
= 35.5
Thus, the mean age of patients is 35.5 years.
Section – C
Question 9.
A spherical glass vessel has a neck 8 cm long and 1 cm in radius. The radius of the spherical part is 9 cm. Find the amount of water (in litres) it can hold, when filled completely.
OR
From a solid cylinder, whose height is 2.4 cm. and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. 4
Answer:
Radius of cylindrical part, r = 1 cm
Height of cylindrical part, h= 8 cm
Volume of cylindrical part = πr2h
= n(1)2 × 8
= 8π cm3
Radius of spherical part = 9 cm
Volume of spherical part = \(\frac{4}{3}\) πr3
= \(\frac{4}{3}\)π(3)3
= \(\frac{4}{3}\)π × 9 × 9 × 9 cm3
= 972π cm3
Total volume of the glass vessel = Volume of cylindrical part + Volume of spherical part
= 8π cm3 + 972π cm3
= π(8 + 972) cm3
= \(\frac{22}{7}\) × 980 cm3
= 22 × 140 cm3 = 3080 cm3
= 3080 × 0.0011 [∵ 1 cm3 = 0.001l]
= 3.08 l
Therefore, the amount of water (in liters) it can hold, when filled completely is 3.08 l cm3.
Question 10.
In figure, the tangent l is parallel to the tangent m drawn at points A and B respectively to a circle centred at O. PQ is a tangent to the circle at R. Prove that ∠POQ = 90°.
Answer:
Since l || m and QP is the transversal, therefore
∠AQR + ∠RPB = 180° ……. (i)
[Since, Sum of the interior angles on the same side of a transversal is 180°]
In ΔAOQ and ΔROQ, we have
QA = QR [Tangents drawn from an external point are equal in length]
QO = QO [Common]
OA = OR [Radii of the same circle]
So, by SSS-congruency criterion, we have
ΔAOQ ≅ ΔROQ ……… (ii)
So, ∠AQO = ∠RQO [CPCT]
Similarly, we can prove
∠RPO = ∠BPO …….. (iii)
Now, (i) can be written as
(∠AQO + ∠RQO) + (∠RPO + ∠BPO) = 180°
(∠RQO + ∠RQO) + (∠RPO + ∠RPO) = 180° [Using (ii) and (iii)]
⇒ 2∠RQO + 2∠RPO = 180°
⇒ 2(∠RQO + ∠RPO) = 180°
⇒∠RQO + ∠RPO = 90° ……. (iv)
In ΔPOQ, we have
∠RQO + ∠RPO + ∠POQ = 180° [Angle sum property of triangle]
⇒ (∠RQO + ∠RPO) + ∠POQ = 180°
⇒ 90° + ∠POQ = 180° [Using (iv)]
⇒∠POQ =180°- 90° = 90°,
Hence, proved.
CASE STUDY 1
Question 11.
Do you know old clothes which are thrown as waste not only fill the landfill site but also produce very harmful greenhouse gas. So, it is very important that we reuse old clothes in whatever way we can.
The picture given below on the right, shows a footmat (rug) made out of old t-shirts yarn. Observing the picture, you will notice that a number of stitches in circular rows are making a pattern : 6, 12,18, 24,…
Based on the above information, answer the following questions:
(A) Check whether the given pattern forms an AP. If yes, find the common difference and the next term of the AP. 2
(B) Write the nth term of the AP. Hence, find the number of stitches in the 10th circular row. 2
Answer:
(A) Given number of stitches in circular rows are making a pattern: 6,12,18, 24,…
Here, first term, a = 6, second term = 12, third term = 18
d = second term – first term = 12 – 6 = 6
d = third term – second term = 18 – 12 = 6
d = fourth term – third term = 24 – 18 = 6
Here common difference is same so terms are in AP.
Common difference = 6,
next term =24 + 6 = 30
(B) nth term of AP = an = a + (n – 1)d
an = 6 + (n – 1)6
∴ an = 6n
Number of stitches in the 10th circular row,
a10 = 6 × 10 = 60
CASE STUDY 2
Question 12.
The following TV Tower was built in 1988 and is located in Pitampura, Delhi. It has an observation deck. Observe the picture given below:
The TV Tower stands vertically on the ground. From a point ‘A’ on the ground, the angle of elevation of top of the tower (point ‘B’) is 60°. There is a point ‘C’ on the tower which is 78 m (approx.) above the ground.
The angle of elevation of the point C from point A is found to be 30°.
(A) Draw a well-labelled figure, based on the information given above. 2
(B) Find the height of the tower and the distance of the tower from point A. 2
Answer:
(A)
(B) In ΔCDA
\(\frac{\mathrm{CD}}{\mathrm{DA}}\) = tan 30°
⇒ \(\frac{78}{\mathrm{DA}}=\frac{1}{\sqrt{3}}\) ……….. (i)
⇒ DA = 78√3 m or 135.09 m
Now, in ΔBDA,
\(\frac{\mathrm{BD}}{\mathrm{DA}}\) = tan 60°
⇒ \(\frac{\mathrm{BD}}{\mathrm{DA}}\) = √3
⇒ BD = 78√3 × √3 m
= 78 × 3 m
= 234 m
The height of the tower is 234 m and the distance of the tower from point A is 135.096 m.